Chem37-38Electrochem

Chem37-38Electrochem - Module 37 — Electrochemistry Calculations In Chemistry Module 37 Electrochemistry Module 38 Electrochemical Cells Module

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Module 37 — Electrochemistry Calculations In Chemistry ***** Module 37: Electrochemistry Module 38: Electrochemical Cells Module 37 – Electrochemistry.................................................................................... 1130 Lesson 37A: Lesson 37B: Lesson 37C: Lesson 37D: Lesson 37E: Lesson 37F: Redox Fundamentals ...................................................................................... 1130 Charges and Electrical Work ......................................................................... 1137 Standard Reduction Potentials ...................................................................... 1141 Non-Standard Potentials: The Nernst Equation ........................................ 1144 Predicting Which Redox Reactions Go......................................................... 1150 Calculating Cell Potential............................................................................... 1157 Module 38 – Electrochemical Cells ........................................................................... 1167 Lesson 38A: Lesson 38B: Lesson 38C: Lesson 38D: Lesson 38E: Cells and Batteries ........................................................................................... 1167 Anodes and Cathodes..................................................................................... 1177 Depleted Batteries and Concentration Cells................................................ 1186 Electrolysis ....................................................................................................... 1194 Amperes and Electrochemical Calculations ................................................ 1197 For additional modules, visit www.ChemReview.Net © 2010 ChemReview.Net v1w Page i Module 37 — Electrochemistry Module 37 — Electrochemistry Timing: Some textbooks begin their electrochemical unit with cells and batteries. These lessons cover cells and batteries in the next module. However, if you finish Module 37 before Module 38, battery calculations should be easier to understand. ***** Lesson 37A: Redox Fundamentals Pretest: This lesson will refresh your memory on the fundamentals of redox reactions that were covered in Modules 15 and 16. Even if you recall those topics well, do at least the last problem on all four problem sets in this lesson. Certain topics are emphasized here that will be especially important in Modules 37 and 38. If you have any difficulty with the problems in this lesson, review Modules 15 and 16. **** Introduction Redox reactions are of particular interest in chemistry because they can both store and release electrical energy. Batteries based on redox reactions have long been used for a range of purposes from flashlights to starting cars. More recently, thanks to chemical research, new types of batteries have been developed with improved safety and re-chargeability. These electrochemical cells power the hybrid vehicles, laptops, cell phones, and personal music systems that define our modern age. Redox Review As a part of this review, you may want to re-run your flashcards from Modules 15 and 16. 1. Oxidation numbers. Oxidation numbers answer the question: if each atom in a particle were an ion, what would be its charge? Most atoms are not ions, but assigning oxidation numbers can help in tracking the movement of electrons that is the key to understanding redox reactions. Rules for assigning oxidation numbers (Ox#) to individual atoms are as follows. a. In an element, each atom has an Ox# of zero. b. In a monatomic ion, the Ox# of the atom is its charge. c. In other types of particles: i. Each O atom is assigned ─2 (except in peroxides, where O = ─1). ii. Each H is +1, except in metallic hydrides where H is ─1. iii. Each alkali metal atom is assigned +1. Each column 2 atom is +2. iv. Other Ox# are assigned so that the ∑ Ox# = the overall charge on the particle. d. The Ox# of an individual atom in a particle is the total of the Ox# for the atoms of that kind, divided by the number of atoms of that kind. © 2010 ChemReview.Net v1w Page 1130 Module 37 — Electrochemistry Practice A: Do and check 1a. Then do every other letter, and more if you need more practice. For a more detailed rule review, see Lesson 15A. 1. Fill in the oxidation number of the atom shown. a. BrO3─ Br = ________ b. Co2+ Co = _______ c. KMnO4 Mn = _______ d. ZnBr2 Zn = ________ e. I2 I = _______ f. NO3─ N = ________ g. S2O82─ S = _______ h. IO─ I = ________ i. LiH H = _______ j. NH3 N = ________ 2. Redox terminology. • Oxidation is the loss of electrons. • Reduction is the gain of electrons. • A redox reaction is an electron transfer: one or more electrons move from one particle to another. In redox reactions, o reducing agents are particles that lose their electrons (and are oxidized) by giving electrons to other particles; o oxidizing agents are particles that gain electrons (and are reduced) by removing electrons from other particles. 3. Labeling redox components. a. Redox reactions are reversible: in a closed system, reactions go to equilibrium. b. Each side of a redox reaction equation has a one reducing agent (RA) and one oxidizing agent (OA). c. If an atom that is changing oxidation number is part of particle that is a reducing agent on one side of a redox reaction equation, that atom is part of the oxidizing agent on the other. d. The RA and OA on each side of a redox equation can be labeled by • assigning oxidation numbers to atoms, • identifying the two atoms that change oxidation number, and • labeling the particles that contain those atoms: particles with atoms that lose electrons as they react are RA, those that gain are OA. d. In redox reactions, assume metals are solid (s) and ions are in aqueous solution (aq) unless otherwise noted. Commit to memory the definitions and behaviors of oxidizing agents and reducing agents in points 2 and 3 above. © 2010 ChemReview.Net v1w Page 1131 Module 37 — Electrochemistry Then, below each of the following reactions, label two particles as oxidizing agents (OA) and two particles as reducing agents (RA). If you are unsure about your answer to Question 1, check it before completing Question 2. Q1. Cu2+ + Pb Cu + Pb+2 Q2. H+ + MnO4─ + Cl─ MnO2 + H2O + Cl2 ***** A1. Assign oxidation numbers, then label the particles. +2 0 (Cu gains 2 electrons) +2 (Pb loses two electrons) Cu + Pb+2 0 Cu2+ + Pb OA RA RA OA Each side must have one reducing agent and one oxidizing agent. The Pb metal atom gives up two electrons in going to the right: Pb is therefore acting as the reducing agent on the left side, and the Pb atom must be part of the oxidizing agent on the other. The Cu metal gives away two electrons when the reaction goes to the left, making Cu the reducing agent on the right side. A2. Identify the atoms that change oxidation number. +7 +4 ─1 H+ + MnO4─ + Cl─ (Mn gains 3 electrons) 0 (each Cl loses one electron) MnO2 + H2O + Cl2 OA RA RA OA Since Mn gains electrons, it is being reduced in going to the right and is an oxidizing agent. Each chloride ion loses an electron in going to the right, so it is being oxidized and is a reducing agent. Going backward, the Mn atom on the right must lose 3 electrons to go left, so it is acting as an RA. Each neutral chlorine atom on the right gains an electron going to the left, so it is acting as an OA. If the atom that is changing its oxidation number is in a reducing agent on one side, it must be in the oxidizing agent on the other. Practice B 1. Define oxidation. 2. Define an oxidizing agent. 3. For each of the following reactions, label two particles as oxidizing agents (OA) and two particles as reducing agents (RA), then circle the reactant particle being oxidized. a. b. Sn4+ + Cu Ba + 2 H+ © 2010 ChemReview.Net v1w Cu2+ + Sn2+ H2 + Ba2+ Page 1132 Module 37 — Electrochemistry c. 2 Al + 3 ZnCl2 2 AlCl3 + 3 Zn d. 4 As + 3 HBrO3 + 6 H2O 4 H3AsO3 + 3 HBrO 4. Half-reactions. A redox reaction can be separated into two half-reactions. One halfreaction shows the gain of electrons for the reactant that is an oxidizing agent, the other shows the loss of electrons for the reactant that is a reducing agent. Half-reactions include the symbol e─ showing the number of electrons gained or lost. Half-reactions must be balanced for atoms and charge. Electrons count when balancing charge, but they do not affect the balancing of atoms. 5. Balancing Half-reactions. Half-reactions can be balanced by trial and error. If H and O atoms must be added to balance half-reactions that are carried out in aqueous solutions, use the CA-WHe! method: a. First balance the central atom (CA), usually one not O or H. Then, b. Add Water to balance oxygen, H+ to balance the hydrogen, and electrons to balance charge. 6. To balance half-reactions using OH─ instead H+ ions, first balance by CA-WHe, then neutralize H+ by adding OH─ equally to both sides, then adjust H O on both sides. 2 7 Half-reactions can be used to label the oxidizing and reducing agent in a redox reaction. If a half-reaction has • electrons on the right, one of the particles on the left is an RA that contains an atom releasing electrons, and the particle on the right that contains that atom is an OA; • electrons on the left, one of the particles on the left is an OA that contains an atom gaining electrons, and the particle on the right that contains that atom is an RA. Practice C: Do every other letter, and more if you need more practice. For a detailed review of redox balancing using half-reactions, see Lessons 16A and 16B. 1. Balance these half-reactions for atoms and charge. a. H2 c. Hg22+ b. Hg2+ S2O32─ d. H+ Pb4+ S4O62─ Pb2+ 2. Balance these half-reactions by the CA-WHe! method. a. Cr3+ b. IO3─ © 2010 ChemReview.Net v1w CrO42─ I2 Page 1133 Module 37 — Electrochemistry c. d. Ag+ Ag SO2 S O4 2 ─ 3. Balance these half-reactions by adding OH─, H2O , and electrons as needed. a. Cr3+ b. Cr2O72─ MnO4─ Mn 2 + 4. Which reactant is the reducing agent in b. Problem 2b. a. Problem 1d. c. Problem 2d. d. Problem 3b. 8. All reactions may be balanced by trial and error. Only one set of ratios will balance an equation. Redox reaction equations may be balanced by two additional methods: by balancing oxidation numbers and by adding balanced half-reactions. Half-reaction addition is the technique used most often in electrochemical calculations. 9. To balance redox reactions by adding half-reactions, the steps are a. Balance each half-reaction. b. Multiply each half-reaction by an LCM to get the same number of electrons in both. The number of electrons lost in one half-reaction must equal the number gained by the other. c. Add the two half-reactions. Cancel like terms on both sides. Like numbers of electrons on each side must cancel. 10. Redox reactions can be divided into half-reactions to assist in balancing. a. Find the two atoms that change their Ox# in the reaction. b. Write two separate half-reactions, one for each atom changing Ox#. c. Balance then add the half-reactions. d. Add spectator ions if needed, then adjust the trial coefficients to balance atoms and charge with the spectators included. Practice D: For a more detailed review of half-reactions, see Module 16. 1. Balance each half-reaction, then add to get a balanced redox reaction. a. Zn Br2 Zn2+ b. Br─ 2. Which reactant is the reducing agent: a. In 1a. © 2010 ChemReview.Net v1w MnO4─ I─ MnO2 I2 b. In 1b. Page 1134 Module 37 — Electrochemistry 3. Separate these redox reactions into two half-reactions, balance and add the halfreactions, then balance the final equation. a. Co + NO3─ Co2+ + NH4+ b. HCl + H2SO4 c. Cd + HNO3 SO2 + Cl2 + H2O Cd(NO3)2 + NO + H2O 4. Which reactant is being reduced a. In problem 3a. b. In 3b. c. In 3c. ANSWERS Practice A 1. Br = +5 Do O first. Each O is ─ 2. 3 x ─ 2 = ─ 6 . Br must be +5 to equal the ─ 1 total charge. Each atom: +5 ─ 2 BrO ― Formula for the particle: 3 Total for those atoms: b. Co = +2 c. Mn = +7 h. I = + 1 +5 ─ 6 d. Zn = +2 i. H = ─ 1 (hydride) must total ─ 1. e. I = 0 f. N = + 5 g. S = + 7 j. N = ─ 3 Practice B 1. Oxidation is the loss of electrons. 2. An oxidizing agent removes electrons from another particle, and in the process the OA is reduced. +4 0 +2 +2 OA 0 +1 b. Ba + 2 H+ Cu2+ + Sn2+ OA 3. a. Sn4+ + Cu RA c. 2 Al + 3 ZnCl2 RA RA RA 2 AlCl3 + 3 Zn OA OA 0 +2 H2 + Ba2+ OA RA OA d. 4 As + 3 HBrO3 + 6 H2O RA RA OA 4 H3AsO3 + 3 HBrO OA Practice C 1. a. c. 2. a. H2 Hg22+ 2 H+ + 2 e─ 2 Hg2+ + 2 e─ 4 H2O + Cr3+ 10 e─ + 12 H+ + 2 IO3─ c. Ag+ + e─ 14 OH─ + 2 Cr3+ b. d. 2 e─ + Pb4+ © 2010 ChemReview.Net v1w Pb2+ I2 + 6 H2O 2 H2O + SO2 Cr2O72─ + 7 H2O + 6 e─ 5 e─ + 4 H2O + MnO4─ Mn2+ + 8 OH─ Ag S4O62─ + 2 e─ 2 S2O32─ CrO42─ + 8 H+ + 3 e─ b. 3. a. b. d. SO42─ + 4 H+ + 2 e─ Page 1135 RA Module 37 — Electrochemistry 4. a. 1d: Pb2+ is releasing 2 e─ b. 2b: I2 is releasing 10 e─ c: 2d: SO2 d: 3b: Mn2+ Practice D 1a. Zn2+ + 2 e─ 2 Br─ Zn 2 e─ + Br2 (Balance central atom, then add e─ to balance charge) (Since 2 e─ on both sides, half-reactions can be added) Zn2+ + 2 Br─ Zn + Br2 Check balance of atoms and charge: 1 Zn, 2 Br atoms on both sides, overall neutral on both sides. 6 3 e─ + 8 4 H+ + 2 MnO4─ 1b. 2 MnO2 + 4 2 H2O 6 2 I─ 8 H+ + 2 MnO4─ + 6 I─ 3 I2 + 6 2 e─ (Use CA-WHe to balance, then 2x) (For e─ to balance, multiply by 3) 2 MnO2 + 4 H2O + 3 I2 Check: 8 H, 2 Mn, 8 O, 6 I atoms on both sides. Overall zero charge on both sides. Balanced. 2a. Zn is the left side RA because in its half-reaction, Zn donates electrons and becomes oxidized. 2b. I─ is the left side RA because in its half-reaction, I─ donates electrons and becomes oxidized. 4Co 3a. 8 e─ + 10H+ + NO3─ 4 Co + 10 H+ + NO3─ 4 Co2+ + 8 2 e NH ++ 3 H O 4 (4x) (1x) 2 4 Co2+ + NH4+ + 3 H2O Check: 4 Co, 10 H, 1 N, 3 O atoms on both sides; +9 charge on both sides. Cl2 + 2 e─ 2 Cl─ 3b. 2 e─ + 2 H+ + H2SO4 SO2 + 2 H2O 2 Cl─ + 2 H+ + H2SO4 SO2 + Cl2 + 2 H2O Substituting the trial coefficients: 2 HCl + H2SO4 SO2 + Cl2 + 2 H2O Check: 4 H, 2 Cl, 1 S, 4 O and zero net charge on each side. 3c. 3 Cd 6 e─ + 8 H+ + 2 NO3─ 3 Cd + 8 H+ + 2 NO3─ 3 Cd2+ + 6 e─ 2 NO + 4 H2O (3x) (2x) 3 Cd2+ + 2 NO + 4 H2O Check: 3 Cd, 8 H, 2 N, 6 O, +6 on both sides. Now add the spectators and finish by trial and error. ***** 3 Cd(NO3)2 + 2 NO + 4 H2O 3 Cd + 8 HNO3 Check: 3 Cd, 8 H, 8 N, 24 O, zero net charge on both sides. Balanced. 4a. NO3─ is the OA on the left because in its half-reaction, its N acquires electrons and becomes reduced. 4b. H SO 4c. NO ─ 2 4 3 ***** © 2010 ChemReview.Net v1w Page 1136 Module 37 — Electrochemistry Lesson 37B: Charges and Electrical Work Measuring Amounts of Charge Electrical charge is a fundamental quantity, like distance, mass, or time. The smallest charge we encounter in chemistry the charge on one electron. That charge is equal to, but opposite, the positive charge on a proton. Charge on one electron (e─) = 1 fundamental or elemental charge (Equation 1) Both electricity and redox reactions involve the movement or transfer of charges. In the SI system, the unit used to measure charge is the coulomb (symbol C). The following definition of the coulomb must be memorized. 96,500 coulombs = Charge on 1 mole of electrons (Equation 2) A more precise relationship is “one mole of fundamental charges = 96,485.3415 coulombs,” but the value to three significant figures will suffice for most calculations. This somewhat awkward definition for a coulomb allows for a simple relationship between the SI units joules and volts. A useful rule is When you see coulombs and electrons in a problem, write in your DATA table 96,500 coulombs = 1 mole of electrons The amount of charge that moves can be measured by either counting the number of electrons that move or the charge in coulombs that moves. Using equation 2, it is easy to convert between these two measures. Try this calculation: Q. 2500 electrons would have how many coulombs of charge? ***** Answer WANT: ? coulombs (C) DATA: (a single unit) 2500 electrons (a single-unit given) 96,500 coulombs = 1 mole of electrons (When you see coulombs and e─) A calculation that includes small particles (such as e─) and moles of those particles will likely need: 6.02 x 1023 e─ = 1 mol e─ (Avogadro’s conversion) ***** SOLVE: (If you want a single unit….) ? C = 2500 e─ • • 1 mol e─ 23 e─ 6.02 x 10 96,500 C 1 mol e─ = 4.0 x 10─16 coulombs ***** © 2010 ChemReview.Net v1w Page 1137 Module 37 — Electrochemistry Measuring the Force of Moving Charges Compare the stream of water from a garden hose, a pressure washer, and a fire hose. List the factors that determine how much work each can do to move an object. ***** The work which the streams of water can do in is determined by both the amount of water and the water pressure. In measuring the work that can be done by moving charges, two similar factors are important: the number of moving charges and the “force” behind them. Electromotive force (emf, symbol ) is defined as the work that moving charges can do per unit of charge. The SI unit that measures emf is the volt. (Equation 3) EMF in volts (V) ≡ Work in joules (J) Charge in coulombs (C) By this definition, one volt is defined as one joule per coulomb. 1 volt ≡ 1 joule/coulomb (Equation 4) Problems that include volts and joules or coulombs can be solved using our usual conversion-factor methods – IF volts are treated as an abbreviation for joules per coulomb, a ratio unit. Just as molarity (M) must be written as moles/liter to solve conversions, the rule is In problems with V and J or C, convert V to J/C to solve using conversions. • If ? V is WANTED, write WANTED = ? Volts = ? joules = coulomb • If X Volts is DATA, write in the data as an equality: X joules = 1 coulomb Using those rules, try this example. Q. In a 25 volt circuit, if a current in a wire can do 3.0 joules of electrical work, how many electrons are flowing in the wire? ***** WANTED: Number of electron charges flowing = ? e─ DATA: 3.0 joules (When volts and joules or coulombs are in a problem, treat V as J/C) 25 V 25 joules = 1 coulomb (V = J/C) (When a problem includes coulombs and a number of electrons, write) 96,500 coulombs = 1 mole of electrons Solve using conversions. ***** © 2010 ChemReview.Net v1w Page 1138 Module 37 — Electrochemistry SOLVE: ? e─ = 3.0 J • 1 C • 1 mol e─ 25 J 96,500 C • 6.02 x 1023 e─ 1 mol e─ = 7.5 x 1017 e─ Practice: Write and/or flashcard the numbered equations above until you can write these fundamental relationships from memory, then use those relationships from memory to solve these problems. Do half now, and half in your next practice session. If you need a review of conversion calculations, see Lessons 4D, 5D, and 11B. 1. Calculate the charge on one electron. 2. One coulomb is the charge on how many electrons? 3. What is the formula for calculating the work that flowing charge can do? 4. If the work that can be done by a current remains constant, but the number of electrons flowing is doubled, what must be true of the voltage in the circuit? 5. 0.35 coulombs of charge at 24 volts can do how much work? 6. If 12.0 joules of work can be done at 400. volts, how much charge is flowing? 7. If 1.20 x 1024 electrons can perform 4.0 joules of work, what is the emf of the electrons? ANSWERS 1. WANTED: ? coulombs electron (The unit of charge is coulombs. When you want something per one something, you want a ratio unit.) DATA: See coulombs and e─? Write SOLVE: ?C = e─ 96,500 coulombs = 1 mol e─ 96,500 C • 1 mol e─ 1 mol e─ 6.02 x 1023 e─ = 1.60 x 10─19 C e─ (When solving for a ratio, your conversions can be in any order, but they must be “right-side up” compared to these. Pick as your given a ratio that has one unit where you want it in the answer.) (You can also solve for the single unit by starting with ? C = 1 e─ , but when you are asked for a unit per one other unit, solving for the ratio is preferred.) 2. WANT: ? e─ C (When you want something per one something, you want a ratio unit) DATA: When you see C and e─, write: 96,500 C = 1 mol e─ SOLVE: ? e─ = C 1 mol e─ • 6.02 x 1023 e─ = 6.24 x 1018 e─ 96,500 C C 1 mol e─ © 2010 ChemReview.Net v1w Page 1139 Module 37 — Electrochemistry 3. Since the fundamental that involves work and charge is EMF in volts = Work in joules Charge in coulombs Work in joules = (Charge in coulombs) times (EMF in volts) 4. EMF in volts = Work in joules Charge in coulombs In this equation, if work is constant but charge is doubled, the voltage must be cut in half. 5. WANTED: DATA: ?J (the unit of work is joules) 0.35 C 24 V 24 J = 1 C (When V and J or C are in a problem, treat V as J/C) SOLVE: ? J = 0.35 C • 24 J • 1C = = 8.4 J 6. WANTED: ? coulombs (C) DATA: 400 V (If you want charge, the unit you want is coulombs) 400. J = 1 C (When V and J or C are in a problem, treat V as J/C) 12.0 J SOLVE: ? C = 12.0 J • 1 C = 0.030 C 400 J 7. WANTED: EMF in V = ? J C DATA: 1.20 x 1024 e─ = 4.0 J (When volts (emf) and J or C are in a problem, treat V as J/C) (equivalent: two measures of the same process) (In conversion calculations, when a ratio is WANTED, all of the data will be in equalities) 96,500 C = 1 mol e─ SOLVE: ( When you see C and e─ … ) (Want a ratio? Start with a ratio – one that has one unit where you WANT it.) ?V = ?J = 4.0 J • 6.02 x 1023 e─ • 1 mole of e─ = 2.1 x 10─5 J/C or volts C 1.20 x 1024 e─ 1 mol e─ 96,500 C ***** © 2010 ChemReview.Net v1w Page 1140 Module 37 — Electrochemistry Lesson 37C: Standard Reduction Potentials Ranking Oxidizing and Reducing Agents Every redox half-reaction has a characteristic tendency to occur. For example, • Molecules of elemental fluorine (F2) are strong oxidizing agents: they have a strong tendency to take electrons away from other substances. Fluorine as an element is difficult to work with in part because it is difficult to find a container that it does not react with. The half-reaction for the reaction of fluorine molecules can be written as F + 2 e─ 2 F─ (strong tendency to go forward) 2 • Neutral alkali metals, such as sodium metal, are strong reducing agents. Water is a relatively stable (non-reactive) compound when mixed with most substances. However, neutral sodium metal has such a strong tendency to give away one electron that it will donate an electron to water, creating hydrogen gas, heat, and flames that can cause the hydrogen produced to burn explosively in air. This sodium half-reaction can be represented as Na Na + + e ─ (strong tendency to go forward) Redox reactions are reversible, and the above reaction can be written in reverse, with the electrons on the left, just as for the F2 reaction above. However, since the formation of Na+ is favored, the half-reaction strongly tends to go backwards. Na+ + e─ Na (strongly tends to go in reverse) This logic holds for any reversible reaction or half-reaction: If a reversible reaction strongly tends to go forward, when written backwards it has a very weak tendency to go forward, and a strong tendency to go backwards. • A third type of redox half-reaction is one without a strong tendency to go in one direction over the other. An example is the hydrogen electrode half-reaction. Adding relatively strong reducing agents to acidic solutions (with H+ ions) tends to cause bubbles of H2 to form. However, H2 gas reacts with many oxidizing agents to form H+ ions. This half-reaction is written with electrons on the left as 2 H+ + 2 e─ H2 (tends to go either way) Every redox half-reaction can be written as particle being reduced (with the particle plus electrons on the left) or as a particle being oxidized (with electrons on the right). By convention, a listing that compares the tendency of half-reactions to go is usually written with the oxidizing agent (OA) being reduced (and the electrons reducing it) on the left. © 2010 ChemReview.Net v1w Page 1141 Module 37 — Electrochemistry Let’s rank the three half-reactions above in that manner, with the reaction that most tends to go forward at the top. Tendency To Be Reduced 2 F─ F + 2 e─ 2 (strong tendency to go forward) 2 H+ + 2 e─ H2 (tends to go in either direction) Na+ + Na (strong tendency to go backwards) e─ We will use these three half-reactions to anchor our understanding of redox reactions. Standard Reduction Potentials In a half-reaction, the tendency of an oxidizing agent to attract electrons (and become reduced) creates an electromotive force (emf, symbol ). Every half-reaction written as a reduction can be assigned a reduction potential, an emf in volts which measures the characteristic tendency of the reduction to go to the right. There is no logical zero (lowest possible value) for the scale for reduction potentials. Instead, we define the reduction potential for the hydrogen electrode reaction as zero volts, and measure all reduction potentials in volts relative to that zero value. At standard thermodynamic conditions (1 atm H2 pressure, 1 M H+ ions, and 25ºC), for the reaction H2 2 H+ + 2 e─ EMF ≡ ≡ 0 volts (an exact definition) At the right is a table of selected half-reactions. Note the position of our three anchor reactions. In standard reduction potential (SRP) tables: • • F2 + 2 e ─ in volts at 25ºC 2 F─ + 2.87 Standard means that all solution concentrations are 1 M and all gas pressures are 1 atm. Au3+ + 3e─ Cl + 2 e ─ Au 1.50 2 Cl─ 1.36 When the half-reaction components are at o . standard conditions, the is an Br2 + 2 e─ 2 Br─ 1.09 Ag+ + e─ I + 2 e─ Ag 2 I─ 0.80 0.54 Cu2+ + 2e─ 2 H+ + 2 e─ Cu 0.34 H2 0.0 Pb+2 + 2e─ Zn+2 + 2e─ Na+ + e─ Pb ─ 0.13 Zn ─ 0.76 Na ─ 2.71 • The highest emf values are at the top. • The half-reactions have electrons on the left side of the arrow, reducing the left-side particle. The form is: OA + electrons • o Standard Reduction Potentials RA The strongest oxidizing agent (the particle most likely to be reduced) is at the top left, with the highest reduction potential. © 2010 ChemReview.Net v1w 2 2 Page 1142 Module 37 — Electrochemistry • The strongest reducing agents (sRA) are at the bottom right of the table. Why? The weakest oxidizing agent is at the bottom left. When it loses its electrons, it becomes the strongest reducing agent. When the highest reduction potential values are listed at the top, the arrangement of a reduction potential table is Strongest OA Weakest OA + electron(s) + electron(s) Weakest RA Strongest RA o o = most positive = most negative Learn the rules above (but assume you can view a table), then try these problems. Practice 1. In a table of reduction potentials, in which corner are the strongest reducing agents? 2. Based on your experience with metals in coins and jewelry, rank these metals with the most likely to be oxidized (tarnished) first: Au, Ag, Cu. 3. Using the reduction potential table, rank these reducing agents from strongest to weakest: Au, Ag, Cu. 4. In the table of standard reduction potential values in this lesson, which particle that contains a halogen atom is a. The strongest oxidizing agent? b. The weakest oxidizing agent? c. The strongest reducing agent? d. The weakest reducing agent? e. Is the most easily reduced? f. Is the most easily oxidized? ANSWERS 1. Bottom right. In the same row as the weakest oxidizing agent. 2. Copper pennies turn dull (tarnish) relatively quickly. Silver needs to be polished, but not often. Gold never needs to be polished. Cu > Ag > Au 3. In the table, the reducing agents are on the right, with the stronger toward the bottom: Cu > Ag > Au 4b. I 4c. I─ 4d. F─ 4e. F has the highest reduction potential (tendency to reduce). 4. a. F 2 2 2 4f. I─ is the strongest reducing agent of the particles in the table containing a halogen atom. The strongest reducing agent by definition is the particle that is most easily oxidized. © 2010 ChemReview.Net v1w Page 1143 Module 37 — Electrochemistry Lesson 37D: Non-Standard Potentials: The Nernst Equation The table of standard reduction potentials lists emf in volts under standard–state conditions, where gas pressures are 1 atm, solution concentrations are 1 M, and the temperature is 25ºC. To use these emf values to solve problems, it is also helpful to know, • how the voltage will change if we multiply the half-reactions; and • how the values will change if conditions are not standard. Let’s address these issues one at a time. EMF: An Intensive Property Extensive properties vary according to the size or amount of matter in an object or system. Mass and energy are extensive properties. Intensive properties are those where the amount does not matter. Measurements of temperature or gas pressure do not depend on the number of particles in a sample. It may help in remember these terms and their meaning by using the chocolate rule. • The calories per gram of chocolate is an intensive property: the same in each piece. • The calories (energy) that you gain from eating chocolate are extensive: the amount that you consume matters. Generally, an intensive property will be a ratio of two extensive properties. In chemical reactions, emf is an intensive property: it has the same value in volts no matter how many electrons are flowing with that emf . One of the implications of intensive emf that we will need in calculations is volts/mole e─ = volts (Equation 1) For calculations of emf in chemical reactions, where charge must be balanced, the moles of electrons producing the emf do not change the voltage. In some cases, such as adding half-reactions in which electrons do not cancel on both sides, emf is extensive, but our focus for emf will be in reactions, where it is intensive. In unit cancellation for reaction calculations, volts and volts/mole are equivalent. Practice A: 1. For these three quantities: charge, emf, and work, a. What is the relationship between the quantities? b. Speculate on whether each property is extensive or intensive. Explain your reasoning. 2. Write the answer to this calculation with its units in three different but equivalent formats. = ? = 0.40 volts ─ 0.084 J C·mol © 2010 ChemReview.Net v1w Page 1144 Module 37 — Electrochemistry The Nernst Equation An equation that calculates emf values at non-standard conditions can be derived based on thermodynamics and our definition of electrical work. In thermodynamics, the maximum work that a system can do is measured by ΔG. ΔG = maximum work possible from a process = wmaximum possible Our definition that involves electrical work is: Emf ( ) = work (w)/charge Emf measures the maximum work that can be done per unit of moving charge. Solving for work: work (w) = charge times emf ( ) Equations in electrochemistry often use the symbol F, termed Faraday’s constant, to represent the charge (number of coulombs) per mole of electrons. F = Faraday’s Constant = 96,500 coulombs/mole of electrons (Equation 2) The symbol and value for Faraday’s Constant (F) must be memorized. In redox reactions, the number of charges involved in a reaction can be calculated as the number of moles of electrons moving (n) times the constant charge per mole of electrons (F). In symbols: Charge = nF so in the work equation above: work (w) = charge times emf ( ) = nF For sign conventions in chemistry, if emf (voltage) does electrical work, the system that includes the moving electrons must lose energy, and the work term is assigned a negative sign. The value for n is positive: the count of moles of electrons without regard to their negative charge. By these rules, the electrical work equations above can be written as ─ w = nF = ─ ΔG or wmaximum possible = ΔG = ─ nF Under standard state conditions: ΔGo = ─ nF o (Equation 3) (Equation 4) Into the thermodynamic equation that calculates non-standard free energy, ΔG = ΔGº + RT ln(Q) (Equation 5) substituting our relationship between free energy and emf above, this equation becomes ΔG = ─ nF = ─ nF º + RT ln(Q) (Equation 6) In calculations using equation 5, we can solve using J or kJ as units for ΔG and in R, as long as the units are consistent. However, in Equation 6, where is a variable, we must solve in joules (J) and not in kJ, because the units of are volts, which are joules/coulomb. Equation 6 can be simplified into the form known as the Nernst equation: © 2010 ChemReview.Net v1w = o ─ RT ln( Q ) nF (Equation 7) Page 1145 Module 37 — Electrochemistry Recall that Q is the reaction quotient: the value calculated by substituting concentrations or pressures into a K expression. For a half-reaction in a SRP table that is in the form: Q = [RA]/[OA] OA + electron(s) RA The charges represented by e─ are not included in a K expression. When conditions are not standard, the emf of an SRP half-reaction in can be calculated by reduction = o reduction ─ RT ln(Q) nF o = reduction ─ RT ln([RA]/[OA]) nF where o reduction = the voltage for the half reaction in the SRP table R = 8.31 J/mol·K (since V = J/C, use this R in calculations with J or V) T = temperature in kelvins = ºC + 273 = n = the moles of electrons in the half-reaction F = 96,500 coulomb/mol e─ = Faraday’s Constant Q = the reaction quotient [RA]/[OA], without units, with solution concentrations in mol/L and [solids, solvents, and pure liquids] = 1 . Let’s apply this equation to two examples. Q1. For the half-reaction Ag+ + e─ at 25 ºC, if the [Ag+] = 1.0 M, o Ag = + 0.80 volts a. What is value of Q ? b. What is the value of ln( Q ) ? c. What is the value of reduction ? ***** a. Qexp. ≡ Kexp. = (product of [products])/(product of [reactants]) = 1/[Ag+] In K expressions, the charges represented by e─ are omitted, and the [solids], including metals, and [pure liquids and solvents] are assigned a value of 1. Since [Ag+] = 1.0 M , Q = 1/[ Ag+] = 1/1.0 = 1.0 b. ln( Q ) = ln(1.0) = ln(e0) = 0 o c. reduction = reduction ─ (recall that units are omitted from K and Q values). RT ( 0 ) = o reduction nF In Q1, all components are at standard conditions, so the emf is the standard emf. Let’s calculate a voltage for the same half-reaction at non-standard conditions. Q2. For the half-reaction Ag+ + e─ If [Ag+] = 0.10 M, at 50. ºC, a. Find Q . Ag b. Calculate ln( Q ) . o = 0.80 volts c. What is the value of reduction ? ***** © 2010 ChemReview.Net v1w Page 1146 Module 37 — Electrochemistry a. Q = 1/[ Ag+] = 1/0.10 = 10. b. ln( Q ) = ln(10.) = 2.30 c. The equation needed is reduction = o reduction ─ RT ln( Q ) nF DATA: R = 8.31 J/mol·K (must use this R in a calculation with joules or volts) T = temperature in kelvins = ºC + 273 = 50.ºC + 273 = 323 K n = the moles of electrons in the half-reaction = 1 mol e─ F = 96,500 coulomb/mol e─ = Faraday’s Constant ln( Q ) = ln(10.) = 2.30 SOLVE: reduction = o reduction ─ RT • 1 ( 2.30 ) = nF 1 • reduction = + 0.80 volts ─ [ 8.31 J • (323 K) • mol·K 1 mol e─ = 0.80 volts ─ 0.064 J C·mol mol e─ • ( 2.30 ) ] = 96,500 C = 0.80 V ─ 0.064 volts = 0.74 V mol SF: The doubt in the hundredth’s place in 0.80 is the place with doubt in the answer. Note that for unit cancellation to work, we must use the rule that since volts are intensive, J/C and V/mol and J/C·mol are the same as volts. The big picture? Cutting the concentration of the silver ion, from 1.0 M at standard conditions to 0.10 M , reduced the half-reaction potential from 0.80 V to 0.74 V: a change of only 0.06 volts. When Q values are between 0.1 and 10, standard and non-standard voltages will be close. But not all Q values are between 0.1 and 10. Let’s try one more half-reaction at nonstandard conditions. Q3. For the half-reaction o 5 e─ + 8 H+ + MnO4─ = 1.51 volts Mn2+ + 4 H2O If [ MnO4─ ] = [ Mn2+ ] = 0.10 M and the pH = 6.0 at 25 ºC, find reduction . ***** The equation needed is reduction = o reduction ─ RT ln( Q ) nF DATA: R = 8.31 J/mol·K (since V = J/C, must use this R in calculations with J or V) T = temperature in kelvins = ºC + 273 = 25ºC + 273 = 298 K © 2010 ChemReview.Net v1w Page 1147 Module 37 — Electrochemistry n = 5 mol e─ = the moles of electrons in the half-reaction (an exact coefficient) F = 96,500 coulomb/mol e─ = Faraday’s Constant Q= = 0.10 [ Mn2+ ] ─ ][ H+ ]8 (0.10)(1.0 x 10─6)8 [ MnO4 = 1.0 x 10+48 If needed, adjust your work then finish. ***** ln( Q ) = ln( 1.0 x 1048 ) = 110.5 SOLVE: reduction = o reduction ─ RT • 1 ( 110.5 ) = nF red. = + 1.51 volts ─ [ 8.31 J • (298 K) • 0.567 J C·mol • 5 mol e─ mol·K = 1.51 volts ─ 1 mol e─ • (110.5 ) ] = 96,500 C = 1.51 V ─ 0.567 volts = 0.94 V mol For this half-reaction, the change in emf is substantial: from 1.51 V at standard conditions to 0.94 V at these non-standard conditions. If we test a variety of half-reactions, our general findings will be the following. In half-reactions, if the ion concentrations and/or partial pressures have values close to 1 , reduction potentials will be close to the standard reduction potentials. In a half-reaction with one ion that has a coefficient of one, if a concentration of the ion drops from 1 M to 0.10 M, its half-reaction potential will drop by 0.06 volts or less. If the coefficients of a half-reaction are more than one, and/or the concentrations or partial pressures have values far from 1, the standard and non-standard reduction potentials can vary substantially. These findings will help us to make some general redox predictions in the next lesson. Alternate Forms of the Nernst Equation This general form of the Nernst equation works for all temperatures and conditions: reduction = o reduction ─ RT ln( Q ) nF In the equation, R and F are constants. For reactions at 25ºC, T is also constant (298 K). Substituting values for those constants simplifies the Nernst equation to reduction = © 2010 ChemReview.Net v1w o reduction ─ 0.0257 V ln( Q ) n (at 25ºC) Page 1148 Module 37 — Electrochemistry If base 10 logs rather than natural logs are used with Q, using the rule ln x = 2.303 log x , the Nernst equation becomes o reduction ─ 0.0591 V log( Q ) n reduction = (at 25ºC) Textbooks may use the Nernst equation in any one of those three forms. All forms result in the same answer, but learning the top form is recommended because it can be used at all temperature conditions. Practice B: Try Problems 1 and 3, and Problem 2 if you need more practice. 1. Calculate the constant value for the term RT/F at 25 ºC, in millivolts. 2. Write the Q expression for a. Cl2(g) + 2e─ 2 Cl─ b. NO3─ + 2 H+ + e─ Zn 3. For the half-reaction Zn2+ + 2e─ 2+] = 0.050 M at 10.ºC, find the value of If [Zn o NO2(g) + H2O(l) = ─ 0.76 volts reduction . o = 0.00 volts 4. For the hydrogen electrode half-reaction: 2 H+ + 2e─ H2(g) If the pH = 6.8 and PH2 = 1.0 atm at 37.ºC, find the value of reduction . ANSWERS Practice A 1a. EMF in volts = Work in joules Charge in coulombs 1b. Your thoughts might include • Work and charge are quantities which depend on the size or amount of an object (extensive). • A ratio of two extensive properties generally results in an intensive property, and emf as defined in answer 1a is a ratio of two extensive properties. • Voltage is analogous to pressure, and pressure is an intensive property. 2. ? = 0.40 volts ─ 0.084 J C·mol = J 0.32 V or volts or mol C·mol Practice B 1. R•T • 1 F = 8.31 J • (298 K) • mol mol·K 96,500 C 2a. Qexpression = Kexpression = [Cl─]2 PCl 2 = 0.0257 J/C = 0.0257 V • 2b. Qexp. = 1 mV = 25.7 mV 10─3 V PNO2 [NO3─ ][ H+ ]2 In 2b, the lack of subscripts after the ions means they are in an aqueous solution, so liquid water is the solvent, and solvent concentrations, being essentially constant during reactions, are omitted from K expressions. © 2010 ChemReview.Net v1w Page 1149 Module 37 — Electrochemistry 3. To find a non-standard , use the Nernst equation: reduction = o reduction ─ RT • ln( Q ) nF DATA: R = 8.31 J/mol·K (must use this R in a calculation with joules or volts (= J/C)) T = kelvins = ºC + 273 = 10.ºC + 273 = 283 K n = the moles of electrons in the half-reaction = 2 mol e─ F = 96,500 coulomb/mol e─ = Faraday’s Constant Q = Kexpression = 1/ [Zn2+] = 1/0.050 = 20. ln( Q ) = ln(20.) = 3.00 reduction = ─ 0.76 V ─ [ 8.31 J • (283 K) • 1 • ─ mol·K 2 mol e 4. To find mol • (3.00 ) ] = 96,500 C = ─ 0.76 V ─ 0.0122 J • (3.00) = ─ 0.76 V ─ 0.037 V/mol = ─ 0.80 volts C·mol o reduction ─ RT • ln( Q ) at non-standard conditions, use reduction = nF DATA: R = 8.31 J/mol·K (use with J or V calcs) T = kelvins = ºC + 273 = 37ºC + 273 = 310. K n = the moles e─ in the half-reaction = 2 mol e─ F = 96,500 C/mol e─ [H+] = 10─pH = 10─6.8 = 1.6 x 10─7 M Q = Kexpression = PH2/ [H+]2 Q = 1/(1.6 x 10─7)2 = 1/(2.56 x 10─14) = 3.91 x 10+13 ln( Q ) = ln(3.91 x 10+13) = 31.30 red. = 0.000 V ─ [ 8.31 J • (310. K) • 1 • mol • (31.30 )] = ─ 0.418 V/mol = ─ 0.418 V mol·K 2 mol e─ 96,500 C ***** Lesson 37E: Predicting Which Redox Reactions Go At Standard Conditions, Which Redox Reactions Go? When oxidizing agents and reducing agents are mixed, some combinations react but some do not. If all particles are at standard conditions, the table of standard reduction potentials will quickly identify which redox reactions go and which do not. Steps to predict which redox side is favored at standard conditions: 1. On each side of a redox reaction, label one particle as the RA and another as the OA. If needed, use an SRP table to identify half-reactions and label the particles. 2. Using the table of standard reduction potentials (SRP), label the stronger oxidizing agent as sOA. Label the other particles that contain atoms that change oxidation number as the stronger reducing agent (sRA), the weaker oxidizing agent (wOA), or the weaker reducing agent (wRA) under standard conditions. We will use the lower case s and w to convey that we are comparing two different particles. For example, in a reaction, two different particles in the reactants and © 2010 ChemReview.Net v1w Page 1150 Module 37 — Electrochemistry products may both be relatively strong or weak oxidizing agents, but one will be stronger (sOA) than the other. 3. Equilibrium favors the side with the wOA and wRA. • If the wOA and wRA in a redox equilibrium are mixed, since they are favored at equilibrium, no substantial change will occur. • If the sOA and sRA are mixed, they will react substantially to form the wRA and wOA. An RA and OA react to form another RA and OA, which can then react to re-form the original RA and OA. Which reaction will win the battle to react more often? The one with the RA and OA that are stronger. That means the wRA and wOA are formed more often. Cover the answer below and, using the steps and table, try this question. Q1. Assuming all particles are under standard-state conditions, label each of the particles in this redox reaction as the sOA, sRA, wOA, and wRA. Then predict which side is favored at equilibrium, and whether the reaction will go to the right. Cu2+ + Pb Cu + Pb2+ ***** Answer First label the OA and RA on each side: Cu2+ + Pb Cu + Pb2+ • OA RA RA OA Pb metal on the left going to the right loses electrons, so Pb is an RA. • Cu metal on the right going to the left loses electrons, so Cu is an RA. In a redox reaction, each side must have one OA and one RA. Now find each particle in the reduction potential table. The particle that is on the left side closest to the top is the stronger oxidizing agent. Label it as the sOA. Using the same logic, label the remaining particles. ***** Cu2+ + Pb sOA sRA Cu + Pb2+ wRA wOA Which side is favored at equilibrium? Will the reaction go to the right? ***** The side with the wRA and wOA is favored. The reaction goes to the right. Add these to the list of rules above. 4. If a particle that contains an atom changing oxidation number is an oxidizing agent on one side, that atom is in the particle that is a reducing agent on the other side. The particle with that atom that is the stronger on one side is the weaker on the other. If an atom is in an sOA particle on one side, that atom is in a wRA on the other. 5. The sOA and sRA are on the same side of the reaction equation. The wOA and wRA are on the same side. © 2010 ChemReview.Net v1w Page 1151 Module 37 — Electrochemistry The SRP table can be used to label each redox component as sOA, wRA, etc., and it is a good idea to label at least two components using the table for safety’s sake. However, rules 4 and 5 mean that once you have labeled one particle correctly, all of the other labels can be done by inspection. Practice A 1. Using the table, below each reaction, label the particle that is the sOA, sRA, wOA, and wRA under standard conditions. After the reaction, write the side favored at equilibrium, left or right. a. Ag + Zn2+ b. Cl2 + I─ c. Ag+ + Pb Cl2 + 2 e─ 2 Cl─ 1.36 Br2 + 2 e─ 2 Br─ 1.09 Ag 0.80 I2 + 2 e ─ + Cl─ 2 I─ 0.54 Cu2+ + 2e─ Pb+2 + 2e─ b. In Problem 1b: Cu 0.34 Pb ─ 0.13 Zn2+ + 2e─ Ag + Pb2+ 2. If the reactants are mixed, will they react: a. In Problem 1a: in V Ag+ + e─ Ag+ + Zn I2 o Std. Reduction Potentials Zn ─ 0.76 c. In Problem 1c: The Diagonal Rule For Redox Reactions A second way to predict whether two particles will redox react, if all particles are under standard conditions, is the Diagonal \ Rule: If a table orders the oxidizing or reducing agents with the strongest at the top left (as in the SRP table), and if all particles are under standard-state conditions, • any particle on the left will react with any particle on the right below it in the table. In the table, diagonals that are \ will redox react. • The products of a reaction will be the particles on the opposite sides of the halfreaction arrows: the weaker agents and the / particles. • Diagonals in the opposite direction / will not redox react. Apply the diagonal rule to the following examples. Q1. Assuming standard conditions, use the reduction potential table in Practice Set A to label each combination as will redox react or won’t redox react when mixed. If the reaction goes, write the products. Check your answers after each part. Br2 + I─ b. c. Ag+ + Zn ***** d. a. © 2010 ChemReview.Net v1w Br2 + Cl─ Zn2+ + Pb Page 1152 Module 37 — Electrochemistry Answers Will redox react. Diagonal \ cases react. The products are a. Br2 + I─ ─, the particles on the opposite sides of the half-reaction arrows. I2 + Br b. Br2 + Cl─ Won’t redox react. Diagonal / cases do not react. c. Ag+ + Zn Will redox react. Ag and Zn2+ will form. d. Zn2+ + Pb Will not react. Zn2+ is on the left but below Pb on the right. The Diagonal Rule At Non-Standard Conditions Practice with the diagonal rule helps in developing your intuition about the patterns of redox behavior for particles. However, when used with a table of standard reduction potentials, the diagonal \ rule only works at standard conditions. The order of the halfreactions in the table is determined by the values of the reduction potentials. If conditions are not standard, those values change and the order of the half reactions changes. As we saw in the Nernst equation calculations above, if values for Q are between 0.1 and 10, reduction potentials change by only small amounts, and most predictions about which reactions go that are based on the quick diagonal rule will be accurate. In the next lesson, we will learn how to predict the direction of redox reactions that is not as quick, but is accurate under all conditions. Redox Reactions and Structure Our prediction rules for redox reactions are logical based on the definitions of strong and weak. The weaker is the particle as an OA or RA, the less likely it is to react. In a reversible redox reaction, the electrons move from side to side, but the particle that is the weaker reducing agent is less likely to give them up, so the electrons tend to be found on the weaker reducing agent at equilibrium. Redox behavior is explained by chemical structure. • A particle that attracts electrons strongly is a strong oxidizing agent without the electrons and a weak reducing agent with the electrons. • A particle that can accept electrons, but does not attract them strongly, is an sRA when it has the electrons and a wOA when it does not. Let’s summarize the new vocabulary and qualitative redox rules learned so far in this module. 1. Each side of a redox reaction equation has a one reducing agent (the electron donor) and one oxidizing agent (the electron acceptor). 2. For a pair of particles, one on each side, that contain the same atom changing oxidation number, • the particle that is an reducing agent on one side of a reaction loses electrons and becomes an oxidizing agent on the other. © 2010 ChemReview.Net v1w Page 1153 Module 37 — Electrochemistry • The particle that is an oxidizing agent on one side gains electrons to become the reducing agent on the other. • The particle that is the stronger on one side becomes the weaker on the other. 3. The sOA and sRA are on the same side of the redox reaction equation. The wOA and wRA are together on the opposite side. 4. Equilibrium favors the side with the wOA and wRA. When an sOA and sRA are mixed, they react. The sRA is oxidized, and the sOA is reduced. The wRA and wOA are formed. When a wOA and wRA are mixed, no substantial change occurs: the side favored at equilibrium already exists. 5. Under standard conditions, the role of particles in a redox reaction can be identified using a table of reduction potentials. In the table, • The format is OA + electrons • The strongest oxidizing agent is at the top left. Its half-reaction has the highest value. • The strongest reducing agent is at the bottom right. RA o 6. The diagonal \ rule: Under standard conditions, a particle in the table on the left of the arrow will redox react with a particle to the right of the arrow that is below it in the table. The opposite diagonals / form. The opposite diagonals / do not redox react. Practice B: Learn the rules in the summary above, then do these problems without looking back at the rules. 1. Use the table at the right to label each combination as will redox react or won’t redox react when mixed under standard conditions. If the reaction goes, write the products. a. Pb + Cu2+ b. F2 + Cl─ + Na+ c.. Ag + Pb2+ + Cl─ d. I2 + Na+ + F─ e. Cl2 + Na+ + Br─ 2. If copper metal is mixed with a mixture of 1.0 M Ag+ and 1.0 M Zn2+ ions, which ions will react with the copper? o Std. Reduction Potentials F2 + 2 e ─ in V 2 F─ + 2.87 Au3+ + 3e─ Au 1.50 Cl2 + 2 e─ 2 Cl─ 1.36 Br2 + 2 e─ 2 Br─ 1.09 Ag+ + e─ Ag 0.80 I2 + 2 e ─ 2 I─ 0.54 Cu 0.34 Pb ─ 0.13 Zn2+ + 2e─ Zn ─ 0.76 Na+ + e─ © 2010 ChemReview.Net v1w Cu2+ + 2e─ Pb2+ + 2e─ Na ─ 2.71 Page 1154 Module 37 — Electrochemistry 3. TRUE or FALSE. In redox equilibria, _____ a. An OA and an RA react to form another OA and RA. _____ b. The stronger reducing agent winds up at equilibrium with the electrons. _____ c. The weaker oxidizing agent winds up at equilibrium with the electrons. _____ d. A strong oxidizing agent will bond strongly to the electrons it acquires. _____ e. The stronger reducing agent has a stronger bond to its electrons than the weaker reducing agent. _____ f. The stronger oxidizing agent is on the same side of the reaction equation as the weaker reducing agent. _____ g. Equilibrium favors the weaker reducing agent. _____ h. The wOA attracts electrons more than the sOA. _____ i. The stronger oxidizing agent reacts to become the weaker reducing agent. 4. For these, you may consult the standard reduction potential table as needed. TRUE or FALSE. In redox equilibria, _____ a. Halogens tend to be strong oxidizing agents. _____ b. Halide ions tend to be strong reducing agents. _____ c. Halide ions can serve as oxidizing agents. _____ d. Alkali metal ions tend to be weak reducing agents. _____ e. Metallic gold is a weak oxidizing agent. _____ f. Metallic gold is a weak reducing agent. _____ g. Alkali metals tend to be strong reducing agents. 5. Use the reduction potential table in Problem 1 for these. If [ions] = Pgases = 1.0 a. Write the formula for a metal ion that will redox react with metallic silver. b. Which particles in the table will redox react with fluoride ion? c. Which particles in the table will redox react with sodium metal? ANSWERS Practice A 1. a. Ag + Zn2+ Ag+ + Zn wRA wOA Ag+ + Zn wRA sRA sRA Cl2 + I─ sRA wOA wRA wOA 2. a. In 1a: No, the reactants are favored. © 2010 ChemReview.Net v1w sOA I2 + Cl─ Ag + Zn2+ sOA c. sOA b. b. In 1b: Yes, the products are favored. c. In 1c: Yes Page 1155 Module 37 — Electrochemistry Practice B Will Redox React forming Pb2+ + Cu 1. a. Pb + Cu2+ (The \ react, the / are the products, the / don’t react.) b. F2 + Cl─ + Na+ Will Redox React and form Cl2 + F─ (Na+ will not react) c. Ag + Pb2+ + Cl─ Will Not Redox React d . I 2 + Na + + F ─ e. Cl + Na+ + Br─ 2 Will Not Redox React Will Redox React and form Br2 + Cl─ (Na+ will not react) 2. Copper metal on the right is below Ag+ ions on the left. Those two will react. Copper on the right is above Zn2+ on the left. Those two will not react. T 3a. An oxidizing and a reducing agent react to form another oxidizing and reducing agent. F b. The stronger reducing agent winds up at equilibrium with the electrons. F c. The weaker oxidizing agent winds up at equilibrium with the electrons. T d. A strong oxidizing agent will bond strongly to the electrons it acquires. F e. The stronger reducing agent has a stronger bond to its electrons than the weaker reducing agent. F f. T g. Equilibrium favors the weaker reducing agent. F h. The wOA attracts electrons more than the sOA. T i. T The stronger oxidizing agent is on the same side of the equation as the weaker reducing agent. The stronger oxidizing agent reacts to become the weaker reducing agent. 4a. Halogens tend to be strong oxidizing agents. F b. Halide ions tend to be strong reducing agents. F c. Halide ions can serve as oxidizing agents. F d. Alkali metal ions tend to be weak reducing agents. F e. Metallic gold is a weak oxidizing agent. T f. Metallic gold is a weak reducing agent. T g. Alkali metals tend to be strong reducing agents. 5. a. Write the formula for a metal ion that will redox react with metallic silver. Au3+ b. Which particles in the table will redox react with fluoride ion? None c. Which particles in the table will redox react with sodium metal? All left column particles above Na+ ***** © 2010 ChemReview.Net v1w Page 1156 Module 37 — Electrochemistry Lesson 37F: Calculating Cell Potential Using a table of reduction potentials, we have learned two ways to predict the side favored in redox equilibria: the weak-side-favored rule, and the diagonal \ rule. Both are accurate at standard conditions, but may not be accurate at non-standard conditions. At nonstandard conditions, reduction potentials change. A third and more systematic method, and one that predict the direction of redox reactions at both standard and non-standard conditions, is to calculate the cell potential. Cell potential measures the potential (emf) difference, in volts, between the two halfreactions that make up a redox reaction. If the cell potential is positive, the products of the redox reaction are favored. Why? We derived in an earlier lesson the relationship between work, free energy, and emf. ─ w = nF = ─ ΔG or wmaximum possible = ΔG = ─ nF These relationships are true under both standard and non-standard conditions. One implication of this equation is: If the emf (voltage) is positive, ΔG is negative. We know from thermodynamics that if ΔG is negative, the reaction is spontaneous: the reaction mixture will shift toward the products. A mixture of the reactants will form products, but a mixture of the products, being the favored side of the equilibrium, will not substantially react. We can summarize this as: If the emf of a redox reaction is positive, ΔG is negative, and the reaction is spontaneous. If a reaction mixture shifts toward the products, it must have a positive emf ( , in volts). To calculate a non-standard cell potential requires finding the standard potential first, so let’s start with calculating standard cell potentials. Cell Potentials At Standard Conditions To find cell potential, we balance and add half-reactions as we have done previously. We also attach the standard cell potentials and add as we did with Hess’s law calculations, with one significant exception. To find the standard cell potential for a redox reaction, use these steps. 1. Write the redox reaction, balanced or unbalanced. Under the reaction, write a dashed line - - - - - - - - - . 2. In the SRP table, find the two half-reactions that contain the particles in the redox reaction. One of the half-reactions in the table will need to be reversed to match the position of the particles in the redox reaction above the - - - - - , and one will not. Write both half-reactions under the - - - - -, one above the other, with one reversed. © 2010 ChemReview.Net v1w Page 1157 Module 37 — Electrochemistry o values after each equation. For the half-reaction reversed, 3. Using the table, list o change the sign of its value. 4. As with previous balancing using half-reactions, multiply each equation by an LCD so that the number of electrons gained in one half-reaction equals the number lost in the o other -- but do not change the value. Why? • When balancing ΔH and ΔG equations to add using Hess’s law, we multiplied the equation coefficients and ΔH or ΔG values by the same factor. However, ΔH and ΔG are extensive properties: the amounts determine their value. • EMF is an intensive property: it has the same value no matter how many particles create the EMF. 5. Add the reactions, canceling like terms on opposite sides of the arrows. Tweak the coefficients if needed to balance spectators, but the final balanced redox reaction should have lowest whole number coefficients If it does not, check the LCD used to balance the electrons in the half-reactions. o o 6. Add the values. The result is the standard cell potential: cell . 7. If the sum is positive, the products side is favored, and the reaction will go (form the products) when the reactants are mixed. To learn these steps, let’s try a problem. Q1. Using the table values at the right, a. calculate the standard cell potential for Ag+ + Zn Ag + Zn2+ b. Under standard conditions, will this reaction go? Std. Reduction Potentials o in volts at 25ºC Ag+ + e─ Ag 0.80 Zn2+ + 2e─ Zn ─ 0.76 ***** Steps 1, 2, and 3: o Ag+ + Zn Ag + Zn2+ cell = ? __ __ __ __ __ __ __ __ __ __ __ __ __ __ o Ag+ + e─ Ag = + 0.80 V o = + 0.76 V Zn Zn + 2 + 2 e ─ (reverse reaction, reverse sign) Steps 4, 5, and 6: o Ag+ + Zn Ag + Zn2+ cell = ? __ __ __ __ __ __ __ __ __ __ __ __ __ __ o o 2 Ag+ + 2 e─ 2 Ag = + 0.80 V (double coefficients, but not ) o = + 0.76 V Zn Zn + 2 + 2 e ─ __ __ __ __ __ __ __ __ __ __ __ __ __ __ o 2 Ag+ + Zn 2 Ag + Zn2+ cell = + 1.56 volts © 2010 ChemReview.Net v1w Page 1158 Module 37 — Electrochemistry To find standard cell potentials, we do not need to balance the final equation, or even the half-reactions, but to find non-standard potentials balancing will be necessary, and balanced equations are always preferred in chemistry. Part B: Since the cell potential is positive, equilibrium favors the products, and the reaction will go to the right. If redox reactions include spectator ions, the steps to find the standard cell potential are the same, but the half-reactions may be more difficult to identify. Try this example that includes spectator ions. Q2. Using the table values at the right, a. calculate the standard cell potential for Cu(NO3)2 + NO + H2O HNO3 Cu + b. Under std. conditions, will this reaction go? o in V (25ºC) Std. Reduction Potentials NO3─ + 4 H+ + 3 e─ NO + 2 H2O 0.96 NO3─ + 2 H+ + e─ NO2(g) + H2O 0.78 Cu2+ + 2e─ Cu 0.34 Cu2+ + e─ Cu+ 0.16 ***** Steps 1, 2, and 3: o Cu(NO3)2 + NO + H2O Cu + HNO3 cell = ? __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ o Cu2+ + 2e─ = + 0.34 V Cu NO + 2 H2O NO3─ + 4 H+ + 3 e─ o = ─ 0.96 V (write backwards, reverse sign) The nitrate ion is a spectator on the left, but is both a spectator and an oxidizing agent on the right. In writing half-reactions, the spectators are omitted, so omit nitrate on the left but not the right. Many particles are listed in more than one half-reaction in tables. Be sure to choose the halfreaction that matches the particles on both sides of the redox equation. Steps 4, 5, and 6: o =? Cu + HNO3 Cu(NO3)2 + NO + H2O __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ o = + 0.34 V (triple coefficients, not o) 3 Cu2+ + 6 2e─ 3 Cu o = ─ 0.96 V 2 NO + 4 2 H O 2 NO ─ + 8 4 H+ + 6 3 e─ (reverse and 2x) 2 3 __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ o 3 Cu2+ + 2 NO + 4 H2O 3 Cu + 8 H+ + 2 NO3─ cell = ─ 0.62 V Be sure that the electron balancing between the two equations is done with lowest common denominators. Check that the bottom reaction is balanced for atoms and charge. Then, to complete the balancing, add in the spectators. The above supplies trial coefficients. ***** © 2010 ChemReview.Net v1w Page 1159 Module 37 — Electrochemistry 3 Cu(NO3)2 + 2 NO + 4 H2O o 3 Cu + 8 HNO3 cell = ─ 0.62 V Check one more time for balanced atoms and charge. Part B: Since the cell potential is negative, equilibrium favors the reactants, and the reaction will not go as written. However, if a redox reaction has a cell potential that is negative, it can be written in reverse. The cell potential is then has the same magnitude, but is positive. The following reaction will go forward. 3 Cu + 8 HNO3 Practice A: o cell = + 0.62 V 3 Cu(NO3)2 + 2 NO + 4 H2O Do not consult tables on other pages to do these. 1. Using the table values at the right, a. calculate the standard cell potential for H+ + MnO4─ + Cl─ MnO2(s) + H2O + Cl2 o in volts at 25ºC Std. Reduction Potentials MnO4─ + 4 H+ + 3 e─ Cl2(g) + 2e─ MnO2(s) + 2 H2O 2 Cl─ MnO4─ + 8 H+ + 5 e─ Mn2+ + 4 H2O 1.68 1.36 1.51 2. Without consulting a table on other pages, given this reaction, o Br2(l) + 2I─ I2(s) + 2 Br─ cell = + 0.55 V o and this half-reaction, I2(s) + 2e─ 2 I─ cell = + 0.54 V o for this half-reaction: Br2(l) + 2e─ 2 Br─ find the Non-Standard Cell Potential If the temperature in a cell is not 25 ºC, or the solution concentrations are not all 1 M, or all gases are not at a partial pressure of 1 atm, conditions are not standard. In such cases, the redox cell potential can be calculated using the Nernst equation. The difference between the use of the Nernst equation with half-reactions and full redox reactions is: • A half-reaction Nernst equation uses the moles of electrons (n) in the one balanced half-reaction of interest; • A full-reaction Nernst equation must use for moles of electrons (n) the lowest whole number coefficient that equalizes the electrons in both of the two half-reactions that make up the redox reaction. © 2010 ChemReview.Net v1w Page 1160 Module 37 — Electrochemistry Steps To Calculate Non-Standard Cell Potential 1. Calculate the standard cell potential using the method above. 2. Adjust the cell potential for non-standard conditions using the Nernst Equation: cell = o cell ─ RT ln( Q ) nF where cell = the non-standard cell potential o cell = the standard cell potential R = 8.31 J/mol·K (use this R in a calculation with joules or volts) T = temperature in kelvins = ºC + 273 = n = the LCD coefficient moles of the electrons that is the same in the two balanced and added half-reactions and results in a balanced equation with lowest whole number coefficients. F = 96,500 C/mol e─ = Faraday’s Constant Q = the reaction quotient, with the terms for solids and pure liquids = 1 Apply those steps to the following example. Q3. For the reaction solved previously for standard potential in Q1, o 2 Ag + Zn2+ 2 Ag+ + Zn cell = 1.56 V what will be the cell potential if [Ag+] = 0.10 M, [Zn2+] = 0.50 M, and the reaction is run at 10. ºC? ***** When data is complex, list a careful DATA table and solve one step at a time. To solve for a non-standard cell potential, the equation is cell = o cell ─ RT ln( Q ) = ? nF DATA: o cell = + 1.56 V R = 8.31 J/mol·K T = 10ºC + 273 = 283 K n = the LCD moles of electrons that is the same in the two balanced and added halfreactions. When we solved to find the standard potential for this reaction in Q1, the electron transfer that was the same in the two added half-reactions was 2 moles of electrons. © 2010 ChemReview.Net v1w Page 1161 Module 37 — Electrochemistry o 2 A g+ + 2 e ─ 2 Ag Zn+2 + 2e─ = +0.80 V o = + 0.76 V Zn This means that in the Nernst equation, n = 2 mol e─ F = 96,500 C/mol = Faraday’s Constant Q=? For this reaction, the equilibrium constant expression is Kexp = Q = [Ag+ ]2/[ Zn+2 ] = ( 0.10 )2/( 0.50 ) = 0.020 = Q ln(Q) = ln( 0.020 ) = ─ 3.91 SOLVE: cell = o o cell ─ RT ln( Q ) = nF cell = + 1.56 V ─ [ 8.31 J • (283 K) • 1 2 mol e─ mol·K = + 1.56 V + cell ─ [ RT • 0.048 J C·mol • mol e─ 1 • ln( Q ) ] = ? nF • (─ 3.91 ) ] = 96,500 C = + 1.56 V + 0.048 volts = + 1.61 V mol The non-standard conditions change the standard emf for the cell slightly, but measurably. Summary: Rules for calculating cell potential 1. First find the standard cell potential, • • • Balance and add two table half-reactions to balance the redox reaction. o values to the half-reactions. Reverse the sign of the one reversed halfAdd o values. reaction, but do not change any o o values to find cell. Add the 2. For non-standard conditions, adjust cell with the Nernst equation. 3. A redox reaction will be spontaneous if its cell potential is positive. If negative, the reverse reaction will be spontaneous. cell is Practice B 1. For the reaction H+ + MnO4─ + Cl─ MnO2(s) + H2O(l) + Cl2(g) o cell = + 0.32 V with the standard cell potential solved in Practice set A above, a. what will be the cell potential if [Cl─] = 0.10 M, [H+] = 0.10 M, [MnO4─] = 0.20 M, PCl = 1.0 atm, and the reaction is run at 50.ºC? 2 b. Is this reaction spontaneous under the non-standard conditions in part a? c. Find ΔG for the reaction under the part a conditions, in kJ. © 2010 ChemReview.Net v1w Page 1162 Module 37 — Electrochemistry ANSWERS Practice A 1. Add two half-reactions to find the standard cell potential. Steps 1, 2, and 3: o MnO2(s) + H2O + Cl2 H+ + MnO4─ + Cl─ cell = ? __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ o = + 1.68 V MnO2(s) + 2 H2O 4 H+ + MnO4─ + 3 e─ o o = ─ 1.36 V (reverse reaction, reverse sign) 2 Cl─ Cl2(g) + 2e─ One half-reaction in the table will always be reversed. Steps 4, 5, and 6: o H+ + MnO4─ + Cl─ MnO2(s) + H2O + Cl2 cell = ? __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ 2 MnO 8 4 H+ + 2 MnO ─ + 6 3 e─ 4 4 2(s) + 2 H2O o 6 2 Cl─ 3 Cl 6 2e─ 2(g) + __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ 8 H+ + 2 MnO4─ + 6 Cl─ o = + 1.68 V = ─ 1.36 V o cell = + 0.32 V 2 MnO2(s) + 4 H2O + 3 Cl2 Check: 8 H, 2 Mn, 8 O, 6 Cl, and zero total charges on both sides. 2. o Br2(l) + 2e─ 2 Br─ cell = ? __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ o Br2(l) + 2 I─ I2(s) + 2 Br─ cell = + 0.55 V o I2(s) + 2e─ 2 I─ cell = + 0.54 V __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ o Br2(l) + 2e─ 2 Br─ cell = + 1.09 V (add reactions and add o values) Practice B 1a. To solve for a non-standard cell potential, use the Nernst equation. o cell ─ RT ln( Q ) = ? cell = nF DATA: o R = 8.31 J/mol·K T = 50ºC + 273 = 323 K cell = + 0.32 V n= the LCD coefficient moles of the electrons in the two added half-reactions that produce lowestwhole-number coefficients in the final balanced equation. n = 6 mol e─ in the half-reaction balancing in Practice A1 above. F = 96,500 C/mol Q = the reaction quotient, with [solids, solvents, and pure liquids] = 1 In this K and Q expression, MnO2(s) and the solvent H2O are assigned values of 1. * * * ** © 2010 ChemReview.Net v1w Page 1163 Module 37 — Electrochemistry (PCl2)3 Kexp = Qexp = [H+ ]8 [MnO4─ ]2 [Cl─]6 = ( 1.0 )3 (0.10)8 (0.20)2 (0.10)6 = 1.0 4.0 x 10─16 = 2.5 x 1015 = Q ln(Q) = ln(2.5 x 1015) = + 35.4 SOLVE: o o cell ─ RT ln( Q ) = cell ─ [ RT • 1 • ln( Q ) ] = ? nF nF cell = + 0.32 V ─ [ 8.31 J • (323 K) • 1 • mol • (+ 35.4 ) ] = ─ mol·K 6 mol e 96,500 C cell = = + 0.34 V ─ 0.164 J = + 0.34 V ─ 0.164 V = + 0.18 V C·mol mol (V = J/C = V/mol = J/C·mol) 1b. Since the cell potential is positive, the reaction is spontaneous: The equilibrium will shift toward the products, and the reaction as written will go. 1c. ΔG = ─ nF = ─ (6 mol e─) · 96,500 C · 0.18 J · 1 kJ = C 103 J mol e─ ─ 104 kJ ***** © 2010 ChemReview.Net v1w Page 1164 Module 37 — Electrochemistry Summary: Electrochemistry 1. Definitions a. Oxidation is the loss of electrons. Reduction is the gain of electrons. b. Reducing agents are particles that lose their electrons and are oxidized. c. Oxidizing agents are particles that gain electrons and are reduced. d. A redox reaction is an electron transfer: one or more electrons move from one particle to another. 2. Redox Reactions and Their Components a. Redox reactions are reversible: in a closed system, reactions go to equilibrium (though the side with the reactants or the products can be strongly favored). b. Each side of a redox reaction equation has a one reducing agent (the electron donor) and one oxidizing agent (the electron acceptor). c. For a pair of particles, one on each side of the reaction equation, that contain the same atom changing oxidation number, • a particle that is a reducing agent on one side of a reaction loses electrons and becomes an oxidizing agent on the other. • A particle that is an oxidizing agent on one side gains electrons to become the reducing agent on the other. • A particle that is the stronger on one side is the weaker on the other. d. In redox reaction equations, the sOA and sRA are on the same side of the arrow. The wOA and wRA are on the other side. e. Under standard conditions, the role of particles in a redox reaction can be identified using a table of Standard Reduction Potentials (SRP). In the SRP table, • The format is OA + electrons • The strongest oxidizing agent is at the top left in the half-reaction with the highest o value. • The strongest reducing agent is at the bottom right. RA . 3. The Direction of Redox Reactions a. Equilibrium favors the side with the wOA and wRA. • When a sOA and sRA are mixed, they react to form wOA and sRA. • When a wOA and wRA are mixed, the favored particles exist, and no substantial change occurs. b. The diagonal \ rule: Under standard conditions, a particle in the table on the left of the arrow will redox react with a particle to the right of the arrow that is below it in the table. Diagonals in the opposite direction / will not redox react. c. If a redox reaction is spontaneous (goes toward the products), it must have a positive emf (voltage). © 2010 ChemReview.Net v1w Page 1165 Module 37 — Electrochemistry 4. Conversions, Equations, Units, and Constants a. Charge on one electron (e─) = 1 fundamental charge b. Charge on 1 mole e─ = 96,500 coulombs c. Faraday’s Constant = F = 96,500 C/mol e─ d. EMF in volts = Work in joules Charge in coulombs e. One volt ≡ 1 joule/coulomb f. Since emf is an intensive property, in terms of unit cancellation, J/C and g. Δ G = ─ n F V/mol and J/C·mol are the same as volts. = maximum work that can be extracted from moving electrons h. When ΔG is negative, cell potential is positive, and equilibrium favors the right side of the reaction equation. i. The non-standard emf for a reaction, half-reaction, or cell is calculated by the Nernst Equation: = o ─ RT ln( Q ) nF where R = 8.31 J/mol·K (use this R in a calculation with joules or volts) T = temperature in kelvins = ºC + 273 = n = the LCD moles of electrons that balance the two half-reactions F = 96,500 C/mol e─ = Faraday’s Constant Q = the reaction coefficient, using solution concentration in M, partial pressures of gases in atm, and [solids, solvents, and pure liquids] = 1 5. Calculating cell potential a. First find the standard cell potential of a redox reaction or electrochemical cell, • • Balance and add two table half-reactions to balance the redox reaction. Add o values to the half-reactions. Reverse the sign of the one reversed • half-reaction, but do not change o values. Add the o values to find o cell. b. For non-standard conditions, adjust cell with the Nernst equation. c. A redox reaction will be spontaneous if its cell potential is positive. If cell is negative, the reverse reaction is spontaneous. ### © 2010 ChemReview.Net v1w Page 1166 Module 38 — Electrochemical Cells Module 38 — Electrochemical Cells Prerequisites: Complete Module 37 before beginning Module 38. ***** Lesson 38A: Cells and Batteries Cell Terminology An electrochemical cell (also known as a galvanic cell or voltaic cell) creates electrical current: the flow of electrons through a conductor. A cell has two sections, called half-cells. In each half-cell, one redox half-reaction takes place. Oxidation occurs in one half-cell and reduction in the other. A cell physically separates the two redox half-reactions so that for the electron transfer in the redox reaction to occur, the electrons must travel in a conductor (usually a wire) between the two half-cells. As the electrons move through the wire, their energy can be employed to do electrical work, such as causing an electric motor to turn. The cell converts chemical to electrical energy. The motor converts electrical energy to mechanical work. Battery is a term applied to either a single cell or a series of connected cells. Connected cells can produce a voltage that is higher than is possible from a single cell. A Cell Example A common type of cell is based on metals and aqueous metal ions. Such a cell consists of • two half-cells containing dissolved metal ions. The two solutions may contain either different metal ions or the same ions with different concentrations. • Into each solution is partially submerged an electrode: a piece of metal that is the same element as the metal ions in the solution. To build one example of such a cell, a strip of silver metal is partially submerged in a solution of silver nitrate, containing silver ions. In a separate container, copper metal is placed into a solution of cupric nitrate, containing copper(II) ions. In each solution, a half-reaction can occur: o Ag+ + e─ Ag red. = + 0.80 V o Cu Cu2+ + 2e─ red. = + 0.34 V Ag ~~ However without extra electrons being added to or removed from a half-cell, the metals and ions in each half-cell are at equilibrium, and no net reaction occurs. Cu ~~ Ag+ soln. ~ ~ ~~ Cu2+ soln. A redox reaction does occur, and current will flow, when the two halves of an electrochemical cell are connected in two ways. © 2010 ChemReview.Net v1w Page 1167 Module 38 — Electrochemical Cells • One connection must be a conductive solid between the two metals through which electrons can move with minimal resistance. This is usually a metal wire. • The other connection must have a way for ions to flow. Examples include a salt bridge (a tube containing a solution of ions) or a porous disk through which ions can move slowly. In a cell, electrons travel through the wire from one half-cell to the other. To maintain balanced charges, ions must also travel between the two half-cells. A salt bridge or porous disk allows the slow movement of positive ions in the same direction as the electrons in the wire and negative ions in the direction opposite the flow of electrons. If a low-voltage incandescent bulb is attached at both of its connectors to the wire, the bulb will glow as the flow of electrons does work: heating the resistant wire in the bulb. If a digital voltmeter (symbol V in the diagram) is placed in the circuit, it will measure the emf difference (also called the potential difference or cell potential) between the two cells. The Direction of the Current at Standard Conditions How can we predict the direction of the electron flow in the wire between the cells? The rules differ for standard and non-standard conditions. Let’s take the standard conditions case first. • The table of reduction potentials lists its half-reactions in order, with the strongest oxidizing agents at the top left corner. • The stronger oxidizing and reducing agents are always on the same side of the redox reaction equation, but they are always in different half-cells. In a cell, the flow of electrons is logical. As the stronger reducing agent (sRA) reacts, it loses electrons and is oxidized. The electrons leave the half-cell with the stronger reducing agent (sRA) and travel through the wire toward the other half-cell. The above rule, focused on the sRA, is all that you need to predict the direction of the electron flow and the chemical changes in the cell. However, we can also view the changes from the opposite perspective. • The electrons flow through the wire toward the half-cell that has the particle that more strongly attracts the electrons: the stronger oxidizing agent (sOA). • wire V Ag Cu Salt Bridge ~~ ~ ~ Ag+ © 2010 ChemReview.Net v1w ~ ~ ~~ As the stronger oxidizing agent (sOA) reacts, it gains electrons and is reduced. Both sets of rules work. These lessons will usually refer to the first set, based on the sRA. Cu2+ Page 1168 Module 38 — Electrochemical Cells In general, if you are given two half-cells at standard conditions and are asked to predict the direction of electron flow: • Find the particle of the 4 that is on the right and lower in the SRP table. That’s the stronger reducing agent. • The electrons will travel through the wire in the direction away from the half-cell with the sRA. Apply the rules to this problem. Q. In the Ag/Cu cell in the diagram above, o Ag+ + e─ Ag red. = + 0.80 V o Cu Cu2+ + 2e─ red. = + 0.34 V In which direction will the electrons flow: to the left or right? ***** The stronger reducing agent in this system is Cu. The electrons travel away from the Cu and through the wire. In this cell as written, the electrons travel to the left. How the Charges Move and Particles Change In a cell, the electrons are a focus, because the electrons can be harnessed to power our MP3 players, etc. Let’s track how and why the electrons travel between the half-cells. Charges travel throughout the cell, so we can begin anywhere. Let’s start with the stronger reducing agent. • The sRA has electrons it has a tendency to give away. If the electrons can travel toward another particle that tends to accept electrons, and if charges can remain balanced, the sRA can give electrons away. A cell fulfills these conditions. • In the Ag/Cu cell above, the sRA is Cu. At the surface of the copper electrode, a neutral Cu metal atom can give away two electrons to the conductive metal around it, becoming a Cu2+ ion. This ion is more stable when dissolved in water, and it moves from the metal surface into the surrounding solution. In this process, the copper metal loses mass, and the solution gains Cu2+ ion, which is blue when dissolved in water. As more Cu oxidizes, the [Cu2+] in the solution increases and blue color of the solution intensifies. • The addition of the two electrons to the surrounding copper metal creates an unbalanced negative charge, and those like charges repel. This repulsion pushes electrons in the metal into the wire. In the wire, electrons can flow away from the copper electrode toward the silver electrode. The silver electrode is in contact with the particle that most attract electrons in the cell: Ag+ ion, the stronger oxidizing agent. • When the copper metal first loses its two electrons, the two extra electrons in the metal create a kind of “pressure” of unbalanced charge throughout the connected metals. The charge can be rebalanced if electrons either return to the Cu2+ ion, or, © 2010 ChemReview.Net v1w Page 1169 Module 38 — Electrochemical Cells at the opposite end of the connected metals, attach to Ag+ ions. The Ag+ ion is a stronger electron attractor than Cu2+ ion, so electrons flow toward the Ag+. The difference between the values in the two half cells is the measure of the “pressure” (the voltage) of the flow. • To maintain a balanced charge in the metals, two electrons on the surface of the silver metal electrode react with two Ag+ ions. This creates two atoms of neutral Ag metal. In the Ag/Ag+ half-cell, as electrons flow in from the wire and react, the concentration of the Ag+ ion decreases, and the mass of Ag metal increases. • In metals, electrons move easily, but ions do not. In solutions, ions move easily. As the negative electrons flow in one direction through the wire, charges on ions must flow slowly through the salt bridge to keep charges balanced in both cells. . • In a cell or redox reaction, the sRA is successful at giving away its electrons, and the sOA is successful at attracting them. As the cell runs, the reactants are gradually used up, the products gradually form, and the voltage gradually drops. When the value for Q (the reaction quotient) reaches K for the redox reaction, the reaction is at equilibrium: ion concentrations no longer change and electrons no longer flow. Unless the ions are replenished in some way, the voltage remains at zero and the cell (a battery) is dead. For electrochemical cells, let’s summarize. A redox reaction is a combination of two balanced half-reactions. An electrochemical cell is a combination of two half-cells. In an electrochemical cell, a redox reaction takes place in which the two half-reactions are carried out in the two separate half-cells. • One half-cell has the particles in one half-reaction. The other half-cell has the particles in the other half-reaction. • One half-cell has the sRA and wOA. The other has the sOA and wRA. In a redox reaction equation, the sRA and sOA are on the same side. In a cell, the sRA and wOA are in the same half-cell. As current flows through the wire connecting the cell electrodes, • the two half-reactions proceed that added together produce the redox reaction. • Reactants and products shift toward the side favored at equilibrium. • sRA is oxidized to become wOA, and sOA is reduced to become wRA. • Electrons flow in the wire from the sRA side toward the side with the sOA. • Ions move through the salt bridge to balance charge. As the sRA and sOA are used up and wRA and wOA build up, the voltage drops gradually. When Q = K for the redox reaction, the voltage is zero. © 2010 ChemReview.Net v1w Page 1170 Module 38 — Electrochemical Cells Practice A: Learn the rules above, then try these problems. 1. Using the table, for these unbalanced redox equilibria under standard conditions, circle the particle that is the stronger reducing agent. a. Ag + Zn2+ b. Cl2 + Br─ Std. Reduction Potentials o in volts at 25ºC Cl2 + 2 e─ 2. Under standard conditions, using the table above and the following representations for two half-cells connected by a porous membrane, state whether electrons would flow through a wire connecting the electrodes to the left or right. 2 Br─ 1.09 Ag+ + e─ Br2 + Cl─ 1.36 Br2 + 2 e─ Ag+ + Zn 2 Cl─ Ag 0.80 Cu2+ + 2e─ Cu 0.34 Pb2+ + 2e─ Pb ─ 0.13 Zn2+ + 2e─ Zn ─ 0.76 2+ 2+ a. ⎩ Cu and Cu ⎭ and ⎩ Zn and Zn ⎭ 2+ 2+ b. ⎩ Pb and Pb ⎭and ⎩ Cu and Cu ⎭ 3. As the electrons flow in the cells in Problem 2 , which particle is oxidized? a. In 2a: b. In 2b: 4. As the electrons flow in Problem 2 above, which metal increases in mass? a. In 2a: b. In 2b: 5. As the electrons flow in Problem 2 above, which particle increases in concentration in its solution? a. In 2a: b. In 2b: Calculating Cell Voltage at Standard Conditions The emf difference between two half-cells (also known as the potential difference or cell potential or cell voltage) is an important property of cells and batteries made from cells. A standard cell voltage is easy to calculate: it’s the cell potential that you calculated in a previous lesson. The cell potential can also be adjusted for non-standard conditions by using the Nernst equation, as you did in Lesson 37F. Our prior calculations of cell potential were based on knowing the redox reaction. Let’s tweak our steps to find cell potential knowing the two half-cells as we solve the following problem. © 2010 ChemReview.Net v1w Page 1171 Module 38 — Electrochemical Cells 2+ + Q. A cell consists of ⎩ Ag and Ag ⎭ and ⎩ Pb and Pb ⎭ . The two half-cells are connected by a wire between the electrodes and a salt bridge between the solutions. a. Under standard conditions, what is the predicted cell voltage? b. In which direction do electrons in the wire flow? c. Which particle is oxidized as the current flows? d. Which particle concentration will increase as the electrons flow? Part a. To find the standard cell potential, add the half-reactions. 1. Using the SRP table, find the two half-reactions that contain the particles in each of the two half-cells. o value. Then copy the higher in the table of the two half-reactions. Attach its o 2. Under the first half-reaction, write the second table half-reaction reversed. Attach with its sign reversed. o 3. Balance the electrons, add the two half-reactions, and add to find cell. Writing the higher half-reaction in the direction shown in the table guarantees that the standard cell potential is always positive. For the problem above, do those steps, then check your answer below. ***** 2 Ag+ + 2 e─ o = + 0.80 V 2 Ag (write higher first, as in table) o = + 0.13 V (write lower reversed, reversing o) Pb P b2 + + 2 e ─ __ __ __ __ __ __ __ __ __ __ __ __ __ __ o 2 Ag+ + Pb 2 Ag + Pb2+ cell = + 0.93 V = the WANTED voltage. For problems in which all you are asked is the cell potential (or emf or voltage) under standard conditions, you can simply subtract the two table values, the higher in the table minus the lower. Using the table, try that rule to find the emf difference for the two halfreactions in this problem, then compare to the answer above. ***** o higher ─ o lower = 0.80 V ─ (─ 0.13 v) = + 0.93 V = the same answer as above. However, in most problems you will be asked other questions about the reaction and/or cell. To answer those questions, you will need to write and balance the two half-reactions. Let’s remember the higher minus lower rule this way: To check a standard cell emf, use o cell = o higher ─ o lower = a positive voltage Part b: In which direction will the electrons flow? ***** © 2010 ChemReview.Net v1w Page 1172 Module 38 — Electrochemical Cells The positive voltage means the reaction equation goes to the right. This means the sRA is on the left side of the reaction equation, since the right side favored must have the wRA and wOA. On each side of the balanced equation is only one RA. The RA on the left side is Pb, so it must be the sRA. The electrons flow from the sRA (Pb) into the wire, traveling to the left in the cell graphic in the problem, in order to react with the Ag+ in the other cell, which is the sOA. Part c. Which particle is oxidized as the current flows? ***** In redox reactions, the stronger reducing agent is oxidized. Here, that’s Pb. Part d: The concentration of a solid metal is constant, so only the ions in solution can change concentration. In this cell, one metal ion concentration increases because as one metal reacts (is used up), and its metal ion forms. The metal that reacts is the sRA: Pb. The ion that increases concentration is therefore Pb2+. ***** Calculating Cell Voltage at Non-Standard Conditions The voltage for a cell is different at non-standard conditions than at standard conditions. At non-standard conditions, the direction of the electron flow may be the same as at standard conditions, or it may be reversed. o To find a non-standard cell voltage, the steps are: find the standard cell, then apply the Nernst equation. Apply those steps to this problem. 2+ + Q. Using the answers found above for the cell ⎩ Ag and Ag ⎭ and ⎩ Pb and Pb ⎭ , if the [Pb2+] = 1.0 M and the [Ag+] = 1.0 x 10─4 M at 25ºC, a. Which components are at lower than standard concentrations: sRA, sOA, wOA, or wRA? b. Will the lower concentration for that component tend to increase or decrease the voltage (pressure) in the cell? c. Calculate the voltage of the cell at these non-standard conditions? d. Was the change in voltage between standard and these non-standard conditions as you predicted in part b? e. In which direction will the electrons flow? ***** Answers a. The Ag+ is at lower than standard (1 M) concentration, and it is the sOA. b. The role of the sOA is to pull electrons toward it. If the sOA has lower concentration, there will be fewer particles pulling the electrons. You might predict that this would lower the pressure of the flowing electrons: their voltage. c. To find the cell potential at non-standard conditions, © 2010 ChemReview.Net v1w Page 1173 Module 38 — Electrochemical Cells • First find the balanced equation and standard reactions (done above), • then apply the Nernst equation: value by adding the half- o cell ─ RT ln( Q ) = ? nF cell = c DATA: o cell = the standard cell potential = + 0.93 V R = 8.31 J/mol·K o (solved above) F = 96,500 C/mol e─ T = 25ºC + 273 = 298 K n = 2 mol e─ in the half-reactions that balanced the equation above. Q=? The balanced equation (solved above) is: 2 Ag+ + Pb 2 Ag + Pb2+ To find the Q value: Kexp = Qexp = [ Pb+2 ] / [Ag+ ]2 = ( 1.0 )/( 1.0 x 10─4 )2 = 1.0 x 10+8 = Q ln(Q) = ln(1.0 x 10+8) = + 18.42 SOLVE: cell = o cell ─ RT ln( Q ) = nF cell = + 0.93 V ─ o 8.31 J • (298 K) • mol·K = + 0.93 V ─ cell ─ RT • 1 2 mol e─ • 1 • ln( Q ) = ? nF mol e─ • (+ 18.42 ) = 96,500 C 0.236 J = + 0.93 V ─ 0.236 V = + 0.69 V C·mol mol These non-standard conditions lower the standard emf for the cell by 0.24 V. d. This lower voltage is as predicted qualitatively in part b. e. In which direction will the electrons flow? ***** The cell voltage, though lower, is still positive. This means the reaction equation as written will still go to the right. The electrons flow through the wire from the sRA Pb on the left side of the balanced equation. The electrons will flow from the Pb in the right half-cell to the left half-cell as drawn in the problem. © 2010 ChemReview.Net v1w Page 1174 Module 38 — Electrochemical Cells Practice B 1. In the cell represented by 2+ 2+ ⎩ Zn and Zn ⎭ and ⎩ Pb and Pb ⎭, o Standard Reduction Potentials if the [Pb2+] = 0.10 M and the [Zn2+] = 4.0 x 10─3 M at 25ºC, F2 + 2 e ─ in volts at 25ºC 2 F─ + 2.87 2 H+ + 2 e─ H2 0.0 a. What is the voltage of the cell? Pb2+ + 2e─ Pb ─ 0.13 b. In which direction will the electrons flow through the wire: To the left or right? Zn2+ + 2e─ Zn ─ 0.76 c. Find the free energy change of the reaction, in kJ. Li+ + e─ ─ 3.05 Li 2. Based on the table above, what is the highest voltage possible from a single electrochemical cell under standard conditions? ANSWERS Practice A 1. a. Ag + Zn2+ 2. a. ⎩ Cu and Cu Ag+ + Zn 2+ b. Cl2 + Br─ Br2 + Cl─ 2+ ⎭ and ⎩ Zn and Zn ⎭ Zn is the sRA. The flow is from the Zn electrode: to the left. 2+ 2+ b. ⎩ Pb and Pb ⎭and ⎩ Cu and Cu ⎭ The flow is from the Pb electrode: to the right. 3. The reaction in a cell is a redox reaction. The particle that is oxidized in a redox reaction is the stronger reducing agent (sRA). a. In 2a: Zn b. In 2b: Pb 4. In these cells, the ion that reacts to form metal is the sOA in each cell. The sOA is the particle in the left column and more toward the top, which in both cells is Cu2+ which reacts to form Cu. In 2a and in 2b: Cu 5. In these cells, ions form as a metal reacts. The metal that is a reactant is the sRA in these cells. The ions that form are a. In 2a: Zn2+ b. In 2b: Pb2+ Practice B 1. The steps to find the non-standard potential based on two half-cells are • Write the half-reaction higher in the table. Under it, write the other half-reaction, reversed. o • Balance and add the two half-reactions, find cell, then apply the Nernst equation. ***** © 2010 ChemReview.Net v1w Page 1175 Module 38 — Electrochemical Cells Pb2+ + 2 e─ o Pb = ─ 0.13 V (write higher first, as in table) o (write lower reversed, reversing ) o Zn Zn+2 + 2e─ = + 0.76 V __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ o Pb2+ + Zn Pb + Zn+2 cell = + 0.63 V Then apply the Nernst equation. cell = o cell ─ RT ln( Q ) = ? nF DATA: o cell = + 0.63 V R = 8.31 J/mol·K n = 2 mol e─ in the balanced half-reactions Q= ? T = 25ºC + 273 = 298 K F = 96,500 C/mol e─ For the balanced equation above, Kexp = Qexp = [ Zn+2 ] / [Pb2+ ] = (4.0 x 10─3 )/( 1.0 x 10─1 ) = 4.0 x 10─2 = Q ln(Q) = ln( 4.0 x 10─2 ) = ─ 3.22 SOLVE: cell = o cell ─ RT ln( Q ) = nF o cell ─ [ RT • 1 • ln( Q ) ] = ? nF cell = + 0.63 V ─ [ 8.31 J • (298 K) • 1 • mol • (─ 3.22 ) ] = ─ mol·K 2 mol e 96,500 coulomb = + 0.63 V + 0.041 J = + 0.63 V + 0.041 V = + 0.67 V C·mol mol In unit cancellation, J/C and V/mol and J/C· mol are the same as volts. 1b. Since the voltage is positive for the balanced equation, the reaction favors the right side products. The electrons will flow from the sRA on the left side of the balanced equation: Zn. In the cell as shown in the problem, the electrons will flow from the Zn electrode on the left to the right. = ─ (2 mol e─) · 96,500 C · 0.67 J · 1 kJ = C 103 J mol e─ 2. Based on our check rule, 1c. ΔG = ─ nF To check a standard cell emf, use o cell = o higher ─ ─ 129 kJ o lower = a positive voltage the largest difference will be for the top and bottom table reactions. o cell = o highest ─ o lowest = 2.87 V ─ (─ 3.05 v) = + 5.92 V ***** © 2010 ChemReview.Net v1w Page 1176 Module 38 — Electrochemical Cells Lesson 38B: Anodes and Cathodes Electrodes Each half-cell has one electrode: a conductive solid component of the half-cell (usually a metal) to which wires can be attached. When two half-cells are connected, one electrode is termed the anode and the other the cathode. The rules to remember are: Labeling Electrodes · Oxidation occurs at the anode (─). Reduction occurs at the cathode (+). · Electrons always move from the anode thru the wire toward the cathode. · In a cell, the sRA is being oxidized, so the anode is in the half-cell with the sRA. The two electrodes on a cell (or a battery containing one or more cells) where wires are attached are termed the two poles. On a battery, the pole labeled (─) is an anode because the anode is the source of the excess electrons that are attracted to a cathode (+). The emf of the half-cell with the sRA and anode has a lower cell with the sOA and the cathode (+). Lower value than the emf of the half- = half reaction and half cell with SRA and anode (─). The emf (or voltage) of a cell that creates a current must be positive, and the emf of a cell is the difference between the emf of its two half-reactions. For a difference to result in a positive number, an equation must have higher minus a lower number. The equation that will give a positive emf is cell = = higher ─ lower half with cathode (+) and SOA ─ half with anode (─) and SRA = + voltage o In the above rules, the values may be either reduction values in an SRP table for reactions at standard conditions or reduction values at non-standard conditions adjusted with the Nernst equation. The terms anode and cathode come from the fact that historically, in the two-metal cells that have long been used as batteries, cations are attracted to the cathode, where they are reduced. However, the “cations are attracted to the cathode” rule does not always predict the labels for the electrodes because, though all cells must contain cathodes, not all cells contain cations that react in the redox reaction. Rules that work in all cases are: • the cathode is the electrode in the half-cell that the electrons the wire flow toward and where reduction occurs. © 2010 ChemReview.Net v1w Page 1177 Module 38 — Electrochemical Cells • The anode is the electrode in the half-cell that the electrons the wire flow from and where oxidation occurs. Memorize one rule of the two and the other can be applied by logic when needed. Platinum Electrodes Not all half-reactions include metals that can serve as electrodes. If a half-reaction does not include a metal or other solid conductor, a non-reactive solid conductor must added to the half-cell solution to serve as an electrode to which a wire can be attached. The metal platinum (Pt) is a very weak reducing agent that will not redox react with most substances. Though platinum is expensive, it is often the electrode of choice in a half-cell that does not include a metal. The transfer of electrons that occurs in the half-reaction takes place at the surface of the platinum. A conductive non-reactant such as Pt can be an anode (─) or a cathode (+). Some solids that are not metals can conduct electricity, but most do not. In problems, unless it is otherwise noted, it should be assumed that if none of the half-cell components are metals, a conductive electrode such as Pt(s) must be added. Labeling the Electrodes In a redox reaction, the sRA gives up its electrons and is oxidized. When a redox reaction occurs in a cell, the sRA gives up its electrons at the anode (─). A cell can be represented either by a redox reaction or by a drawing of the two half-cells. The drawing shows the platinum electrodes that the reaction equation does not. An example of a cell drawing is e─ ↑ anode cathode ↓ 2+ ⎩ Pb and Pb ⎭ and ⎩ Ag sRA wOA and wRA Ag+⎭ sOA In labeling the cell components, key rules are: • Each half-cell must include the particles in one half-reaction: two particles in which the same central atom has different oxidation numbers. • The anode (─) is the electrode in the half-cell that the electrons flow from. • In a cell, the electrons flow from the sRA. In drawing a cell, the method we will use is to • identify the sRA. • Label as the anode the electrode in the half-cell with the sRA. • Label the other electrode as the cathode. • Show the flow of electrons from the anode through the wire toward the cathode. Let’s try an example. © 2010 ChemReview.Net v1w Page 1178 Module 38 — Electrochemical Cells Q1. For the reaction Cu + H+ + MnO4─ state conditions, a. draw the two half-cells. b. Label the anode and cathode. c. Label the direction of electron flow. Apply the following steps. Cu2+ + Mn2+ carried out at standardo Standard Reduction Potentials MnO4─ + 8 H+ + 5e─ Cu2+ + 2e─ Zn2+ + 2e─ Mn2+ + 4 H2O in V 1.51 0.34 Cu ─ 0.76 Zn Steps in Labeling the Electrodes 1. If the redox reaction is given, under the reaction, skip a line, then draw the two halfcells: ⎭ and ⎩ ⎩ ⎭. In each half-cell, with the two cells in any order. • Write the particles in one half reaction. • Include two particles that have the same central atom. • Leave out H2O and e─. • If a half-cell does not include a solid metal, add Pt(s) . 2. Under the particle formulas in the reaction (if supplied) and the half-cells, label the sRA, sOA, wOA, and wRA. Assign the labels using any one of the following methods. a. In an SRP table, the sOA is higher on the left, and the sRA is lower on the right. o b. If is positive or the K is > 1, the wRA and wOA are on the right in the reaction; c. The favored side of the reaction contains the wRA and wOA. 3. Find the metal in the half-cell with the sRA. Above the metal, draw a ↑ to show electrons leaving that half-cell. Try those steps, then check below. ***** Your paper should look like this: ↑ 2+ 2+ and MnO ─ and Pt + ⎩ Cu and Cu ⎭ and ⎩ H and Mn 4 ⎭ sRA wOA wRA sOA (labels based on table) (The half-cells can be drawn in the reverse order.) 4. Draw a ↓ above the metal in the half-cell with the sOA. 5. Between the arrows, draw a ← or → arrow showing the direction of electron flow between the half-cells. © 2010 ChemReview.Net v1w Page 1179 Module 38 — Electrochemical Cells 6. Next to ↑ write anode. Next to ↓ write cathode. Finish those steps, then check below. ***** e─ anode ↑ ↓ cathode 2+ 2+ and MnO ─ and Pt + 4 ⎭ ⎩ Cu and Cu ⎭ and ⎩ H and Mn sRA wOA wRA sOA Done. If the half-cells can be drawn in the reverse order, each particle must have the same label, and electrons must flow toward the cathode. The anode is the electrode in the half-cell that contains the sRA. In this problem, the sRA is Cu, and Cu is the metal in the half-cell with the sRA, so Cu is the anode (─). In the other half-cell must be the cathode (+). The electrode normally must be a solid metal, and the only metal in that half-cell is Pt. Apply the six steps above to one more example. Q2. Using the SRP table above, identify the anode and cathode, and show the direction of electron flow, in a cell containing these half-cells at standard-state conditions. 2+ ⎩ Cu and Cu 2+ ⎭ and ⎩ Zn and Zn ⎭ ***** e─ cathode ↓ 2+ ⎩ Cu and Cu ⎭ and wRA sOA ↑ anode ⎩ Zn and sRA Zn2+⎭ wOA At standard conditions, the stronger reducing agent (sRA) is the particle in the right SRP column closer to the bottom: Zn. Since Zn must be in the half-cell with the anode (─), and Zn is the metal in that half-cell, it is the anode. In the other half-cell must be a cathode (+). The cathode must be a metal, so it must be Cu. Note that in Q1, Cu is the anode (─), and in Q2, Cu is the cathode (+). Any metal can be an anode or cathode, depending on the direction of the flow of current in the cell. Summary Steps: Labeling the Electrodes 1. Draw the two half-cells. Each half-cell has the particles in half-reaction, leaving out H2O and e─. Add Pt to any half-cell without a solid metal. 2. Below the symbols, label the sRA, sOA, wRA, and wOA. 3. Draw a ↑ above the metal in the half-cell with the sRA. 4. Draw a ↓ above the one metal in the half-cell with the sOA. © 2010 ChemReview.Net v1w Page 1180 Module 38 — Electrochemical Cells 5. Between the arrows, draw a ← or → arrow showing the direction of electron flow. 6. Next to ↑ write anode. Next to ↓ write cathode (+). Practice A: Do not use an SRP table for these problems. 1. Will this half-cell require an added non-reactive electrode? 2+ + ⎩ H and Mn and MnO2(s) ⎭ o cell = ─ 0.90 V Pb + Fe3+ 2. Based on this cell reaction: Pb2+ + Fe2+ label the anode, cathode, and electron flow in a cell connecting these half-cells under standard conditions. 2+ ⎩ Pb and Pb ⎭ and ⎩ Pt and Fe 2+ and Fe3+ ⎭ 3. Assuming standard conditions, draw the half-cells showing the anode, cathode (+), and electron flow in a cell based on this reaction. o Cu + Ni2+ Cu2+ + Ni cell = + 0.59 V 4. In a cell based on this reaction under standard conditions, draw the half-cells showing the anode, cathode (+), and electron flow. Ag+ + Fe2+ Ag + Fe3+ that favors the right. 5. In a cell based on the redox equilibrium: Ni2+ + Sn Ni + Sn2+ if Ni is the anode under the reaction conditions, which side is favored at equilibrium? Cell Diagrams A cell diagram is another method that is frequently used to represent an electrochemical cell. An example of a cell diagram is ⏐ Zn(s) ⏐ Zn2+ (1 M) ⏐ Pb2+ (1 M) ⏐ Pb(s) In a cell diagram, 1. A vertical double line ⏐ represents a salt bridge or porous disk between two half-cells. ⏐ 2. The components of the half-cell that includes the sRA are written on the left side of the ⏐ . The metal anode is written first, at the far left. ⏐ 3. On each side, a single vertical line ⏐ is written between the symbols of particles reacting in the half-cell that are in different phases (solid, liquid, gas, or aqueous). Values that can vary, such as concentration in mol/L for ions and pressure in atm for gases, are written in parentheses after each symbol. 4. If there is no metal in the left half-cell, Pt(s)⏐ representing a platinum anode (─) is written at the far left. If there is no conductor in the half-cell with the sOA, ⏐Pt(s) is written at the far right. © 2010 ChemReview.Net v1w Page 1181 Module 38 — Electrochemical Cells 5. Except for the anode (─) at the far left and cathode (+) at the far right, the phases on each side of the ⏐ may be listed in any order. ⏐ The general order of the components in a cell diagram is always direction of electron flow in wire anode(s) ⏐ phase ⏐ phase ⏐ phase ⏐ phase ⏐ cathode(s) ⏐ ⏐ ⏐ sRA and wOA sOA and wRA Let’s apply those rules to an example. Q. For a cell based on this reaction: 2 H+ + Zn Zn2+ + H2(g) o cell = + 0.76 V At standard-state conditions, represent the cell with a cell diagram. * * ** * 1. First identify the anode and write its symbol on the far left. * * ** * The anode (─) is in the half-reaction that includes the sRA. The Zn on the left is acting as an RA. Since the cell potential is positive, the Zn on the left side must be the sRA. Since Zn is the solid metal in the half-reaction with the sRA, Zn is the anode. Write Zn(s) ⏐ first. 2. Next write the formulas for remaining particles in the half-reaction that includes the sRA. Separate particles in different phases by a⏐ line. The phases may be listed in any order. After ions, write (concentrations in M) and after gases write (partial pressures in atm). At standard-state conditions, [ions] are (1 M) and gas pressures are (1 atm). 3. Write the double line representing the salt bridge or porous cell separating the half-cells and half-reactions. * * ** * So far: Zn(s) ⏐ Zn2+(1 M) ⏐ ⏐ 4. Write the components in the other half-reaction and half-cell the same way. List the cathode after a single vertical line at the far right. If no component in this half-cell and half-reaction is a metal, write ⏐ Pt(s) to the far right as the cathode. * * ** * In this example: Zn(s) ⏐ Zn2+ (1 M) ⏐ H+ (1 M) ⏐ H2(g) ⏐ Pt(s) ⏐ The H+ and H2(g) may be written in reverse order. © 2010 ChemReview.Net v1w Page 1182 Module 38 — Electrochemical Cells Practice B: Do not use an SRP table for these problems. 1. Write the cell diagram for cells based on these reactions under standard conditions. a. Pb2+ + Zn Pb + Zn2+ b. Pb + 2 H+ Pb2+ + H2(g) o cell = + 0.63 V o cell = + 0.13 V 2. For this cell at standard conditions, ⏐ Ni(s) ⏐ Ni2+ (1 M) ⏐ H2(g) (1 atm) ⏐ H+ (1 M) ⏐ Pt(s) a. Write the formula for the cathode (+). b. What is the sign of the cell potential for the cell as written? c. Which particle is the sOA? d. Which particle is the wOA? o o 2+ + 2e─ e. If cell = + 0.25 V, what is reduction for Ni Ni 3. Write the cell diagram for a cell based on this reaction at standard conditions. Br2(l) + 2 I─ I2(s) + 2 Br─ o cell = + 0.55 V ANSWERS Practice A 1. Though the half-cell includes a solid, it is not a metal. Metals are electrically conductive, but most other solids are not. Assume this half-cell needs a non-reactive electrode. o is negative, the left side has the wRA and wOA. 2. Since o cell = ─ 0.90 V Pb2+ + Fe2+ Pb + Fe3+ wOA wRA sRA sOA e─ Based on those labels: anode ↑ ↓ cathode Pb and Pb2+ ⎭ and ⎩ Fe2+ and Fe3+ and Pt ⎭ ⎩ sRA wOA wRA sOA The half-cells may be reversed, so long as the component labels remain the same and electrons leave the anode. In the half-cell that includes Fe2+ and Fe3+, ions dissolved in water cannot be connected to a wire to serve as an electrode. Pt is therefore added. Since it is on the side the electrons flow toward, Pt is the cathode. Once one electrode is labeled, the metal in the other half-cell must be the opposite electrode. © 2010 ChemReview.Net v1w Page 1183 Module 38 — Electrochemical Cells 3. Cu2+ + Ni sOA sRA o cell = + 0.59 V Cu + Ni2+ wRA (positive V = right side favored) wOA e─ Based on those labels: anode ↑ ↓ cathode 2+ 2+ ⎭ ⎩ Ni and Ni ⎭ and ⎩ Cu and Cu sRA wOA wRA sOA The half-cells may be reversed as long as the component labels remain the same and e─ leave the anode. Since the voltage is positive, the RA on the left side of the reaction (Ni) must be the sRA. The half-cell with the sRA contains the anode, and the metal is Ni. The metal in the other half-cell must be the cathode (+). Electrons flow from the anode to the cathode. 4. Ag+ + Fe2+ sOA sRA Ag + Fe3+ wRA wOA that favors the right. e─ cathode ↓ ↑ anode 2+ and Fe3+ and Pt + ⎩ Ag and Ag ⎭ and ⎩ Fe ⎭ wRA sOA sRA wOA The half-cells may be reversed as long as the component labels remain the same and e─ leave the anode. Based on those labels: 5. Ni is the anode. The anode is the metal in the same half-cell as the sRA. Since Ni must be the sRA, and the sRA and sOA are on the same side of the reaction, and the wOA and wRA are on the favored side of the reaction, the left side of this reaction is favored at equilibrium. Practice B 1. a. The anode must be the Zn metal on the left side of the reaction. If the cell voltage is positive, the RA on the left side of the reaction must be the sRA. Once the anode is identified, the position of the other particles can be determined by inspection. ⏐ Zn(s) ⏐ Zn2+ (1 M) ⏐ Pb2+ (1 M) ⏐ Pb(s) b. The anode must be the Pb metal on the left side of the reaction, since it is the sRA. Once the anode is identified, the position of the other particles is set. Since the other half-reaction does not include a metal, a platinum cathode is added. The order of the H and H+ in the cell diagram below can be switched, but Pt must be at the right. 2 At standard conditions, all [ions] are labeled (1 M) and all gas pressures are labeled (1 atm). ⏐ Pb(s) ⏐ Pb2+ (1 M) ⏐ H2(g) (1 atm) ⏐ H+ (1 M) ⏐ Pt(s) 2. a. Pt(s) . The cathode (+) is written at the right. b. A cell diagram represents a working cell, and cells run in a direction that produces a positive voltage. c. The sOA? Ni is the anode and RA in its half-cell. It must be the sRA because it is in the half-cell with the anode, and the other particle in its half-cell must be the wOA. The sOA must be in the other half-cell, where H+ is the OA. H+ is the sOA. © 2010 ChemReview.Net v1w Page 1184 Module 38 — Electrochemical Cells d. The wOA? The wOA is the particle formed when the sRA (Ni) gives up its electrons in the same half-cell as the sRA, which is Ni2+. o e. The cell reaction is: Ni + 2 H+ Ni2+ + H2(g) cell = + 0.25 V sRA sOA wOA wRA Once the sRA is identified, the labels are automatic. When the cell potential is positive, the sRA must be on the left side of the redox equation. One half-reaction is the hydrogen electrode reaction under standard conditions o (Write the OA on the left) 2 H+ + 2e─ H2(g) red. = + 0.0 V o . Use a process that adds reactions and half-reactions to find the unknown ***** o WANTED: Ni2+ + 2e─ Ni cell = ? __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ o = ─ 0.25 V Ni2+ + H2(g) Ni + 2 H+ o 2 H+ + 2e─ H2(g) = + 0.0 V __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ o Ni2+ + 2e─ Ni cell = ─ 0.25 V 3. First label the reaction components. o Br2(l) + I ─ I2(s) + Br─ cell = + 0.55 V sOA sRA wOA (write backwards, reversing (add reactions and o o ) values) wRA Once one component is labeled, the other labels are automatic. The anode must be in the same half-cell and half-reaction as the sRA. That half-cell must include I ─ and I2(s) I2(s) is a solid, but it is not in the metal section of the periodic table, so assume it will not be a good conductor and a non-reactive electrode must be added. The half-cell with the sRA is always on the left. So far, this gives us Pt(s) ⏐ I2 (s) ⏐ I ─ (1 M) ⏐ ⏐ For the same reasons, the other half-cell will need a non-reactive metal electrode. The cell diagram is ⏐ Pt(s) ⏐ I2 (s) ⏐ I ─ (1 M) ⏐ Br2 (s) ⏐ Br─ (1 M) ⏐ Pt(s) The order of the solid and aqueous components may be reversed in either or both half-cells * * * * *. © 2010 ChemReview.Net v1w Page 1185 Module 38 — Electrochemical Cells Lesson 38C: Depleted Batteries and Concentration Cells Dead Batteries and K Once the half-cells are connected in an electrochemical cell, the redox reaction will proceed spontaneously and the cell will transfer electrons in the direction that produces a positive voltage. However, as the cell runs, the amount of reactants decrease and products increase, the value for Q in the Nernst equation for the reaction increases. This makes the negative term in the Nernst equation more negative, cell = o cell ─ RT ln( [increasing products]/[decreasing reactants] ) nF and the emf of the cell gradually decreases. The redox reaction continues with declining voltage until the cell emf drops to zero. When the emf of the cell is zero, since ΔG = ─ nF , ΔG for the redox reaction is zero. Recall that by the laws of thermodynamics, when ΔG is zero, the reaction is at equilibrium, and the value of Q = K at the reaction temperature. When cell = 0 , ΔG = 0, the cell redox reaction is at equilibrium, and Q = K . At equilibrium, when the cell voltage is zero, the Nernst equation becomes cell = 0 = o o cell ─ RT ln( K ) nF cell = RT ln( K ) nF At 25ºC: which simplifies to o cell = 0.0257 V ln(K) = 0.0591 V log(K) n n These equations mean that for the redox reaction in the cell, if we know either one of the values for the standard cell potential ( ocell) or the equilibrium constant ( K ), we can solve for the other. A key step in the calculation is writing the volts of the standard cell potential as joules per coulomb. Use the rules above on this problem. o cell = + 0.93 volts Q. For the reaction in this cell: 2 Ag+ + Pb 2 Ag + Pb2+ under standard conditions, find the equilibrium constant for the reaction. ***** Answer o cell = RT ln( K ) When cell = 0 , Q = K and the Nernst equation becomes: nF Solve that equation in symbols for ln(K). ***** © 2010 ChemReview.Net v1w Page 1186 Module 38 — Electrochemical Cells ln( K ) = ( o cell )(nF) Solve for K. RT ***** o DATA: cell = the standard cell potential = + 0.93 volts or V/mol or J/C·mol R = 8.31 J/mol·K n = 2 mol e─ T = std. 25ºC + 273 = 298 K F = 96,500 C/mol e─ in the half-reactions that previously balanced the equation. ***** ln( K ) = ( o cell )(nF) • 1 • 1 = + 0.93 J • 2 mol e─ • 96,500 C • mol·K • 1 = R T C·mol mol e─ 8.31 J 298 K = 72.48 R units must include J (not kJ) if volts (=J/C) is in the data. Often in calculations, volts must be written as volts/mole or J/C·mol for units to cancel properly. Find K. ***** K = e(ln K) = e(72.48) = 3.0 x 1031 o The positive cell value means that the reaction as written favors the right-side products. The large K also indicates a reaction that strongly favors the products. Practice A: First learn the rules above, then do the problems. 1. For the reaction Cl2(g) + I─ using the SRP table, assuming standard-state conditions, a. Write the products and balance the equation by adding half-reactions. b. Calculate the cell potential. c. Calculate K for the reaction. 2. For the reaction in this cell: 2 Ag+ + Cu 2 Ag + Cu2+ a. Find the cell potential under standard conditions. b. If the value of K is found to be 5.5 x 1013 when the solution is at a different temperature, what is that temperature in ºC? Concentration Cells A concentration cell is a special type of cell composed of two half-cells that contain the same components, but different concentrations for the ions in the solutions in the two halfcells. In a concentration cell, current flows between the half-cells until the Q value in both halfcells is the same. For two half-cells that contain only one ion formula that redox reacts, the current flows © 2010 ChemReview.Net v1w Page 1187 Module 38 — Electrochemical Cells • in a direction that leads toward the ion concentrations being equal, and • until the concentration of the ion is equal in both cells. As the concentration cell transfers electrons, the ion concentrations gradually equalize, and the voltage gradually drops, until the concentrations are equal and the voltage is zero. wire Cu Cu Salt Bridge ~~ Consider the two half-cells shown at the right. ~ ~ 0.10 M Cu2+ ~ ~ ~~ 1M Cu2+ Q1. Is this a concentration cell? ***** The left cell has a lower ion concentration. Since that is the only difference between the cells, this is a concentration cell. Q2. In this cell, in which direction will the electrons flow? ***** The electrons will flow in the direction that leads to equalized ion concentrations. If electrons leave the left electrode and flow to the right, Cu metal atoms at the electrode surface are converted to Cu2+ ions, increasing the [Cu2+] and becoming closer to the higher ion concentration in the right cell. Q3. As electrons flow to the right, what changes will take place in the right side halfcell? ***** As electrons flow into the right side electrode, they react with the Cu2+ ion to form Cu metal at the surface of the electrode. This reaction lowers the [Cu2+] in the rightside half-cell, taking it closer to the lower ion concentration in the left half-cell. The reactions in both half-cells tend to equalize the ion concentrations in the two solutions. As the electrons flow, the ion concentrations move toward being equal, and the voltage drops, until the concentrations are equal and the voltage is zero. Labeling Concentration-Cell Electrodes An anode is defined as the electrode that electrons are leaving to enter the wire between the electrodes. To label the electrodes in a concentration cell, find the direction in which the electrons travel in the wire. That will be the direction of electron flow that will equalize the [ions] in the two solutions. • In a concentration cell composed of a metal and its metal ion, the electrons leave the cell that has the lower [reacting metal ion]. The electrode that electrons are leaving is the anode (─). © 2010 ChemReview.Net v1w Page 1188 Module 38 — Electrochemical Cells Concentration-Cell Voltage The emf of a concentration cell at any point in its reaction can be calculated using the Nernst equation. For a concentration cell, o • concentration cell = o cathode ─ o anode ≡ 0 Since both half-cells have the same half-reaction particles, they have the same standard reduction potential. • The Nernst equation must result in a positive voltage, since reactions are spontaneous in the direction that results in a positive voltage. • In the Nernst equation, negative. concentration cell = positive when the value of ln(Q) is concentration cell = ─ RT ln( Q ) nF Ln(Q) is negative when the Q ratio has a value less than one. • • 0 In a cell that involves just one kind of redox-reactive ion, Q is < 1 when the two ion concentrations are written as Q = [lower]/[higher] , which is the same as • Q = [diluted]/[concentrated] n = the moles of electrons that balance the half-reaction. Using those rules, try this problem. Q. In the Cu/ Cu2+ concentration cell above, [Cu2+] = 1.0 M and 0.10 M . a. Which half-cell contains the cathode (+)? b. Solve for the initial cell voltage at 20.ºC. ***** Part a. The electrons leave the half-cell with the lower [reacting metal ion], so that half-cell contains the anode (─). The cathode (+) must be in the half-cell with the higher ion concentration. In the diagram above, that is the half-cell on the right. Part b. To find concentration-cell voltage, use the Nernst equation. cell = with o o cell ─ RT ln( Q ) = ? nF cell = 0 and the Q ratio arranged to have a value <1. DATA: o cell = 0 (exact) F = 96,500 C/mol e─ R = 8.31 J/mol·K T = 20.ºC + 273 = 293 K n = 2 mol e─ in the Cu/ Cu2+ half-reaction Q = [lower]/[higher] = 0.10/1.0 = 0.10 ln(Q) = ln(0.10) = ─ 2.303 SOLVE: © 2010 ChemReview.Net v1w Page 1189 Module 38 — Electrochemical Cells cell = + 0.0000 V ─ [ 8.31 J • (293 K) • mol·K = + 0.0000 V ─ 1 2 mol e─ mol e─ • • (─ 2.303 ) ] = 96,500 C ─ 0.0291 J = + 0.0291 V = + 0.0291 V C·mol mol All concentration cells have relatively low voltage. Summary for Concentration Cells 1. Concentration cells have the same half-cell components but differing [ions]. 2. In a concentration cell, current flows, with steadily dropping voltage, until Q = 1. 3. Electrons leave the half-cell with the lower [redoxing ions] (or lower product of [redoxing ions]. 4. The half-cell which electrons are leaving contains the anode (─). 5. To solve for cell at any point, use the Nernst equation, with and the Q ratio arranged to have a value < 1 : o cell = 0 (exact) Q = [lower]/[higher] . Fundamentals for Electrochemical Calculations 1. The fundamental relationship between work, free energy, and emf is wmaximum possible = ΔG = ΔGº + RT ln(Q) = ─ nF 2. Under standard conditions, Q = 1 , and ΔG = ΔGo = ─ nF o 3. Combining equations 1 and 2 leads to the Nernst equation: cell = o cell ─ RT ln( Q ) = ? nF 3. In depleted cells (dead batteries) , cell = 0 (and ΔG = 0 ) and Q = K. o 4. In concentration cells, cell = 0 and Q = [lower]/[higher] , © 2010 ChemReview.Net v1w Page 1190 Module 38 — Electrochemical Cells Practice B: Learn the two summaries above, then do the problems by writing the needed relationships from memory. 1. Two half-cells contain silver electrodes. In the two solutions, [Ag+] = 1.0 M in one and 0.050 M in the other. a. What will be the initial cell potential at 15ºC? b. Which half-cell contains the anode (─)? 2. Two half-cells contain lead electrodes. In one, [Pb2+] = 1.0 M. If the cell voltage is 0.040 V at 25ºC, a. find [Pb2+] in the second half-cell. b. Which half-cell contains the cathode (+)? 3. A cell is composed of two half cells: one composed of Ag and Ag+ and the other an unknown metal M and its ions M2+. For the reaction that occurs in the cell, ΔGº = ─ 301 kJ. o a. Find K at 25ºC. b. Find cell o values in the tables in Modules 37 and c. If Ag is the cathode in the cell, using the 38, identify the unknown metal. ANSWERS Practice A 1a,b. The reactants are diagonal \, so the reaction will go. The products are the particles in the SRP table that are opposite each reactant. (The \ react, the / are the products, the / don’t react.) Cl2(g) + I─ I2(s) + Cl─ To balance: o Cl2(g) + I─ I2(s) + Cl─ cell = ? __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ o 2 Cl─ Cl2(g) + 2e─ red. = + 1.36 V o 2 I─ I (s) + 2e─ red. = ─ 0.53 V __ __ __ __ __ __2__ __ __ __ __ __ __ __ __ __ __ __ o o Cl2(l) + I─ values) I2(s) + Cl─ (add reactions and add cell = + 0.83 V o 1c. The equation that relates and K is o cell = RT ln( K ) Solve this equation in symbols for ln(K), then find K. nF ***** o ln( K ) = ( cell )(nF) RT © 2010 ChemReview.Net v1w Page 1191 Module 38 — Electrochemical Cells o DATA: cell = the standard cell potential = + 0.83 V or V/mol or J/C·mol R = 8.31 J/mol·K T = std. 25ºC + 273 = 298 K F = 96,500 C/mol e─ n = 2 mol e─ in the two half-reactions that balanced the equation. ***** o cell )(nF) • 1 • 1 = + 0.83 J • 2 mol e─ • 96,500 C • mol·K • 1 = 64.69 ─ 8.31 J 298 K RT C·mol mol e Often, volts must be written as volts/mole or J/C·mol for units to cancel properly. Solve for K. ln( K ) = ( ***** ln( K ) = 64.69 ; K = e(ln K) = e(64.69) = 1.2 x 1028 = K o values attached. 2a. To find the standard cell potential, balance then add the half-reactions with their o 2 Ag+ + 2 e─ 2 Ag = 0.80 V (write higher first, as in table) o o 2+ ─ ─ 0 __ __ (write lower backwards, reversing ) __ __ __Cu __ __ Cu __ __ 2e __ __ __ __=__ __.34 V __ __ + __ o 2 Ag+ + Cu 2 Ag + Cu2+ cell = + 0.46 V o and K is 2b. K is in the Nernst equation only when = 0 and K = Q. The equation that relates o cell = RT ln( K ) Solve this equation in symbols for T nF ***** o ( cell )(nF) = T R ln( K ) DATA: o R = 8.31 J/mol·K T =? F = 96,500 C/mol cell = + 0.46 V n = 2 mol e─ , in the two LCD half-reactions above that balanced the equation. ln( K ) = ln( 5.5 x 1013 ) = 31.64 ***** T =( o cell )(nF) • 1 • 1 = + 0.46 J • 2 mol e─ • 96,500 C • mol· K • 1 R ln( K ) C·mol 8.31 J 31.64 mol e─ = = 338 K ─ 273 = 65 ºC Practice B o 1a. To find concentration-cell voltage, use the Nernst equation with cell = 0 and the Q ratio arranged to have a value < 1. o cell ─ RT ln( Q ) = ? cell = nF o DATA: R = 8.31 J/mol·K T = 15ºC + 273 = 288 K cell = 0 V (exact) F = 96,500 C/mol e─ n = 1 mol e─ in the Ag/Ag+ half-reaction Q = [lower]/[higher] = 0.050/1.0 = 0.050 © 2010 ChemReview.Net v1w ln(Q) = ln(0.050) = ─ 3.00 Page 1192 Module 38 — Electrochemical Cells SOLVE: o cell ─ RT ln( Q ) nF cell = + 0.0000 V ─ [ 8.31 J • (288 K) • 1 • mol e─ • (─ 3.00 ) ] = mol·K 1 mol e─ 96,500 C cell = = + 0.0 V ─ ─ 0.0744 J C·• mol = + 0.074 V = + 0.074 V mol 1b. The electrons leave the half-cell with the lower ion concentrations, and that half-cell contains the anode. o 2a. For concentration-cell voltage, use the Nernst equation with cell = 0 and Q arranged [lower]/[higher] so that Q <1. o cell ─ RT ln( Q ) = ? cell = nF DATA: = 0.040 V cell o cell = 0 (exact) F = 96,500 C/mol e─ R = 8.31 J/mol·K T = 15ºC + 273 = 288 K n = 1 mol e─ in the Ag/Ag+ half-reaction Q = [lower]/[higher] = ?/1.0 = ? SOLVE for the wanted variable in symbols first. First find ln(Q), then Q. cell )(nF) = ln( Q ) RT ln( Q ) = ─ ( cell )(nF) • 1 • 1 = ─ 0.040 J • 1 mol e─ • 96,500 C • mol· K • 1 = ─ 1.61 RT C·mol ─ 8.31 J 288 K mol e Q = eln(Q) = e(─ 1.61) = 0.200 = [lower]/[higher] = 0.200 / 1.0 ; 0.20 M = lower [Ag+] ─( 2b. The electrons leave the half-cell with the lower ion concentrations, and that half-cell contains the anode. The cathode (+) is the electrode in the other half-cell, where [Ag+] = 1.0 M . 3a. There are several ways to solve. You may use any that result in the correct answer. One is: ΔG = ΔGº + RT ln(Q) = ─ nF When = 0 , ΔG must = 0 and and K = Q when =0. ΔGo = ─ RT(ln K) ΔGº = ─ 301,000 J ; convert to J if you use R = 8.31 J/mol·K ln(K) = ─ΔGº = RT T = 273 +25 = 298 K, ─ΔGº ·· 1 = + 301,000 J · mol·K · 1 = 121.5 RT 8.31 J 298 K K = e(ln K) = e(121.5) = 5.8 x 1052 = ? x 1052 At these high powers of e , a small change in ln(K) will make a large change in K. For that reason, any ? x 1052 would be within the range of experimental uncertainty. o 3b. The equation that relates ΔGº and cell is ΔGo = ─ nF o . The reaction must be © 2010 ChemReview.Net v1w Page 1193 Module 38 — Electrochemical Cells 2 Ag+ + 2 e─ 2 Ag o = 0.80 V o 2+ ─ =? 2 __ __ __ __ __ M __ __ M __+__ e __ __ __ __ __ __ __ __ __ o 2 Ag+ + M 2 Ag + M2+ cell = ? o = ─ ΔGº = ─ (─ 301 kJ) nF (2 mol e─)(96,500 C/mol) o ) o (write lower backwards, reversing ) (half cell with cathode has higher = 301,000 J = + 1.56 Volts 193,000 C For this simple equation, we omitted the data table. But note that to find the unit volts, kJ must be converted to joules, since V = J/C . If you do not include the units, you may not get the right answer. When in doubt, do a DATA table for each equation, and make the units in the table consistent. 3c. We know 2 Ag+ + 2 e─ 2 Ag o = 0.80 V o ) o (write lower backwards, reversing ) (half cell with cathode has higher o 2+ ─ =? 2 __ __ __ __ __ M __ __ M __+__ e __ __ __ __ __ __ __ __ __ o = + 1.56 V from part b 2 Ag+ + M 2 Ag + M2+ cell The second reaction must therefore have a table o of ─ 0.76 V. That half reaction is Zn2+ + 2e─ Zn ***** Lesson 38D: Electrolysis Reversing the Flow of Electrons In a cell, electrons travel from the sRA through the wire toward the sOA. However, by inserting a battery or other device that produces a voltage higher than the cell voltage into the circuit between the electrodes, the flow of electrons in the cell can be reversed, and the redox reaction in the cell can be reversed. This process is called electrolysis. Electrolysis can either add electrons to or remove electrons from a half-cell. • In a standard electrochemical cell, electrons leave the sRA. In electrolysis, electrons leave the wRA. • In a standard cell, the sRA and sOA used up and the wRA and wOA formed. In electrolysis, the wRA and wOA are used up and the sRA and sOA are formed. Some applications of electrolysis are: • In a rechargeable battery, a voltage supplied from outside the cell reverses the redox reaction in the battery, regenerating the reactants. This allows a depleted battery to be restored to its original concentrations and voltage. In theory, any battery can be recharged. In practice, the physical changes that take place during redox reactions can make recharging, and especially repeated recharging, difficult. Battery designs that can store and release energy repeatedly are valuable to society and are a major topic of current scientific research. © 2010 ChemReview.Net v1w Page 1194 Module 38 — Electrochemical Cells • Strong oxidizing and reducing agents can be produced by electrolysis. Examples Fluorine gas (F2) is the strongest chemical oxidizing agent. In the SRP table, no table reaction is above o F2 + 2 e ─ 2 F ─ reduction = + 2.87 V Because of the high reduction potential of F2, there is no redox reaction or cell in which substantial amounts of F─ react and produce F2 as its product. In a cell, this half-reaction can only go to the right. Electrolysis, however, can remove electrons from F─ and drive the half-reaction to the left. Alkali metals are strong reducing agents at the bottom right of the SRP table, so they tend to be reactants (used up) in redox reactions. Electrolysis can reverse that tendency, driving half-reactions such as o Na+ + e─ Na reduction = ─ 2.71 V which strongly favors the left, to the right. The metal Na is then a product. Water can react with some strong oxidizing and reducing agents. For this reason, the electrochemical cells that are used to produce SRA and SOA species by electrolysis usually employ melted (molten) ionic compounds, instead of in aqueous solutions, in their half-cells. When an ionic compound is either dissolved or melted, its ions can flow and react in the cell. • In electroplating, driving a cell backward can be used to convert metal ions in a half-cell solution to solid metal on the surface of a metal electrode. In electrolysis, • the spontaneous flow of electrons in a cell is reversed. • The wRA and wOA react, and the sRA and sOA are formed. • The electrons move from the half-cell with the wRA toward the wOA. Labeling Electrodes in Electrolysis In electrolysis, anode remains the term for the electrode where oxidation occurs and electrons flow from. However, in electrolysis, the electron flow is opposite that of a cell with the same components. This reverses the electrode labels when compared to the cell. To name the electrodes in electrolysis, use one of these rules. • As in cells, oxidation occurs at the anode, and electrons flow toward the cathode. • Unlike cells, in electrolysis the anode is the half-cell with the wRA, not the sRA. • In electrolysis the sRA, sOA, wRA, and wOA labels are attached to the same particles, but compared to the cell labels, the labels for the anode, cathode and electron flow direction are reversed. © 2010 ChemReview.Net v1w Page 1195 Module 38 — Electrochemical Cells Try the following problem. Q. For the cell reaction: Zn + Cu2+ o Cu + Zn2+ cell = + 1.10 V the cell drawing is e─ cathode ↓ ↑ anode 2+ ⎩ Cu and Cu ⎭ and wRA Write the reaction, sOA ⎩ Zn Zn2+⎭ and sRA wOA o , and cell drawing for electrolysis using these two cells. ***** Answer: When the reaction is written backwards, the sign on the cell potential is reversed. o In electrolysis: Cu + Zn2+ Zn + Cu2+ cell = ─ 1.10 V In electrolysis, the current flow is reversed. The electrode where oxidation occurs remains termed the anode (─), but the wRA is oxidized. The cell drawing becomes e─ anode ↑ 2+ ⎩ Cu and Cu ⎭ and wRA Practice: sOA ↓ cathode 2+ ⎩ Zn and Zn ⎭ sRA wOA Check answers as you go. 1. In the reaction: 2 Ag+ + Cu 2 Ag + Cu2+ o cell = + 0.46 V a. Which side is favored in the redox reaction? b. Which metal would increase in mass during a redox reaction? c. Which metal would be the anode (─) in a cell based on the above reaction? d. Which metal would increase in mass during electrolysis? e. Which metal would be the anode (─) during electrolysis? 2. A single cell of molten NaCl has two electrodes inserted and a current introduced that causes electrolysis. a. Write the balanced equation for the electrolysis reaction. o for this electrolysis reaction. b. Using an SRP table earlier in this module, calculate c. What is the minimum voltage that will need to be supplied in reverse for electrolysis to take place? d. Which half-reaction will occur at the anode, and what product is produced? e. Which half-reaction will occur at the cathode, and what product is produced? © 2010 ChemReview.Net v1w Page 1196 Module 38 — Electrochemical Cells ANSWERS 1. a. The right side. b. Silver. Ag is a product. c. The anode is in the half-cell with the sRA. The sRA is Cu. d. Cu. In electrolysis, the reaction runs backwards. The reactants of the redox reaction are formed. e. Oxidation occurs at the anode. Going backward, Ag is being oxidized. Ag is the anode. 2. a. Na+ + Cl─ ? Find SRP table half-reactions that include these two particles. ***** a. 2 Na+ + 2 Cl─ 2 Na + Cl2 o WANTED: 2 Na+ + 2 Cl─ 2 Na + Cl2 cell = ? __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ o = ─ 2.71 V 2 Na+ + 2e─ 2 Na o o = ─ 1.36 V 2 Cl─ Cl2 + 2e─ (write backwards, reversing ) __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ o o values) 2 Na+ + 2 Cl─ 2 Na + Cl2 cell = ─ 4.07 V (add reactions and c. + 4.07 V d. Oxidation occurs at the anode. The particle being oxidized is Cl─ . The half-reaction that is oxidiation is Cl─ losing its electrons: 2 Cl─ Cl + 2e─ . The product is Cl . 2 2 e. The other half-reaction must take place at the cathode. That half-reaction is 2 Na+ + 2e─ The particle reduced at the cathode (+) is Na+. The product is Na metal. 2 Na . ***** Lesson 38E: Amperes and Electrochemical Calculations Amperes The ampere (amp) is an SI unit that measures the rate of flow of charge past a point. All rates are quantities per unit of time. An ampere is defined as a flow of one coulomb of charge per second. Ampere = coulomb/second As a unit, ampere is similar to molar: ampere is an abbreviation for a ratio unit. Just as the key to solving concentration problems is to treat molarity as moles/liter, the key to solving problems that include amperes is: In calculations, treat amperes as coulombs/second: a ratio unit. • If amperes is wanted, write WANTED: ? amps = ? C/s • If 5 amperes is data, write DATA: 5 amps © 2010 ChemReview.Net v1w 5C=1s a ratio unit. as ratio data. Page 1197 Module 38 — Electrochemical Cells Using this strategy, calculations involving amperes can be solved by conversion factor methods without having to recall memorized equations. Use the rules above for this problem. Q. If a 5.00 amp current flows past a cross-sectional point in a circuit for 6.50 minutes, a. How much charge flows past the point? b. How many electrons pass the point? ***** a. WANTED: ? coulombs 5.00 amps 5.00 C = 1 s DATA: (write the unit wanted) (in conversions, translate amps to C/s) 6.5 minutes SOLVE: (Want a single unit? Start with a single unit. See Lessons 5B & 5C.) ? C = 6.50 min ● 60 s ● 5.00 C 1 min 1s = 1,950 C ***** WANT: ? e─ DATA: b. see above. 96,500 C = 1 mol e─ SOLVE: (answers from earlier steps may shorten solving later steps.) ? e─ = 1,950 C ● Practice A: (Add to DATA when mixing charge and e─ ) 1 mol e─ ● 6.02 x 1023 e─ = 1.22 x 1022 e─ 96,500 C 1 mol e─ If your conversion recall is rusty, try all three. 1. How many minutes are required for an electrolysis cell to gain 4.50 x 104 coulombs of charge from a 12.0 amp current? 2. If a 15.0 amp current enters an electrolysis cell for 2.00 hours, how many electrons have entered the cell? 3. If 6.02 x 1021 electrons enter a cell in 2.00 minutes, what is the current in amperes? Electrolysis and Stoichiometry In electrolytic processes, we often needed to calculate how much charge is required to produce a given amount of reaction product. A calculation involving chemical reaction and two different particles is stoichiometry. Electrolytic calculations can be solve by combining the ampere conversions above with the Seven steps to solve stoichiometry (from Lesson 12C) • WDBB: Write the Wanted, Data, Balance, Bridge steps, then • Convert units given to moles given to moles WANTED to units WANTED. © 2010 ChemReview.Net v1w Page 1198 Module 38 — Electrochemical Cells The differences between electrolytic stoichiometry and other types are • Conversions must be done between time, amperage, and electrons (as done above). The Balanced equation is a half-reaction (a reaction with an e─ term). • • The Bridge conversion includes moles of electrons in one term and moles of WANTED or given particles in the other. With those rules in mind, try the following problem. Q. To convert Cu2+ ions in a half-cell solution to copper metal on an electrode surface, a 3.50 amp current is supplied for 5.00 minutes. a. How many grams of copper will be deposited on the electrode? b. Is the electrode being plated an anode or a cathode? ***** Part a Answer WANTED: ? g Cu DATA: 3.50 amp 3.50 C = 1 s 5.00 min. (in conversions, treat amps as C/s) (the single-unit given) 63.5 g Cu = 1 mol Cu 96,500 C = 1 mol e─ (grams prompt) (Add to DATA when mixing C and e─) Balance: Cu2+ + 2 e─ Cu (electrolysis balancing uses a half-reaction) Bridge: ***** 1 mol Cu = 2 mol e─ (moles of WANTED or given = moles e─) SOLVE: (In stoichiometry, head for the bridge: convert to the given to the unit in the bridge (mol e─) that is not part of the answer unit. ***** ↓ Goal 1 ↓ Bridge ↓ Final goal ? g Cu = 5.00 min. ● 60 s ● 3.50 C ● 1 mol e─ ● 1 mol Cu ● 63.5 g Cu = 0.345 g Cu 1 min 1s 96,500 C 1 mol Cu 2 mol e─ Part b Answer Electrons are going toward the electrode where they are donated to Cu2+ to form Cu. Where reduction occurs, the electrode is a cathode (+). Practice B: Commit to memory the electrolysis stoichiometry rules above, then try these. 1. How many coulombs would be needed to deposit 32.4 mg Ag at an electrode in an Ag+ ion solution? 2. How many minutes are needed for an electrolysis apparatus with a 500. amp current to convert molten Al3+ ions into 0.540 kg of aluminum metal? 3. In an electrolysis apparatus, to convert molten KF to 4.48 L F2 gas at STP in 4.00 minutes, what amperage is needed? © 2010 ChemReview.Net v1w Page 1199 Module 38 — Electrochemical Cells 4. In 16.0 minutes, a 4.00 amp current forms 4.12 grams of metal M from an M2+ solution. a. What is the molar mass of the metal? b. What is the metal? 5. Is the half-cell producing the stated product at the anode (─) or cathode (+) a. in Problem 1? b. In Problem 2? c. In Problem 3? ANSWERS Practice A 1 WANTED: ? minutes DATA: 4.50 x 104 C (start with a single unit) 12.0 C = 1 s (in conversions, treat amps as C/s) SOLVE: 2. WANTED: DATA: ? minutes = 4.50 x 104 C ● (wanted single unit) 1 s ● 1 min = 62.5 min 12.0 C 60 s ? e─ 15.0 C = 1 s (in conversions, translate amps to C/s) 2.00 hours 96,500 C = 1 mol e─ (Add to DATA when mixing coulombs and electrons) SOLVE: ? e─ = 2.00 hrs ● 60 min ● 60 s ● 15.0 C ● 1 mol e─ ● 6.02 x 1023 e─ = 6.74 x 1023 e─ 1 hr 1 min 1s 96,500 C 1 mol e─ 3. WANT: DATA: ? amperes = ? C s 6.02 x 1021 e─ = 2.00 min 96,500 C = 1 mol e─ SOLVE: (Write wanted ratio units as ratios) (2 measures of same process. When a ratio unit is WANTED, all of the data is in ratios) (Add to DATA when mixing charge and electrons) (Want a ratio? Start with a ratio. Time on the bottom is one option to arrange your given right-side up. See Lesson 11B.) ? C = 6.02 x 1021 e─ ● 1 min ● 1 mol e─ ● 96,500 C s 2.00 min 60 s 6.02 x 1023 e─ 1 mol e─ © 2010 ChemReview.Net v1w = 8.04 C s = 8.04 amp Page 1200 Module 38 — Electrochemical Cells Practice B 1. WANTED: DATA: ? coulombs (C) 32.4 mg Ag (the single unit given) 107.9 g Ag = 1 mol Ag 96,500 C = 1 mol e─ Balance: Ag+ + 1 e─ Ag 1 mol Ag = 1 mol e─ (grams prompt) (Add to DATA when mixing C and e─) (electrolysis balancing includes electrons) (equate moles of WANTED or given with e─ moles) Bridge: ***** SOLVE: (In stoichiometry, head for the bridge: convert to the given unit in the bridge conversion) ↓ Bridge ─3 g ● 1 mol Ag ● 1 mol e─ ● 96,500 C = 29.0 C ? C = 32.4 mg Ag ● 10 1 mg 107.9 g Ag 1 mol Ag 1 mol e─ 2. WANTED: ? min. DATA: 500. amp 500. C = 1 s 0.540 kg Al (the single unit given) 27.0 g Al = I mol Al 96,500 C = 1 mol e─ Balance: (in conversions, translate amps to C/s) Al3+ + 3 e─ Al 1 mol Al = 3 mol e─ (grams prompt) (Add to DATA when mixing C and e─) (electrolysis balancing includes electrons) (equate moles of WANTED and give; with e─ term) Bridge: ***** SOLVE: (In stoichiometry, head for the bridge: convert to the moles of given in the bridge conversion) ? min = 0.540 kg Al ● 103 g ● 1 kg 3. WANT: DATA: ↓ Bridge 1 mol Al ● 3 mol e─ ● 96,500 C ● 1 s ● 1 min 500. C 60 s 27.0 g Al 1 mol Al 1 mol e─ ? amperes = ? C s 4.48 L F2 at STP = 4.00 min 22.4 L any gas at STP = 1 mol F2 Balance: 96,500 C = 1 mol e─ 2 F─ F + 2 e─ 2 = 193 min (Write wanted ratio units as ratios) (2 measures of same process, all data in ratios) (STP prompt, Lesson 17B) (Add to DATA when mixing charge and electrons) (electrolysis balancing includes electrons) (electrolysis can either add or remove half-cell electrons.) (equate moles of WANTED or given with moles of e─ charges ) Bridge: 1 mol F = 2 mol e─ 2 ***** SOLVE: (Want a ratio? Either solve in parts, or start with a ratio. Time on the bottom is one option as a way to arrange your given right-side up. Head for the bridge conversion in the middle.) ***** © 2010 ChemReview.Net v1w Page 1201 Module 38 — Electrochemical Cells ↓ Bridge 1 mol F2 ● 2 mol e─ ● 96,500 C = 161 C = 161 amps ? C = 4.48 L F2 STP ● 1 min ● s 4.00 min 22.4 L F2 STP 1 mol F2 s 60 s 1 mol e─ (write the unit wanted) 4a. WANT: ? g of M mol M DATA: Balance: Bridge: ***** SOLVE: 4.12 g M = 16.00 min 4.00 amp 4.00 C = 1 s 96,500 C = 1 mol e─ M2+ + 2 e─ M 1 mol M = 2 mol e─ (equivalency: 2 measures of same process) (treat amperage as C/s) (Add to DATA when mixing charge and electrons) (electrolysis balancing includes electrons) (an electrolysis bridge conversion will usually have an e─ term.) You can start with a ratio and solve, but since this is stoichiometry in which a ratio unit is wanted, an easier strategy is to solve for the WANTED unit in two parts, then divide. Solve first for the unit that is not moles, using as your single-unit given the side of the equivalency that includes the WANTED unit. Then solve for moles using the other side of the equivalency as a given (see Lesson 12C). ***** WANTED unit not moles = ? g of M = 4.12 g M ↓ Bridge ─ ● 1 mol M = 0.01990 mol M WANTED = ? mol M = 16.0 min. ● 60 s ● 4.00 C ● 1 mol e 1 min 1s 96,500 C 2 mol e─ = 207 g/mol WANT: ? g of M = 4.12 g M mol M 0.01990 mol M 4b. Which metal has a molar mass close to 207 g/mol and an M2+ ion? The most likely metal is lead: Pb. 5. a. At the electrode, electrons are being donated to Ag+ to make Ag. Electrons are flowing toward this electrode. The 1/2 cell with the Ag+ and Ag is therefore a cathode (+). b. By the same logic as part a, the electrode in the half-cell with the Al components is a cathode. c. At this electrode, electrons are being taken away from F─ to form F2. Oxidation is occurring, and electrons are therefore flowing away from this electrode. Where the F─ forms F is therefore an anode. 2 ***** © 2010 ChemReview.Net v1w Page 1202 Module 38 — Electrochemical Cells Summary: Electrochemical Cells 1. Cells and Batteries a. A redox reaction is a combination of two balanced half-reactions. An electrochemical cell is a combination of two half-cells. Battery is a term for a single cell or for multiple cells that are connected to produce higher voltage than a single cell can provide. b. In an electrochemical cell, a redox reaction takes place. The each half-cell, one of the half-reactions takes place. • One half-cell contains the particles in one half-reaction. The other half-cell contains the particles in the other half-reaction. • One half-cell has the sRA and wOA. The other has the sOA and wRA. c. As electrons flows through the wire connecting the cell electrodes, • the two half-reactions proceed that added together produce the net redox reaction. • The reaction mixture shifts toward the side favored at equilibrium. • sRA is oxidized to become wOA, and sOA is reduced to become wRA. • Electrons flow in the wire from the sRA side toward the side with the sOA. • Ions move through the salt bridge to balance charge. d. As the wRA and wOA products build up, the voltage drops gradually. When Q = K for the redox reaction, the voltage is zero. 2. Cell Components a. Each half-cell must have a conductive electrode. b. To label the electrodes, the rules are: • Oxidation occurs at the anode (─). Reduction occurs at the cathode (+). • Electrons move from the anode thru the wire toward the cathode (+). • In a cell, the anode (─) is in the half-cell with the sRA. c. For electrons to flow, the electrodes are connected by a conductor (usually a metal wire) and the half-cell solutions are connected by a salt bridge or porous disk. d. As the stronger reducing agent (sRA) reacts, it loses electrons and is oxidized. e. In a redox reaction equation, the sRA and sOA are on the same side. In a cell, the sRA and wOA are in the same half-cell. 3. Conversions, Equations, Units, and Constants a. The fundamental relationship between work, free energy, and emf is wmaximum possible = ΔG = ΔGº + RT ln(Q) = ─ nF b. Under standard conditions, Q = 1 , and © 2010 ChemReview.Net v1w ΔG = ΔGo = ─ nF o Page 1203 Module 38 — Electrochemical Cells c. Combining equations 1 and 2 leads to the Nernst equation: cell = o cell ─ RT ln( Q ) nF d. To check a standard cell emf, use e. f. o cell = o higher ─ o lower = a positive V half rxn. with anode = a positive V cell = half rxn. with cathode ─ o concentration cell = 0 and Q = [lower]/[higher] g. In dead batteries, cell = 0 and Q = K. 4. Depleted Batteries When a cell is run until it is dead (no more voltage flows), cell = 0 = ΔG , the cell redox reaction is at equilibrium, Q = K , and the Nernst equation simplifies to o cell = RT ln( K ) nF 5. Concentration Cells a. Concentration cells have the same half-cell components but differing [ions]. b. In a concentration cell, current flows, with gradually dropping voltage, until Q = 1. c. Electrons move in the direction that equalizes the [redoxing ions] (or lower product of the redoxing and variable concentrations and gas pressures). d. The half-cell which electrons are leaving contains the anode (─). e. To solve for concentration cell at any point, use the Nernst equation, with o concentration cell = 0 (exact) and the Q ratio arranged to have a value < 1 : Q = [lower]/[higher] 6. Volts and Amps a. Volts ≡ joules/coulomb b. In conversions, treat volts as joules/coulomb: a ratio unit. c. Ampere ≡ coulomb/second d. In conversions, treat amperes as coulombs/second: a ratio unit. 7. Electrolysis a. In electrolysis, voltage supplied in the circuit makes the electrons flow in reverse. The redox reaction goes in the direction that is not spontaneous. b. In electrolysis, the wRA and wOA react, and the sRA and sOA are formed. Electrolysis can be used to form strong oxidizing and reduction agents. c. In a standard electrochemical cell, electrons leave the sRA. In electrolysis, electrons are taken away from the wRA. © 2010 ChemReview.Net v1w Page 1204 Module 38 — Electrochemical Cells d. To label the electrodes in electrolysis, • As in cells, oxidation occurs at the anode; electrons flow toward the cathode. • Unlike cells, in electrolysis electrons leave the cell with the wRA, not the sRA. • Compared to the cell, in electrolysis the sRA, sOA, wRA, and wOA are the same, but the labels for the electrodes and electron flow direction are reversed. 8. Electrochemical Stoichiometry For reaction calculations (stoichiometry) involving electrolysis, use the Seven steps to solve stoichiometry: • WDBB: Wanted, Data, Balance, Bridge, then • Convert units given to moles given to moles WANTED to units WANTED Solving a reaction calculation for a ratio unit, solve for the WANTED unit in two parts, then divide. • Solve for the unit that is not moles first, using as your single-unit given the side of the equivalency that includes the WANTED unit. • Then solve for moles using the other side of the equivalency as a given. The differences between electrolytic stoichiometry and other types are • • • Conversions must be done between time, amperage, and electrons. The reaction is a half-reaction (one with a term for e─). The bridge conversion will include moles of electrons and moles of WANTED or given particles. ##### © 2010 ChemReview.Net v1w Page 1205 ...
View Full Document

This note was uploaded on 11/16/2011 for the course CHM 1025 taught by Professor J during the Summer '09 term at Santa Fe College.

Ask a homework question - tutors are online