Unformatted text preview: Module 37 — Electrochemistry Calculations In Chemistry
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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
NonStandard 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 rechargeability.
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 rerun 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. Halfreactions. A redox reaction can be separated into two halfreactions. 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.
Halfreactions include the symbol e─ showing the number of electrons gained or lost.
Halfreactions must be balanced for atoms and charge. Electrons count when balancing
charge, but they do not affect the balancing of atoms.
5. Balancing Halfreactions. Halfreactions can be balanced by trial and error. If H
and O atoms must be added to balance halfreactions that are carried out in aqueous
solutions, use the CAWHe! 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 halfreactions using OH─ instead H+ ions, first balance by CAWHe, then
neutralize H+ by adding OH─ equally to both sides, then adjust H O on both sides.
2 7 Halfreactions can be used to label the oxidizing and reducing agent in a redox
reaction. If a halfreaction 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 halfreactions, see Lessons 16A and 16B.
1. Balance these halfreactions for atoms and charge.
a. H2 c. Hg22+ b. Hg2+ S2O32─ d. H+ Pb4+ S4O62─
Pb2+ 2. Balance these halfreactions by the CAWHe! 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 halfreactions 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 halfreactions. Halfreaction
addition is the technique used most often in electrochemical calculations.
9. To balance redox reactions by adding halfreactions, the steps are
a. Balance each halfreaction.
b. Multiply each halfreaction by an LCM to get the same number of electrons in both.
The number of electrons lost in one halfreaction must equal the number gained by
the other.
c. Add the two halfreactions. Cancel like terms on both sides. Like numbers of
electrons on each side must cancel.
10. Redox reactions can be divided into halfreactions to assist in balancing.
a. Find the two atoms that change their Ox# in the reaction.
b. Write two separate halfreactions, one for each atom changing Ox#.
c. Balance then add the halfreactions.
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 halfreactions, see Module 16. 1. Balance each halfreaction, 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 halfreactions, 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, halfreactions 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 CAWHe 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 halfreaction, Zn donates electrons and becomes oxidized.
2b. I─ is the left side RA because in its halfreaction, 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 halfreaction, 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 singleunit 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
conversionfactor 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 “rightside 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 halfreaction 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 halfreaction 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 (nonreactive) 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 halfreaction 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 halfreaction strongly tends to go backwards.
Na+ + e─ Na (strongly tends to go in reverse) This logic holds for any reversible reaction or halfreaction:
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 halfreaction is one without a strong tendency to go in one
direction over the other. An example is the hydrogen electrode halfreaction.
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 halfreaction is written with electrons on the left as
2 H+ + 2 e─ H2 (tends to go either way) Every redox halfreaction 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 halfreactions 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 halfreactions 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 halfreactions to anchor our understanding of redox reactions. Standard Reduction Potentials
In a halfreaction, the tendency of an oxidizing agent to attract electrons (and become
reduced) creates an electromotive force (emf, symbol ).
Every halfreaction 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 halfreactions.
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 halfreaction 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 halfreactions have electrons on the
left side of the arrow, reducing the leftside
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: NonStandard 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 halfreactions; 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 halfreactions 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 nonstandard 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 nonstandard 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 halfreaction 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 halfreaction 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 halfreaction
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 halfreaction
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 halfreaction at nonstandard conditions.
Q2. For the halfreaction 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 halfreaction = 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 halfreaction 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 nonstandard
voltages will be close.
But not all Q values are between 0.1 and 10. Let’s try one more halfreaction at nonstandard conditions.
Q3. For the halfreaction
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 halfreaction (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 halfreaction, the change in emf is substantial: from 1.51 V at standard conditions to
0.94 V at these nonstandard conditions.
If we test a variety of halfreactions, our general findings will be the following.
In halfreactions, 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 halfreaction with one ion that has a coefficient of one, if a concentration of the ion
drops from 1 M to 0.10 M, its halfreaction potential will drop by 0.06 volts or less.
If the coefficients of a halfreaction are more than one, and/or the concentrations or
partial pressures have values far from 1, the standard and nonstandard 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 halfreaction
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 halfreaction: 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 nonstandard , 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 halfreaction = 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 nonstandard 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 halfreaction = 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 halfreactions 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 reform 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 standardstate 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 standardstate 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 halfreaction 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 NonStandard 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 halfreaction 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 weaksidefavored rule, and the diagonal \ rule. Both are accurate
at standard conditions, but may not be accurate at nonstandard 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 nonstandard 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 nonstandard 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 nonstandard 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 halfreactions 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 halfreactions that contain the particles in the redox
reaction.
One of the halfreactions 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 halfreactions 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 halfreaction reversed,
3. Using the table, list
o
change the sign of its
value.
4. As with previous balancing using halfreactions, multiply each equation by an LCD so
that the number of electrons gained in one halfreaction 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 halfreactions.
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
halfreactions, but to find nonstandard 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 halfreactions 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 halfreactions, the spectators are omitted, so omit nitrate on the left but
not the right.
Many particles are listed in more than one halfreaction 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 halfreaction, I2(s) + 2e─
2 I─
cell = + 0.54 V
o
for this halfreaction:
Br2(l) + 2e─
2 Br─
find the NonStandard 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 halfreactions and full redox reactions is:
• A halfreaction Nernst equation uses the moles of electrons (n) in the one balanced
halfreaction of interest; • A fullreaction Nernst equation must use for moles of electrons (n) the lowest whole
number coefficient that equalizes the electrons in both of the two halfreactions that
make up the redox reaction. © 2010 ChemReview.Net v1w Page 1160 Module 37 — Electrochemistry Steps To Calculate NonStandard Cell Potential
1. Calculate the standard cell potential using the method above. 2. Adjust the cell potential for nonstandard conditions using the Nernst Equation:
cell = o cell ─ RT ln( Q )
nF where
cell = the nonstandard 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 halfreactions 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
nonstandard 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 halfreactions 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 nonstandard 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 halfreactions to balance the redox reaction.
o
values to the halfreactions. 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 nonstandard 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 nonstandard 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 halfreactions 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 halfreaction 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 nonstandard 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 halfreactions that produce lowestwholenumber coefficients in the final balanced equation.
n = 6 mol e─ in the halfreaction 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 halfreaction 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 nonstandard emf for a reaction, halfreaction, 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 halfreactions
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 halfreactions to balance the redox reaction.
Add o values to the halfreactions. Reverse the sign of the one reversed • halfreaction, but do not change o values.
Add the o values to find o cell. b. For nonstandard 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 halfcells. In each halfcell, one redox halfreaction takes
place. Oxidation occurs in one halfcell and reduction in the other.
A cell physically separates the two redox halfreactions 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 halfcells. 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 halfcells 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 halfreaction 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 halfcell, the
metals and ions in each halfcell 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 halfcell to the other. To maintain
balanced charges, ions must also travel between the two halfcells. 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 lowvoltage 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 nonstandard conditions. Let’s take the standard
conditions case first.
• The table of reduction potentials lists its halfreactions 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 halfcells. 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 halfcell with the stronger reducing agent (sRA) and travel
through the wire toward the other halfcell.
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 halfcell 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 halfcells 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 halfcell
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 halfcells.
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+ halfcell, 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 halfreactions.
An electrochemical cell is a combination of two halfcells.
In an electrochemical cell, a redox reaction takes place in which the two halfreactions
are carried out in the two separate halfcells.
• One halfcell has the particles in one halfreaction. The other halfcell has the
particles in the other halfreaction. • One halfcell 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 halfcell.
As current flows through the wire connecting the cell electrodes,
• the two halfreactions 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 halfcells 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 halfcells (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 nonstandard 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 halfcells 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 halfcells 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 halfreactions.
1. Using the SRP table, find the two halfreactions that contain the particles in each of
the two halfcells.
o
value.
Then copy the higher in the table of the two halfreactions. Attach its
o
2. Under the first halfreaction, write the second table halfreaction reversed. Attach
with its sign reversed.
o
3. Balance the electrons, add the two halfreactions, and add to find cell.
Writing the higher halfreaction 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 halfreactions.
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 NonStandard Conditions
The voltage for a cell is different at nonstandard conditions than at standard conditions.
At nonstandard conditions, the direction of the electron flow may be the same as at
standard conditions, or it may be reversed.
o
To find a nonstandard 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 nonstandard conditions?
d. Was the change in voltage between standard and these nonstandard 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 nonstandard 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 halfreactions 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 nonstandard 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 halfcell to the left halfcell 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 nonstandard potential based on two halfcells are
• Write the halfreaction higher in the table. Under it, write the other halfreaction, reversed.
o
• Balance and add the two halfreactions, 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 halfreactions
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 halfcell has one electrode: a conductive solid component of the halfcell (usually a
metal) to which wires can be attached. When two halfcells 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 halfcell 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 halfcell 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 halfreactions. 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 nonstandard conditions adjusted
with the Nernst equation.
The terms anode and cathode come from the fact that historically, in the twometal 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 halfcell 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 halfcell 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 halfreactions include metals that can serve as electrodes. If a halfreaction does not
include a metal or other solid conductor, a nonreactive solid conductor must added to the
halfcell 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 halfcell
that does not include a metal. The transfer of electrons that occurs in the halfreaction takes
place at the surface of the platinum. A conductive nonreactant 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 halfcell 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 halfcells.
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 halfcell must include the particles in one halfreaction: two particles in which
the same central atom has different oxidation numbers. • The anode (─) is the electrode in the halfcell 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 halfcell 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 halfcells.
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 halfcell, 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 halfcell does not include a solid metal, add Pt(s) . 2. Under the particle formulas in the reaction (if supplied) and the halfcells, 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 halfcell with the sRA. Above the metal, draw a ↑ to show
electrons leaving that halfcell.
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 halfcells can be drawn in the reverse order.)
4. Draw a ↓ above the metal in the halfcell with the sOA.
5. Between the arrows, draw a ← or → arrow showing the direction of electron flow
between the halfcells. © 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 halfcells 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 halfcell that contains the sRA. In this problem, the
sRA is Cu, and Cu is the metal in the halfcell with the sRA, so Cu is the anode (─).
In the other halfcell must be the cathode (+). The electrode normally must be a
solid metal, and the only metal in that halfcell 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 halfcells at standardstate 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 halfcell with the
anode (─), and Zn is the metal in that halfcell, it is the anode.
In the other halfcell 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 halfcells. Each halfcell has the particles in halfreaction, leaving out
H2O and e─. Add Pt to any halfcell without a solid metal.
2. Below the symbols, label the sRA, sOA, wRA, and wOA.
3. Draw a ↑ above the metal in the halfcell with the sRA.
4. Draw a ↓ above the one metal in the halfcell 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 halfcell require an added nonreactive 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 halfcells under
standard conditions.
2+ ⎩ Pb and Pb ⎭ and ⎩ Pt and Fe 2+ and Fe3+ ⎭ 3. Assuming standard conditions, draw the halfcells 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 halfcells 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 halfcells.
⏐
2. The components of the halfcell 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 halfcell 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 halfcell, Pt(s)⏐ representing a platinum anode (─) is
written at the far left. If there is no conductor in the halfcell 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 standardstate 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 halfreaction 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 halfreaction with the sRA,
Zn is the anode. Write Zn(s) ⏐ first.
2. Next write the formulas for remaining particles in the halfreaction 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 standardstate 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 halfcells
and halfreactions.
* * ** *
So far: Zn(s) ⏐ Zn2+(1 M) ⏐
⏐ 4. Write the components in the other halfreaction and halfcell the same way. List the
cathode after a single vertical line at the far right. If no component in this halfcell and
halfreaction 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 halfcell includes a solid, it is not a metal. Metals are electrically conductive, but most other
solids are not. Assume this halfcell needs a nonreactive 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 halfcells may be reversed, so long as the component labels remain the same and electrons leave the
anode.
In the halfcell 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 halfcell 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 halfcells 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 halfcell with
the sRA contains the anode, and the metal is Ni. The metal in the other halfcell 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 halfcells 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 halfcell 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 halfreaction 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 halfcell. It must be the sRA because it is in the halfcell
with the anode, and the other particle in its halfcell must be the wOA. The sOA must be in the other
halfcell, 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
halfcell 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 halfreaction 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 halfreactions 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 halfcell
and halfreaction as the sRA. That halfcell 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 nonreactive electrode must be added. The halfcell 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 halfcell will need a nonreactive 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 halfcells
* * * * *. © 2010 ChemReview.Net v1w Page 1185 Module 38 — Electrochemical Cells Lesson 38C: Depleted Batteries and Concentration Cells
Dead Batteries and K
Once the halfcells 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 halfreactions 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 rightside 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 standardstate conditions,
a. Write the products and balance the equation by adding halfreactions.
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 halfcells 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 halfcells until the Q value in both halfcells is the same. For two halfcells 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 halfcells 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 halfcell, taking it closer to the lower ion concentration in the left halfcell.
The reactions in both halfcells 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 ConcentrationCell 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 ConcentrationCell 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 halfcells have the same halfreaction 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 redoxreactive 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 halfreaction. Using those rules, try this problem.
Q. In the Cu/ Cu2+ concentration cell above, [Cu2+] = 1.0 M and 0.10 M .
a. Which halfcell contains the cathode (+)?
b. Solve for the initial cell voltage at 20.ºC.
*****
Part a. The electrons leave the halfcell with the lower [reacting metal ion], so that halfcell
contains the anode (─). The cathode (+) must be in the halfcell with the higher
ion concentration. In the diagram above, that is the halfcell on the right. Part b. To find concentrationcell 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+ halfreaction 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 halfcell components but differing [ions].
2. In a concentration cell, current flows, with steadily dropping voltage, until Q = 1.
3. Electrons leave the halfcell with the lower [redoxing ions] (or lower product of
[redoxing ions].
4. The halfcell 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 halfcells 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 halfcell contains the anode (─)? 2. Two halfcells 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 halfcell.
b. Which halfcell 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 halfreactions 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 halfreactions 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 halfreactions 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 concentrationcell 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+ halfreaction
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 halfcell with the lower ion concentrations, and that halfcell contains the anode.
o
2a. For concentrationcell 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+ halfreaction 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 halfcell with the lower ion concentrations, and that halfcell contains the anode.
The cathode (+) is the electrode in the other halfcell, 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 halfcell.
• 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 halfreaction can only go to the right.
Electrolysis, however, can remove electrons from F─ and drive the halfreaction
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 halfreactions 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 halfcells. 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
halfcell 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 halfcell 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 halfcell 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 halfreaction will occur at the anode, and what product is produced?
e. Which halfreaction 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 halfcell 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 halfreactions 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 halfreaction
that is oxidiation is Cl─ losing its electrons: 2 Cl─
Cl + 2e─ . The product is Cl .
2 2 e. The other halfreaction must take place at the cathode. That halfreaction 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 crosssectional 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 halfreaction (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 halfcell 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 singleunit 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 halfreaction) 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 halfcell 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
rightside 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 halfcell 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 rightside 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 singleunit 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 halfcell 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 halfreactions.
An electrochemical cell is a combination of two halfcells.
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 halfcell, one of the
halfreactions takes place.
• One halfcell contains the particles in one halfreaction. The other halfcell
contains the particles in the other halfreaction. • One halfcell 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 halfreactions 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 halfcell 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 halfcell with the sRA. c. For electrons to flow, the electrodes are connected by a conductor (usually a metal
wire) and the halfcell 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 halfcell.
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 halfcell 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 halfcell 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 singleunit 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 halfreaction (one with a term for e─).
The bridge conversion will include moles of electrons and moles of
WANTED or given particles.
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