Electrochemistry 1 Oxidation and Reduction Reactions Oxidation-Reduction Reactions In our previous studies, we examined reactions that occurred in aqueous solutions that involved formation of water (acid-base neutralization) and formation of a solid (precipitation) as the driving force of the chemical reactions. During these reactions, all substances maintained their charge after the reactions were completed. Another type of reaction, which occurs in both aqueous solutions and in reactions where substances are burned in the presence of oxygen gas, involves a transfer of electrons as the driving force of the chemical reaction. The reactants in these reactions will lose or gain electrons and change their charge as they form the products. The driving force of these chemical changes is electron transfer and they are called oxidation and reduction reactions. Oxidation-reduction reactions are a very important class of chemical reactions. They occur all around us and even within us. The bulk of the energy needed for the functioning of all living organisms, including humans, is obtained from food through oxidation-reduction processes. Such diverse phenomena as the electricity obtained from a battery to start a car, the use of natural gas to heat a home, iron rusting, and the functioning of antiseptic agents to kill or prevent the growth of bacteria all involve oxidation-reduction reactions. The knowledge of this type of reaction is fundamental to understanding many biological and technological processes. Historically, the word oxidation was first used to describe the reaction of a substance with oxygen. According to this historical definition, each of the following reactions involves oxidation: 4 Fe (s) + 3 O2 (g) 2 Fe2O3 (s) S (s) + O2 (g) SO2 (g) CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g) Originally, the term reduction referred to processes where oxygen was removed from a compound. A particularly common type of reduction reaction, according to the original definition, is the removal of oxygen from a metal oxide to produce free metal. CuO (s) + H2 (g) Cu (s) + H2O (g) 4 Fe (s) + 3 CO2 (g) 2 Fe2O3 (s) + 3 C (s) Today the words oxidation and reduction are used in a much broader sense. Current definitions include the previous examples but also include reactions with numerous non-oxygen containing substances. Reactions that involve the transfer of electrons from one reactant to another reactant, regardless of the substances involve, are collectively called oxidation-reduction reactions (redox). The reactants in these reactions will lose or gain electrons and change their charge as they form the products. Oxidation is the process in which a substance in a chemical reaction loses electrons. Reduction is the process in which a substance in a chemical reaction gains electrons. For all redox reactions, the reactants must lose electrons and gain electrons during the chemical process. OIL RIG OIL oxidation is loss of electrons (metals in salts) RIG reduction is gain of electrons (nonmetals in salts) The name given to the charge on an element, ion or element in a covalent bond is called the oxidation state or oxidation numbers. For example:
2 Na+ has an oxidation state of +1. F has an oxidation state of 1.
To determine what substance is oxidized and which is reduced, we must determine the original and final oxidation states of each substance during the chemical reaction. To determine the oxidation states of all elements in molecules or ionic compounds, it is necessary to follow a few rules. Oxidation Numbers To assign oxidation numbers to elements in compounds, there are a few simple rules: 1. The oxidation number of any free, uncombined element is zero. The oxidation number (or charge on each atom) is written above the atom Na (s), Mg (s), Cu (s), Fe (s) and Zn (s) 2. All naturally occurring diatomic molecules have zero oxidation states. Br2 I2 N2 Cl2 H2 O2 F2 This makes sensethere is no dipole (difference in Electronegativity) between the shared electrons in the diatomic molecules. Electron Transfer Nonmetals, in their natural uncombined state, have zero oxidation states. E.g., P4 (s) and S8 (s)
0 0 0 0 0 0 0 0 0 0 0 0 0 0
3. For simple Group A binary ionic compounds (salts), the oxidation state is the charge of the element: Metals (Group I A, IIA, and IIIA) are assigned a positive oxidation state determined by the number of electrons the element has lost. Nonmetals (Group VA, VIA, VIIA) are assigned a negative oxidation state determined by the number of electrons the element has gained. Use the periodic table to help with assigning oxidation numbers to elements. IA metals have oxidation numbers of +1. IIA metals have oxidation numbers of +2. IIIA metals have oxidation numbers of +3. There are a few rare exceptions. 4. VA elements have oxidation numbers of 3 in binary compounds with H, metals or NH4+. 5. VIA elements below O have oxidation numbers of 2 in binary compounds with H, metals or NH4+. 6. VIIA elements have oxidation numbers of 1 in binary compounds with H, metals or NH4+. 7. The oxidation number of an element as a simple (monatomic) ion is the charge on the ion. Group IA metals +1 (Li+, Na+, K+, Rb+, Cs+) Group IIA metals +2 (Mg2+, Ca2+, Sr2+, Ba2+) Group IIIA metals +3 (Al3+, Ga3+, In3+) Group V nonmetals 3 (N3, P3, As3, Sb3, Bi3)
3 Group VI nonmetals Group VII nonmetals 2 (O2, S2, Se2, Te2) 1 (F, Cl, Br, I)
8. In the formula for any compound, the sum of the oxidation numbers of all elements in the compound is zero. Na2SO4 = zero charge on formula unit 9. In a polyatomic ion, the sum of the oxidation numbers of the constituent elements is equal to the charge on the ion. SO42- = -2 on the ion. 10. For simple binary ionic compounds that are comprised of transition metals (with more than one possible oxidation state), the oxidation state of the metal is determined by balancing the charge: For our purpose, all transition metals in salts (except zinc and silver) have oxidation states that are determined by the balancing the charge of the anion. e.g. FeO vs. Fe2O3
-2
FeO
Since the oxidation state of oxygen is 2 in the compound, the iron atom must have a charge of + 2 to give the formula unit an overall net charge of zero. x + (-2) = 0 x = +2 2 Fe2O3
Since the oxidation state of oxygen is 2 in the compound and there are three oxygen atoms, the iron must have a charge of +3 to give the formula unit an overall net charge of zero. 2x + 3(-2) = 0 x = +3 1. Zinc always has an oxidation state of +2 when combined in a salt. ZnCl2, ZnS, ZnSO4, Zn3(PO4)2 2. Silver always has an oxidation state of +1 when combined in a salt. AgCl, Ag2O, Ag2SO4, Ag3PO4 11. Fluorine has an oxidation number of 1 in its compounds. 12. Hydrogen, H, has an oxidation number of +1 unless it is combined with metals, where it has the oxidation number -1. Examples LiH, BaH2 13. Oxygen usually has the oxidation number 2. Exceptions: 3. In peroxides O has oxidation number of 1. Examples: H2O2, CaO2, Na2O2 4. In OF2, O has oxidation number of +2. (Fluorine is the most electronegative element and has the greater share of electrons.) Example: Assign oxidation numbers to each element in the following compounds: NaNO3 Na = +1 O = 2
4 N=? Calculate using rule 3. +1 + 3(-2) + x = 0 x = +5 N = +5 K2Sn(OH)6 K = +1 O = -2 H = +1 Sn = ? 2(+1) + 6(-2) + 6(+1) + x = 0 x = +4 Sn = +4 H3PO4 H = +1 O = -2 P=? 3(+1) + 4(2) + x = 0 =+5 P = +5 SO32O = -2 S=? 3(-2) + x = -2 x = +4 S = +4 HCO3 O = -2 H = +1 C=? +1 + 3(-2) + x = -1 x = +4 C = +4 Cr2O72O = 2 Cr = ? 7(2) + 2(x) = 2 x = +6 Cr = +6 Determine oxidation the number of the underlined element in FeSO4. The oxidation state of iron must be +2 in order to balance the 2 of the sulfate ion. Fe = + 2 O = 2 +2 + x + 4(2) = 0
5 x = +6 OR The oxidation state of sulfur can be determined by the sulfate ion: SO42 x + 4(2) = 2 x = +6 Determine the oxidation number of the underlined element in Mg(ClO4)2. Mg = + 2 O = 2 +2 + 2x + 8(2) = 0 x =+7 OR ClO4 x + 4(2) = 1 x = +7
Electron Transfer OxidationReduction Reactions: REDOX 5. Oxidizing agents are those elements that cause loss of electrons (cause oxidation) oxidizing agents are reduced by gaining electrons. 6. Reducing agents are those elements that give up electrons (cause reduction) reducing agents are oxidized by giving up electrons. 7. Electron Transfer In redox reactions, the reactants transfer electronsoxidation and reduction occurs between the reactants. 0 0 +1 1 2 Na (s) + Cl2 (g) 2 NaCl (s) To determine the oxidizing agent and reducing agent, we can divide the reaction of the reactants into half-reactions: Since we cannot either create or destroy electrons, the numbers of electrons transferred in the reaction must be the same on both sides of the chemical equation: 0 0 +3 2 4 Fe (s) + 3 O2 (g) 2 Fe2O3 (s) 0 +3 oxidation reaction 4 Fe 4 Fe3+ + 12 e 0 2 reduction reaction 3 O2 + 12 e 6 O2 + 0 2 NaBr (aq) + Cl2 (l) + 0 2 NaCl (aq) + Br2 (s)
We can remove the spectator ion (Na+) before we set up the half-reactions. A more complicated redox reaction requires that we must identify what is oxidized and what is reduced by assigning oxidation states to all the elements in the reactants and products.
6
+1 +6 2
H2SO4 (aq) + 2 HI (g)
+1 1
+4 2
SO2 (g) + I2 (s) + 2 H2O
0
+1 2
We must identify in the reaction which elements changed oxidation states. The oxidation state of hydrogen is the same on both sides of the equation, so it is not involved in transfer of electrons. The oxidation state of oxygen is the same on both sides of the equation, so it is not involved in transfer of electrons. The oxidation state of sulfur changed and the oxidation state of iodine changed. oxidation reaction 2 I
+6 1
I2
0
+ 2 e
+4
red agent ox agent
reduction reaction SO42 + 2 e
SO2
It is possible in a few reactions that an element in the reactant is oxidized and it is also the element that is reduced. This is called a disproportionation oxidation-reducion reaction.
Balancing Redox Equations
S lide 25
p.148
It is not always obvious how to balance a redox equation. For these more complex oxidation-reduction reactions, we have a system by which to balance the equation by mass and by charge. The half-reaction method is useful in that it is the method that is used in electrolysis. Balance the equation as much as possible without the spectator ions. Set up half-reactions (oxidation and reduction) as we did in the previous reactions. Balance the half-reactions by the following rules: First, determine if the reaction has taken place in acidic (H+) conditions or basic (OH) conditions. The rules for balancing oxygen and hydrogen are different depending on whether the reaction takes place in acidic or basic solutions. Half reaction method rules: 1. Write the unbalanced reaction. 2. Break the reaction into 2 half reactions: One oxidation half-reaction and One reduction half-reaction Each reaction must have complete formulas for molecules and ions. 3. Mass balance each half reaction by adding appropriate stoichiometric coefficients. 4. Charge balance the half reactions by adding appropriate numbers of electrons. 5. Multiply each half reaction by a number to make the number of electrons in the oxidation half-reaction equal to the number of electrons reduction half-reaction. 6. Add the two half reactions. 7. Eliminate any common terms and reduce coefficients to smallest whole numbers.
7 In acidic conditions: Balance the oxygen by adding H2O molecules to the opposite side of the equation. Balance the hydrogen by adding H+ to the opposite side of the equation. Balancing Redox Equations Example:
0
C (s) +
+1+5 2
HNO3 (aq)
+4 2
CO2 (g) + NO2 (g) + H2O
+42
+1 2
Set up the half-reactions: C
+5 0 +4 +4
CO2
+ 4 e
oxidation reduction
NO3 + 1e
NO2
Balance the oxygen by adding H2O to the opposite side. (One H2O for each oxygen) 2 H2O + C NO3 + 1e CO2 + 4 e NO2 + H2O
Balance the hydrogen by adding H+ to the opposite side. 2 H2O + C 2 H + NO3 + 1e
+
CO2 + 4 e + NO2 + H2O
4 H+
The number of e must be the same on both sides of the equation. Balance the charge by multiplying the half-reaction with the appropriate number to give the same number of electrons on both sides of the half-equations. 4 (2 H
+
2 H2 O + C + NO3 + 1e
CO2 NO2
+ 4 e + + H2O)
4 H+
2 H2O + C 8 H + 4 NO3 + 4 e
+
CO2 + 4 e + 4 NO2 + 4 H2O
4 H+
Cancel e, water, and H+ from opposite sides of the equations, and sum the remaining reactants and products. 2 H2O + C CO2 + 4 e + 4 H+ 4 2 8 H+ + 4 NO3 + 4 e 4 NO2 + 4 H2O 4 H+ + 4 NO3 + C 4 NO2 + CO2 + 2 H2O
Add the H+ back to the NO3 ions to complete the equation: 4 HNO3 +C 4 NO2 + CO2 + 2 H2O
8 In basic conditions: Balance the oxygen by adding 2 OH molecules to the side that needs oxygen and adding one H2O to the other side. Balance the hydrogen by adding H2O to the side that needs hydrogen and adding one OH to the other side. Example: In basic solution, hypochlorite ions (ClO) oxidize chromite ions (CrO2) to chromate ions (CrO42) and are reduced to chloride ions. Write a balanced net ionic equation for this reaction.
+3
CrO2 + ClO
+1
+6
CrO42 + Cl CrO42 + 3 e Cl oxidation reduction
1
Set up the half-reactions: CrO2 ClO + 2 e
Balance the oxygen by adding 2 OH to the opposite side. (2 OH for each oxygen) CrO2 + 4 OH ClO + 2 e CrO42 + 3 e Cl + 2 OH
Balance the hydrogen by adding one H2O for every two OH- on the opposite side. CrO2 + 4 OH H2O + ClO + 2 e CrO42 + 2 H2O + 3 e Cl + 2 OH
Balance the charge by multiplying the half-reaction with the appropriate number to give the same number of electrons on both sides of the equation. 2 (CrO2 + 4 OH 3 (H2O + ClO + 2 e CrO42 + 2 H2O + 3 e) Cl + 2 OH )
2 CrO42 + 4 H2O + 6 e 2 CrO2 + 8 OH 3 H2O + 3 ClO + 6 e 3 Cl + 6 OH Add the two half reactions. Eliminate any common terms and reduce coefficients to smallest whole numbers. 2 1 2 CrO2 + 8 OH 2 CrO42 + 4 H2O + 6 e 3 H2O + 3 ClO + 6 e 3 Cl + 6 OH 2 CrO2 + 2 OH + 3 ClO
-
2 CrO42 + H2O + 3 Cl
Example: Dichromate ions oxidize iron (II) ions to iron (III) ions and are reduced to chromium (III) ions in acidic solution. Write and balance the net ionic equation for the reaction. We will solve this one in class. Example: Balance the following redox reaction: NO2 (g) + OH (aq) NO3 + NO2 + H2O
9 Stoichiometry of Redox Reactions Redox titrations: Titrations can be performed to determine the presence of substances that are oxidized or reduced. In this procedure, the standard is a solution of a known oxidizing or reducing agent. Example. What volume of 0.200 M KMnO4 is required to oxidize 35.0 mL of 0.150 M HCl? The balanced reaction is:
2 KMnO 4 + 16 HCl 2 KCl + 2 MnCl2 + 5 Cl2 + 8 H 2 O
( 35
m L H C l )( 0.150 M H C l ) = 5.25 m m ol H C l 2 m m ol K M nO 4 = 0.656 m m ol K M nO 4 16 m m ol H C l
( 5.25 m m ol H C l ) ( 0.656
1 mL m m ol K M nO 4 ) = 3.28 m L 0.200 m m ol K M nO 4
Example. The iron in a 5.675 g sample containing some Fe2O3 is reduced to Fe2+. The Fe2+ is titrated with 12.02 mL if 0.1467 M K2Cr2O7 in acid solution. What is the mass of Fe and what is the percentage of Fe in the sample? Fe2+ + Cr2O72 Fe3+ + Cr3+