1. Chapter 18: Electrochemistry
CHE 124: General Chemistry II
Dr. Jerome Williams, Ph.D.
Saint Leo University

2. Overview
Balancing Redox Reactions: Half-Reaction Method

3. Balancing Redox Reactions
The main principle is that electrons are transferred – so if we find a method to keep track of the electrons it will allow us to balance the equation
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4. Half-Reaction Method
We split the redox reaction into two separate half-reactions – a reaction just involving oxidation or reduction
oxidation half-reaction has electrons as products
reduction half-reaction has electrons as reactants
3 Cl2 + I− + 3H2O → 6 Cl− + IO3− + 6 H+
− − − −2 +1
oxidation: I− → IO3− + 6 e−
reduction: Cl2 + 2 e− → 2 Cl−
Tro: Chemistry: A Molecular Approach, 2/e 4

5. Balancing Redox Reactions by the Half-Reaction Method
Reaction is separated into two half-reactions; one for oxidation, the other reduction
Each half-reaction includes electrons
electrons go on the product side of the oxidation half-reaction – loss of electrons
electrons go on the reactant side of the reduction half-reaction – gain of electrons
Each half-reaction is balanced for its atoms
Then the two half-reactions are adjusted so that the electrons lost and gained will be equal when combined
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6. Example 18.2: Balancing redox reactions
in acidic solution
1. assign oxidation states and determine element oxidized and element reduced 2. separate into oxidation & reduction half- reactions
Fe2+ + MnO4– → Fe3+ + Mn2+ +2 +7 −2 +3 +2
oxidation
reduction
ox: Fe2+ → Fe3+
red: MnO4– → Mn2+
Tro: Chemistry: A Molecular Approach, 2/e 6 6

7. Example 18.2: Balancing redox reactions
in acidic solution
3. balance half-reactions by mass
a) first balance atoms other than O and H
b) balance O by adding H2O to side that lacks O
c) balance H by adding H+ to side that lacks H
Fe2+ → Fe3+
MnO4– → Mn2+
MnO4– → Mn2+ + 4H2O
MnO4– + 8H+ → Mn2+ + 4H2O
Tro: Chemistry: A Molecular Approach, 2/e 7 7

8. Example 18.2: Balancing redox reactions
in acidic solution
4. balance each half-reaction with respect to charge by adjusting the numbers of electrons
a) electrons on product side for oxidation
b) electrons on reactant side for reduction
Fe2+ → Fe e−
MnO4– + 8H+ → Mn2+ + 4H2O +7 +2
MnO4– + 8H+ + 5 e− → Mn2+ + 4H2O
Tro: Chemistry: A Molecular Approach, 2/e 8

9. Example 18.2: Balancing redox reactions
in acidic solution
5. balance electrons between half-reactions 6. add half-reactions, canceling electrons and common species 7. Check that numbers of atoms and total charge are equal
Fe2+ → Fe e−
} x 5
MnO4– + 8H+ + 5 e− → Mn2+ + 4H2O
5 Fe2+ → 5 Fe e−
MnO4– + 8H+ + 5 e− → Mn2+ + 4H2O
5 Fe2+ + MnO4– + 8H+ → Mn2+ + 4H2O + 5 Fe3+
reactant side
Element
product side 5 Fe 1 Mn 4 O 8 H
+17
charge
Tro: Chemistry: A Molecular Approach, 2/e 9

10. Balancing Redox Reactions
1. assign oxidation states
a) determine element oxidized and element reduced
2. write ox. & red. half-reactions, including electrons
a) ox. electrons on right, red. electrons on left of arrow
3. balance half-reactions by mass
a) first balance elements other than H and O
b) add H2O where need O
c) add H+ where need H
d) if reaction done in Base, neutralize H+ with OH−
4. balance half-reactions by charge
a) balance charge by adjusting electrons
5. balance electrons between half-reactions
6. add half-reactions
7. check by counting atoms and total charge
Tro: Chemistry: A Molecular Approach, 2/e 10 10

11. Practice – Balance the following equation in acidic solution I– + Cr2O72− → Cr3+ + I2
Tro: Chemistry: A Molecular Approach, 2/e 11

12. Practice – Balancing redox reactions
1. assign oxidation states and determine element oxidized and element reduced
2. separate into oxidation & reduction half- reactions
I− + Cr2O72– → I Cr3+ −1 +6 −2 +3
oxidation
reduction
ox: I− → I2
red: Cr2O72– → Cr3+
Tro: Chemistry: A Molecular Approach, 2/e 12 12

13. Practice – Balancing redox reactions
3. balance half-reactions by mass
a) first balance atoms other than O and H
b) balance O by adding H2O to side that lacks O
c) balance H by adding H+ to side that lacks H
ox: I− → I2
ox: 2 I− → I2
red: Cr2O72– → Cr3+
red: Cr2O72– → 2 Cr3+
red: Cr2O72– → 2Cr3+ +7H2O
Cr2O72– +14H+ → 2Cr3+ +7H2O
Tro: Chemistry: A Molecular Approach, 2/e 13 13

14. Practice – Balancing redox reactions
4. balance each half-reaction with respect to charge by adjusting the numbers of electrons
a) electrons on product side for oxidation
b) electrons on reactant side for reduction
2 I− → I2 + 2e−
2 I− → I2
Cr2O72– +14H+ → 2Cr3+ +7H2O
Cr2O72– +14H+ + 6e−→ 2Cr3+ +7H2O
Tro: Chemistry: A Molecular Approach, 2/e 14

15. Practice – Balancing redox reactions
5. balance electrons between half-reactions
6. add half-reactions, canceling electrons and common species
7. check
6 I− → 3 I2 + 6e−
2 I− → I2 + 2e−
} x 3
Cr2O72– +14H+ + 6e−→ 2Cr3+ +7H2O
Cr2O72– +14H+ + 6 I−→ 2Cr3+ +7H2O + 3 I2
reactant side
Element
product side 2 Cr 6 I 7 O 14 H +6
charge
Tro: Chemistry: A Molecular Approach, 2/e 15

16. Practice – Balance the following equation in basic solution I– + CrO42− → Cr3+ + I2
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17. Practice – Balancing redox reactions
1. assign oxidation states and determine element oxidized and element reduced
2. separate into oxidation & reduction half- reactions
I− + CrO42– → I Cr3+ −1 +6 −2 +3
oxidation
reduction
ox: I− → I2
red: CrO42– → Cr3+
Tro: Chemistry: A Molecular Approach, 2/e 17 17

18. Practice – Balancing redox reactions
3. balance half-reactions by mass
a) first balance atoms other than O and H
b) balance O by adding H2O to side that lacks O
c) balance H by adding H+ to side that lacks H
d) if in basic solution, neutralize the H+ with OH−
ox: I− → I2
ox: 2 I− → I2
red: CrO42– → Cr3+
red: CrO42– → Cr3+ +4H2O
CrO42– +8H+ → Cr3+ +4H2O
CrO42– +8H2O → Cr3+ +4H2O +8OH−
CrO42– +8H+ +8OH− → Cr3+ +4H2O +8OH−
CrO42– +4H2O → Cr3+ +8OH−
Tro: Chemistry: A Molecular Approach, 2/e 18 18

19. Practice – Balancing redox reactions
4. balance each half-reaction with respect to charge by adjusting the numbers of electrons
a) electrons on product side for oxidation
b) electrons on reactant side for reduction
2 I− → I2 + 2e−
2 I− → I2
CrO42– +4H2O+ 3e− → Cr3+ +8OH−
CrO42– +4H2O → Cr3+ +8OH−
Tro: Chemistry: A Molecular Approach, 2/e 19

20. Practice – Balancing redox reactions
5. balance electrons between half-reactions
6. add half-rxns; cancel electrons and common species
7. check
6 I− → 3 I2 + 6e−
2 I− → I2 + 2e−
} x 3
2CrO42– +8H2O+ 6e− → 2Cr3+ +16OH−
CrO42– +4H2O+ 3e− → Cr3+ +8OH−
} x 2
2CrO42– +8H2O + 6 I− → 2Cr3+ +16OH− + 3 I2
reactant side
Element
product side 2 Cr 6 I 16 O H
−10
charge
Tro: Chemistry: A Molecular Approach, 2/e 20