1. Chapter 15 Oxidation and Reduction
15.3 Balancing Oxidation−Reduction Reactions Using Half-Reactions
Learning Goal Balance oxidation− reduction reactions using the half-reaction method.
A dichromate solution (yellow) and an iodide solution (colorless) form a brown solution of Cr3+ and iodine (I2).

2. Balance Oxidation and Reduction Half-Reaction Method
In the half-reaction method for balancing oxidation−reduction reactions
the reaction is written as two half-reactions
the loss and gain of electrons needs to be equalized for the half-reactions
the two balanced half-reactions are added

3. Balancing Redox, Guide to Using Half-Reactions

4. Learning Check
Balance the following reaction. Zn(s) + Fe3+(aq)  Zn2+(aq) + Fe2+(aq)

5. Solution
Balance the following reaction. Zn(s) + Fe3+(aq)  Zn2+(aq) + Fe2+(aq) Step 1 Write two half-reactions for the equation. Zn(s)  Zn2+(aq) Fe3+(aq)  Fe2+(aq)

6. Solution
Balance the following reaction. Zn(s) + Fe3+(aq)  Zn2+(aq) + Fe2+(aq) Step 2 For each half-reaction, balance the elements other than H and O. If necessary, balance O by adding H2O and H by adding H+. Zn(s)  Zn2+(aq) Fe3+(aq)  Fe2+(aq)

7. Solution
Balance the following reaction. Zn(s) + Fe3+(aq)  Zn2+(aq) + Fe2+(aq) Step 3 Balance each half-reaction for charge by adding electrons. Zn(s)  Zn2+(aq) + 2 e− Fe3+(aq) + 1 e−  Fe2+(aq)

8. Solution
Balance the following reaction. Zn(s) + Fe3+(aq)  Zn2+(aq) + Fe2+(aq) Step 4 Multiply each half-reaction by factors that equalize the loss and gain of electrons. Zn(s)  Zn2+(aq) + 2 e− 2 × [Fe3+(aq) + 1 e−  Fe2+(aq)] 2Fe3+(aq) + 2 e−  2Fe2+(aq)

9. Solution
Balance the following reaction. Zn(s) + Fe3+(aq)  Zn2+(aq) + Fe2+(aq) Step 5 Add half-reactions, cancel electrons and any identical ions or molecules.

10. Solution
Balance the following reaction. Zn(s) + Fe3+(aq)  Zn2+(aq) + Fe2+(aq) Step 5 Check balance of atoms and charge. Reactants Products Zn Zn 2Fe 2Fe 2 e− 2 e−

11. Balancing Redox Equations in Acidic Solution
When balancing oxidation−reduction reactions by the half-reaction method in acidic solution, we add
H2O to balance oxygen atoms
H+ to balance H atoms

12. Learning Check
Balance the following reaction in acidic solution. I−(aq) + SO42−(aq)  I2(s) + S(s)

13. Solution
Balance the following reaction in acidic solution. I−(aq) + SO42−(aq)  I2(s) + S(s) Step 1 Write two half-reactions for the equation. I−(aq)  I2(s) SO42−(aq)  S(s)

14. Solution
Balance the following reaction in acidic solution. I−(aq) + SO42−(aq)  I2(s) + S(s) Step 2 For each half-reaction, balance the elements other than H and O. If necessary, balance O by adding H2O, and H by adding H+. 2I−(aq)  I2(s) SO42−(aq) + 4H+(aq)  S(s) + 4H2O(l)

15. Solution
Balance the following reaction in acidic solution. I−(aq) + SO42−(aq)  I2(s) + S(s) Step 3 Balance each half-reaction for charge by adding electrons. 2I−(aq)  I2(s) + 2 e− SO42−(aq) + 8H+(aq) + 6 e−  S(s) + 4H2O(l)

16. 6I−(aq)  3I2(s) + 6 e− Solution
Balance the following reaction in acidic solution.
I−(aq) + SO42−(aq)  I2(s) + S(s)
Step 4 Multiply each half-reaction by factors that equalize the loss and gain of electrons.
3 × [2I−(aq)  I2(s) + 2 e−]
6I−(aq)  3I2(s) + 6 e−
SO42−(aq) + 8H+(aq) + 6 e−  S(s) + 4H2O(l)

17. Solution
Balance the following reaction in acidic solution. I−(aq) + SO42−(aq)  I2(s) + S(s) Step 5 Add half-reactions, cancel electrons and any identical ions or molecules.

18. Balance the following reaction in acidic solution.
6I−(aq) + SO42−(aq) + 8H+(aq) 
3I2(s) + S(s) + 4H2O(l)
Step 5 Check balance of atoms and charge Reactants Products
6I I
S S
8H 8H
4O 4O
6 e− 6 e−

19. Balancing Redox Equations in Basic Solution
When balancing oxidation−reduction reactions by the half-reaction method in basic solution, we
balance the equation using H+ and H2O
neutralize the H+ by adding OH− to both sides of the reaction to form H2O

20. Learning Check
Balance the following equation in basic solution. NO(g) + ClO3−(aq)  NO3−(aq) + Cl−(aq)

21. Solution
Balance the following equation in basic solution. NO(g) + ClO3−(aq)  NO3−(aq) + Cl−(aq) Step 1 Write two half-reactions for the equation. NO(g)  NO3− (aq) ClO3−(aq)  Cl−(aq)

22. Solution
Balance the following equation in basic solution. NO(g) + ClO3−(aq)  NO3−(aq) + Cl−(aq) Step 2 For each half-reaction, balance the elements other than H and O. If necessary, balance O by adding H2O, and H by adding H+. 2H2O(l) + NO(g)  NO3− (aq) + 4H+(aq) 6H+(aq) + ClO3−(aq)  Cl−(aq) + 3H2O(l)

23. Solution
Balance the following equation in basic solution. NO(g) + ClO3−(aq)  NO3−(aq) + Cl−(aq) Step 3 Balance each half-reaction for charge by adding electrons. 2H2O(l) + NO(g)  NO3− (aq) + 4H+(aq) + 3 e− 6 e− + 6H+(aq) + ClO3−(aq)  Cl−(aq) + 3H2O(l)

24. Balance the following equation in basic solution.
NO(g) + ClO3−(aq)  NO3−(aq) + Cl−(aq)
Step 4 Multiply each half-reaction by factors that equalize the loss and gain of electrons.
2 × [2H2O(l) + NO(g)  NO3− (aq) + 4H+(aq) + 3 e−]
4H2O(l) + 2NO(g)  2NO3− (aq) + 8H+(aq) + 6 e−
6 e− + 6H+(aq) + ClO3−(aq)  Cl−(aq) + 3H2O(l)

25. Solution
Balance the following equation in basic solution. NO(g) + ClO3−(aq)  NO3−(aq) + Cl−(aq) Step 5 Add half-reactions, cancel electrons and any identical ions or molecules. 2

26. Balance the following equation in basic solution.
NO(g) + ClO3−(aq)  NO3−(aq) + Cl−(aq)
Step 5 To convert the equation to an oxidation– reduction reaction in basic solution, we neutralize H+ with OH− to form H2O.
H2O(l) + 2NO(g) + ClO3−(aq) + 2OH−(aq) 
2NO3−(aq) + 2H+(aq) + 2OH−(aq) + Cl−(aq)
2NO3−(aq) + 2H2O(l) ) + Cl−(aq)
2NO(g) + ClO3−(aq) + 2OH−(aq) 
2NO3−(aq) + H2O(l) ) + Cl−(aq)

27. Step 5 Check balance of atoms and charge. Reactants Products 2N 2N
Solution
Balance the following equation in basic solution. 2NO(g) + ClO3−(aq) + 2OH−(aq) 
2NO3−(aq) + H2O(l) ) + Cl−(aq)
Step 5 Check balance of atoms and charge Reactants Products
2N N
7O O
Cl Cl
2H H
6 e− e−