Verified answer

PbO + Co(NH3)6^3+ ↔ PbO2 + Co(NH3)6^2+

Oxidation numbers

.......Reactant......Product

Pb.....+2..............+4...........Oxidation

Co.....+9..............+8...........Reduction

Half Reactions

PbO -------> PbO2 (oxidation)

Co(NH3)6^3+ ---------> Co(NH3)6^2+ (reduction)

1) Pb and Co are already balanced

2) Add H2O to balance O:

PbO + H2O -------> PbO2

Co(NH3)6^3+ ---------> Co(NH3)6^2+

3)Add H+ to balance H:

PbO + H2O -------> PbO2 + 2H+

Co(NH3)6^3+ ---------> Co(NH3)6^2+

4) Add e- to balance charge

PbO + H2O -------> PbO2 + 2H+ + 2e-

Co(NH3)6^3+ + e- ---------> Co(NH3)6^2+

5 Equalize e-

PbO + H2O -------> PbO2 + 2H+ + 2e-

2Co(NH3)6^3+ + 2e- ---------> 2Co(NH3)6^2+

6 Combine and simplify

PbO + H2O + 2Co(NH3)6^3+ -----> PbO2 + 2H+ + 2Co(NH3)6^2+

7) For basic solution, add OH- to neutralize H+

PbO + H2O + 2Co(NH3)6^3+ + 2OH- -----> PbO2 + 2H2O+ + 2Co(NH3)6^2+

8) solution

PbO + 2Co(NH3)6^3+ + 2OH- -----> PbO2 + H2O+ + 2Co(NH3)6^2+