Question

8.) Balance the following redox reaction in acidic solution: PbO2(s) + I-(aq) → Pb2+(aq) + I2(s)

Answer 1

PbO2(s) + I -(aq) --------->Pb2+(aq) + I2(s)

To balance this reaction equation, let's see individual redox reaction:

Reduction: as PbO2 is gaining 2 e- it is reduced

PbO2(s) + 4H+ + 2e- ---------> Pb2+(aq) + 2H2O

Oxidation: as I- is giving electrons it is oxidized

2 I-(aq) -----------> I2 (s) +2 e-

Thus total reaction is

PbO2(s) + 4H+ + 2e- + 2 I-(aq)    ---------> Pb2+(aq) + 2H2O  + I2 (s) +2e-

PbO2(s) + 4H+ + 2 I-(aq)    ---------> Pb2+(aq) + 2H2O  + I2 (s)