Complete and balance the following redox reaction in the basic solution Pb2+(aq) +IO3- (aq) ----> PbO2(s)+I2(s)

Question

Complete and balance the following redox reaction in the basic solution

Pb²⁺(aq) +IO₃^- (aq) ----> PbO₂(s)+I₂(s)

Answer 1

Pb²⁺(aq) +IO₃^- (aq) ----> PbO₂(s) + I₂(s)

Pb²⁺ ==> PbO₂ ... oxidation half reaction (Pb goes from 2+ to 4+)

Pb²⁺ + 2H2O ==> PbO₂ ... balance for Pb and O

Pb²⁺ + 2H2O + 4OH- ==> PbO₂ + 4H2O ... balanced for Pb, O and H (using base, OH-)

Pb²⁺ + 2H2O + 4OH- ==> PbO₂ + 4H2O + 2e- ... balanced for Pb, O, H and charge = balanced oxid. rxn

IO₃^- ==> I₂ ... reduction half reaction (I goes from 5+ to zero)

2IO₃^- ==> I₂ + 6H2O ... balanced for I and O

2IO₃^- + 12H2O ==> I₂ + 6H2O + 12OH- ... balanced for I, O and H (using base, OH-)

2IO₃^- + 12H2O + 10e- ==> I₂ + 6H2O + 12OH- ... balanced for I, O, H and charge = balanced reduction rxn

Multiply oxidation half reaction by 5 to equalize electrons; add the 2 equations and combine/cancel to get..

5Pb²⁺ + 10H2O + 20OH- ==> 5PbO₂ + 20H2O + 10e-

2IO₃^- + 12H2O + 10e- ==> I₂ + 6H2O + 12OH-

______________________________________________

5Pb²⁺ + 2IO₃^- + 8OH- => 5PbO₂ + 4H2O + I₂ ... BALANCED REDOX EQUATION