Balance the following redox reactions by ion-electron method.
`MnO_(4)^(-)(aq)+I^(-)(aq)toMnO_(2)(s)+I_(2)(s)` (in basic medium)

Explanation

a. Balancing of reduction half equation
i.
ii. Balance atoms by adding `2H_(2)O` to RHS of equation (i)
`3e^(-)+MnO_(4)^(Θ)rarrMnO_(2)+2H_(2)O`
iii. Balance `H` atoms by adding `4H_(2)O` to LHS and add simultaneously of `4overset(Θ)OH` to RHS (basic medium)
`overset(2)cancel(4H_(2)O)+3e^(-)+MnO_(4)^(Θ)rarrMnO_(2)+cancel(2H_(2)O)+4overset(Θ)OH`
iv. Balancing of oxidation half equation
`2I^(Θ)rarrI_(2)e^(-)`
v. Multiply equation (iii) by `2` and equation (iv) by `3` and add them
`{:(4H_(2)O+cancel(6e^(-))+2MnO_(2)+8overset(Θ)OH),(6I^(Θ)rarr3I_(2)+cancel(6e^(-))),(ulbar(4H_(2)O+6I^(Θ)+2MnO_(4)^(Θ)rarr2MnO_(2)+8overset(Θ)OH)):}`
b. The balanced half reactions are:
Oxidation half equation:
`SO_(2)(g)+2H_(2)O(l)rarrHSO_(4)^(Θ)(aq)+3H^(o+)(aq)+2e^(-)`....(i)
Reduction half equation:
`MnO_(4)^(Θ)(aq)+8H^(o+)(aq)+5e^(-) rarr Mn^(2+)(aq)+4H_(2)O(l)` ...(ii)
Multiply equation (i) by `5` and add it to equation (ii) by `2` and add, we have,
`2MnO_(4)^(Θ)(aq)+5SO_(2)(g)+2H_(2)O(l)+H^(o+)rarr2Mn^(2+)(aq)+5HSO_(4)^(Θ)(aq)`
c. Oxidation half equation: ltbr `Fe^(2+)(aq)rarrFe^(3+)(aq)+e^(-)`...(i)
Reduction half equation: `H_(2)O_(2)(aq)+2Fe^(+2)(aq)+2H^(o+)(aq)+2e^(-)rarr2H_(2)O(l)`.....(ii)
Multiply equation (i) by `2` and add it to equation (ii), we have,
`H_(2)O_(2)(aq)+2Fe^(+2)(aq)+2H^(o+)(aq)rarr2Fe^(3+)(aq)+2H_(2)O(l)`
d. Oxidation half equation:
`SO_(2)(g)+2H_(2)O(l)rarrSO_(4)^(2-)(aq)+4H^(o+)(aq)+2e^(-)`.....(i)
Reduction half equation
`Cr_(2)O_(7)^(2-)(aq)+14H^(o+)(aq)+6e^(-)rarr2Cr^(3+)(aq)+7H_(2)O(l)`..(ii)
Multiply equation (i) and add it to equation (ii), we have,
`Cr_(2)O_(7)^(2-)(aq)+3SO_(2)(q)+2H^(o+)rarr2Cr^(3+)(aq)+3SO_(4)^(2-)(aq)+H_(2)O(l)`