Balance the following redox reactions by ion-electron method :
`MnO_(4)^(-)(aq)+SO_(2)(g)to Mn^(2+)(aq)+HSO_(4)(aq)` (in acidic solution )

Explanation

`MnO_4^(-)(aq)+I^(-)(aq) rarr MnO_2(s)+I_2` (in basic medium)
Step-1 : Indication of O.N. of each atom
`underset(+7)(Mn)underset(2)(O_4^(-))+underset(-1)(I^(-))rarrunderset(+4)(Mn)underset(-2)(O_2(s))+underset(0)(I_2(s))`
Thus Mn in `MnO_4^(-)` and I in `I^(-)` changed their oxidation numbers.
Step - 2: Writing the oxidation and reduction half reactions
`MnO_4^(-) rarr MnO_2` Reduction half
`I^(-) rarr I_2` Oxidation half
Step-3: Addition of electrons to make up the difference in O.N.
`MnO_4^(-) +3e^(-) rarr MnO_2`
`I^(-) rarr I_2+e^(-)`
Step - 4: Balance o atoms by adding `H_2O` molecules to this side deficient in .O. atoms.
`MnO_4^(-)+3e^(-)+2H_2O rarr MnO_2+2OH^(-)`
Step - 5 : Balance H atoms by adding OH ions to the side deficient in H atoms.
Since there are no, H atoms in oxidation half reactions no need to balance.
Step -.6 : Equalise the atoms in oxidation half reaction on both sides.
`2I^(-) rarr I_2+2e^(-)`
Step - 7: Equalise the electrons lost and gained by multiply the reduction half reaction with 2 and oxidation half reaction with 3