8.
​pls answer

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$$\begin{gathered} \mathrm{Step} 1: \mathrm{Divide} \mathrm{the} \mathrm{equation} \mathrm{into} \mathrm{two} \mathrm{half} \mathrm{cells} \\ \mathrm{Oxidation} : {{\mathrm{SO}}_2}^{2-} \to {{\mathrm{SO}}_4}^{2-} \\ \mathrm{Reduction} : {{\mathrm{MnO}}_4}^{-} \to {\mathrm{MnO}}_2 \\ \mathrm{Step} 2: \mathrm{Balance} \mathrm{all} \mathrm{atoms} \mathrm{other} \mathrm{than} \mathrm{H} \mathrm{and} \mathrm{O} \\ \mathrm{Oxidation} : {{\mathrm{SO}}_2}^{2-} \to {{\mathrm{SO}}_4}^{2-} \\ \mathrm{Reduction} : {{\mathrm{MnO}}_4}^{-} \to {\mathrm{MnO}}_2 \\ \mathrm{Step} 3: \mathrm{Balance} \mathrm{O} \mathrm{by} \mathrm{adding} 2{\mathrm{OH}}^{-} \mathrm{per} \mathrm{O} \mathrm{atom} \\ \mathrm{Oxidation} : {{\mathrm{SO}}_2}^{2-} +4{\mathrm{OH}}^{-}\to {{\mathrm{SO}}_4}^{2-} \\ \mathrm{Reduction} : {{\mathrm{MnO}}_4}^{-} \to {\mathrm{MnO}}_2 +4{\mathrm{OH}}^{-} \\ \mathrm{Step} 4: \mathrm{Balance} \mathrm{H} \mathrm{by} \mathrm{adding} {\mathrm{H}}_2\mathrm{O} \\ \mathrm{Oxidation} : {{\mathrm{SO}}_2}^{2-} +4{\mathrm{OH}}^{-}\to {{\mathrm{SO}}_4}^{2-}+2{\mathrm{H}}_2\mathrm{O} \\ \mathrm{Reduction} : {{\mathrm{MnO}}_4}^{-} +2{\mathrm{H}}_2\mathrm{O}\to {\mathrm{MnO}}_2 +4{\mathrm{OH}}^{-} \\ \mathrm{Step} 5: \mathrm{Balance} \mathrm{charge} \mathrm{by} \mathrm{adding} \mathrm{electrons} \\ \mathrm{Oxidation} : {{\mathrm{SO}}_2}^{2-} +4{\mathrm{OH}}^{-}\to {{\mathrm{SO}}_4}^{2-}+2{\mathrm{H}}_2\mathrm{O}+4{\mathrm{e}}^{-} \\ \mathrm{Reduction} : {{\mathrm{MnO}}_4}^{-} +2{\mathrm{H}}_2\mathrm{O}+3{\mathrm{e}}^{-}\to {\mathrm{MnO}}_2 +4{\mathrm{OH}}^{-} \\ \mathrm{Step} 6: \mathrm{Balance} \mathrm{electrons} \mathrm{on} \mathrm{both} \mathrm{side}, \mathrm{so} \mathrm{multiply} \mathrm{oxidation} \mathrm{half} \mathrm{by} 3 \mathrm{and} \mathrm{reduction} \mathrm{half} \mathrm{by} 4 \mathrm{and} \mathrm{add} \mathrm{both} \\ 3{{\mathrm{SO}}_2}^{2-} +4{{\mathrm{MnO}}_4}^{-} +2{\mathrm{H}}_2\mathrm{O}\to 3{{\mathrm{SO}}_4}^{2-}+4{\mathrm{MnO}}_2 +4{\mathrm{OH}}^{-} \end{gathered}$$
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