`Mn^(3+)` ions are unstable in solution and undergo disproportionation to give `Mn^(2+)`, `MnO_(2)` and `H^(+)` ions. What will be the balanced equation for the reaction ?

Explanation

The skeletal equation is:
`Mn^(3+)(aq)rarr Mn^(2+)(aq)+MnO_(2)(s)+H^(o+)(aq)`
Oxidation half equation:
`overset(+3)(Mn_(+3))(aq) rarr overset(+4)(MnO_(2)(s)`
Balance oxidation number by adding electrons,
`Mn^(3+)(aq) rarr MnO_(2)(s)+e^(-)`
Balance charge by adding `H^(o+)` ions,
`Mn^(3+)(aq)rarr MnO_(2)(s)+4H^(o+)(aq)+e^(-)`
Balance `O` atoms by adding `H_(2)O:`
`Mn^(3+)+2H_(2)O(l) rarr MnO_(2)(s)+4H^(o+)(aq)+e^(-) ...(i)`
Reduction half equation:
`overset(+3)(Mn^(+3)) rarr overset(+2)(Mn^(2+))`
Balance oxidation number by adding electron:
`Mn^(3+) (aq)+e^(-) rarr Mn^(2+)(aq)`
Adding equation reaction is
`2Mn^(3+)(aq)+2H_(2)O(l) rarr MnO_(2)(s)+Mn^(2+)(aq)+4H^(o+)(aq)`