The `Mn^(3+)` ion is unstable in solution and undergoes disproportionation to give `Mn^(2+)` , `MnO_(2)` and `H^(+)` ion. Write a balanced ionic equation for the reaction.

Explanation

The skeletal equation is:
`Mn_(aq)^(3+) rarr Mn_(aq)^(2+) + MnO_(2_((s))) + H_((aq))^(+)`
Oxidation half equation:
`Mn_((aq))^(3+) rarrMnO_(2_((s)))`
Balance O.N. by adding electrons,
`Mn_((aq))^(3+) rarrMnO_(2((s))) + e^(-)`
Balance charge by adding `4H^(+)` ions
Balance O atoms by adding `2H_(2)O`
`Mn_((aq))^(3+)+2H_(2)OrarrMnO_(2((s)))+4H_((aq))^(+)+e^(-)`
Balance O atoms by adding `2H_(2)O`
`Mn_((aq))^(3+)+2H_(2)O_((l))rarrMnO_(2((s)))+4H_((aq))^(+)+e^(-)` .......................(1)
Reduction half equation:
`Mn^(3+)rarrMn^(2+)`
Balance O.N. by adding electrons:
` Mn_((aq))^(3+) + e^(-) rarr Mn_((aq))^(2+)` ..............(2)
Adding Equation (1) and (2), the balanced equation for the disproportionation reaction is
`2Mn_((aq))^(3+)+2H_(2)O_((l)) rarr MnO_(2_((s)))+Mn_((aq))^(2+)+4H_((aq))^(+)`