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.

(N/A) The given reaction can be represented as:
$Mn_{(aq)}^{3+} \to Mn_{(aq)}^{2+} + MnO_{2_{(s)}} + H_{(aq)}^{+}$
The oxidation half-reaction is:
$Mn_{(aq)}^{3+} \to MnO_{2_{(s)}}$
Balancing $O$ atoms by adding $2H_2O$ and $H$ atoms by adding $4H^{+}$:
$Mn_{(aq)}^{3+} + 2H_2O_{(l)} \to MnO_{2_{(s)}} + 4H_{(aq)}^{+}$
Balancing charge by adding $e^{-}$:
$Mn_{(aq)}^{3+} + 2H_2O_{(l)} \to MnO_{2_{(s)}} + 4H_{(aq)}^{+} + e^{-} \dots (i)$
The reduction half-reaction is:
$Mn_{(aq)}^{3+} + e^{-} \to Mn_{(aq)}^{2+} \dots (ii)$
Adding equations $(i)$ and $(ii)$ gives the balanced ionic equation:
$2Mn_{(aq)}^{3+} + 2H_2O_{(l)} \to Mn_{(aq)}^{2+} + MnO_{2_{(s)}} + 4H_{(aq)}^{+}$