8.21 :  The ${\mathrm{Mn}}^{3+}$ ion is unstable in solution and undergoes disproportionation to give  ion. Write a balanced ionic equation for the reaction.

The given reaction can be represented as:
${{\mathrm{Mn}}^{3+}}_{\left(\mathrm{aq}\right)}\to {{\mathrm{Mn}}^{2+}}_{\left(\mathrm{aq}\right)}+{{\mathrm{MnO}}_{2}}_{\left(\mathrm{g}\right)}+{{\mathrm{H}}^{+}}_{\left(\mathrm{aq}\right)}$
The oxidation half equation is:
$\stackrel{+3}{\mathrm{M}}{{n}^{3+}}_{\left(aq\right)}\to \stackrel{+4}{\mathrm{M}}{\mathrm{nO}}_{2\left(\mathrm{s}\right)}$
The oxidation number is balanced by adding one electron as:
$\stackrel{}{\mathrm{M}}{{n}^{3+}}_{\left(aq\right)}\to \stackrel{}{\mathrm{M}}{\mathrm{nO}}_{2\left(\mathrm{s}\right)}+{\mathrm{e}}^{-}$
The charge is balanced by adding 4${\mathrm{H}}^{+}$ ions as:
$\stackrel{}{\mathrm{M}}{{n}^{3+}}_{\left(aq\right)}\to \stackrel{}{\mathrm{M}}{\mathrm{nO}}_{2\left(\mathrm{s}\right)}+4{{\mathrm{H}}^{+}}_{\left(\mathrm{aq}\right)}+{\mathrm{e}}^{-}$
The O atoms and ${\mathrm{H}}^{+}$ ions are balanced by adding $2{\mathrm{H}}_{2}\mathrm{O}$ molecules as:
$\stackrel{}{\mathrm{M}}{{n}^{3+}}_{\left(aq\right)}+2{H}_{2}{O}_{\left(l\right)}\to \stackrel{}{\mathrm{M}}{\mathrm{nO}}_{2\left(\mathrm{s}\right)}+4{{\mathrm{H}}^{+}}_{\left(\mathrm{aq}\right)}+{\mathrm{e}}^{-}........\left(\mathrm{i}\right)$
The reduction half equation is:
$\stackrel{}{\mathrm{M}}{{n}^{3+}}_{\left(aq\right)}\to \stackrel{}{\mathrm{M}}{{\mathrm{n}}^{2+}}_{\left(aq\right)}$
The oxidation number is balanced by adding one electron as:
The balanced chemical equation can be obtained by adding equation (i) and (ii) as:
$2{{\mathrm{Mn}}^{3+}}_{\left(\mathrm{aq}\right)}+2{\mathrm{H}}_{2}{\mathrm{O}}_{\left(\mathrm{l}\right)}\to {\mathrm{MnO}}_{2\left(\mathrm{s}\right)}+2{\mathrm{Mn}}^{2+}\left(\mathrm{aq}\right)+4{{\mathrm{H}}^{+}}_{\left(\mathrm{aq}\right)}$