Given below are the half-cell reactions 

$\begin{array}{l}M{n}^{2+}+2e^{-}⟶Mn;E^{\circ }=-1.18\mathrm{V}\\ M{n}^{2+}⟶M{n}^{3+}+e^{\mathit{-}};E^{\circ }=-1.51\mathrm{V}\end{array}$

The ${\mathrm{E}}^{\mathrm{o}}$ For the reaction $3M{n}^{2+}⟶M{n}^{\mathit{0}}+2M{n}^{3+}$,and possibility of forward reaction are, respectively

Given,
$\begin{array}{l}{\mathrm{Mn}}^{+2}+2{\mathrm{e}}^{-}\to {\mathrm{Mn}}_{,}{\mathrm{E}}^{\circ }=-1.18\mathrm{V}\dots ...\left(1\right)\\ {\mathrm{Mn}}^{+2}\to {\mathrm{Mn}}^{+3}+{\mathrm{e}}^{-},{\mathrm{E}}^{\circ }=-1.5\mathrm{V}\dots ...\left(2\right)\\ \left(1\right)\Rightarrow {\mathrm{Mn}}^{+2}+2{\mathrm{e}}^{-}\to \mathrm{Mn},{\mathrm{E}}_1^{\circ }=-1.18\mathrm{V}\\ \left(2\right)\Rightarrow {\mathrm{Mn}}^{+2}\to {\mathrm{Mn}}^{+3}+{\mathrm{e}}^{-},{\mathrm{E}}_2^{\circ }=-1.51\mathrm{V}\\ 3{\mathrm{Mn}}^{+2}\to \mathrm{Mn}+{\mathrm{Mn}}^{+3}\dots \dots ..\left(3\right)\\ {\mathrm{E}}_3^{\circ }={\mathrm{E}}_1^{\circ }+{\mathrm{E}}_2^{\circ }\\ {\mathrm{E}}_3=-1.18+(-1.51)\\ =-2.69\mathrm{V}\end{array}$
If ${\mathrm{E}}_{\mathrm{cell}}^{\circ }=-\mathrm{ve}$ then $∆$G = +ve
Hence, the reaction is not possible.