Given below are the half-cell reactions, 
         $\begin{array}{l}{\mathrm{Mn}}^{2+}+2{\mathrm{e}}^{-}⟶\mathrm{Mn};{\mathrm{E}}^{\circ }=-1.18\mathrm{eV}\\ 2\left({\mathrm{Mn}}^{3+}+{\mathrm{e}}^{-}⟶{\mathrm{Mn}}^{2+}\right);{\mathrm{E}}^{\circ }=+1.51\mathrm{eV}\end{array}$
The ${\mathrm{E}}^{\circ }\text{ for }3{\mathrm{Mn}}^{2+}⟶\mathrm{Mn}+2{\mathrm{Mn}}^{3+}$ will be 

If ${\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}$ is positive, reaction is spontaneous, otherwise it is non-spontaneous.
                 $\begin{array}{l}{\mathrm{Mn}}^{2+}+2{\mathrm{e}}^{-}⟶\mathrm{Mn};{\mathrm{E}}^{\circ }=-1.18\mathrm{V}\\ 2{\mathrm{Mn}}^{2+}⟶2{\mathrm{Mn}}^{3+}+2{\mathrm{e}}^{-};{\mathrm{E}}^{\circ }=-1.51\mathrm{V}\\ 3{\mathrm{Mn}}^{2+}⟶\mathrm{Mn}+2{\mathrm{Mn}}^{+3}\\ {\mathrm{E}}^{\circ }={\mathrm{E}}_{\mathrm{oxi}}^{\circ }+{\mathrm{E}}_{\mathrm{red}}^{\circ }=-1.51+(-1.18)=-2.69\mathrm{V}\end{array}$
Negative ${\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}$ of the reaction indicates that reaction is non-spontaneous.