Reduction of ${\mathrm{MnO}}_4^{-}$ , in strongly basic solution produces

 Strong oxidant ${\mathrm{KMnO}}_4$ with a vivid violet colour is reduced to $M{n}^{2+}$ in extremely acidic solution. It is reduced to ${\mathrm{MnO}}_2$ in a neutral alkaline solution and to${\mathrm{MnO}}_4^{2-}$  in a strongly basic solution. Therefore, the strongly basic solution of ${\mathrm{MnO}}_4^{-}$ yields${\mathrm{MnO}}_4^{-}+{\mathrm{e}}^{-}⟶{\mathrm{MnO}}_4^{2-}$  after reduction.