How to balance the disproportionation reaction of manganate anion in basic conditions?

Question

$$\ce{MnO4^2- (aq) -> MnO2(s) + MnO4^-(aq)}$$

How would I balance this in basic conditions? I know we need to separate this into two half reactions but how do we do this if there is only one reactant?

Answer 1

You are referring to the disproportionation reaction of $\ce{Mn(VI)}$ to $\ce{Mn(IV)}$ and $\ce{Mn(VII)}$, which is:

$\ce{MnO4^{2-}(aq) -> MnO2 (s) + MnO4^{-}(aq)}$

Reduction: $\ce{MnO4^{2-} + 2e^{-} + 2H2O -> MnO2 + 4OH-}$

Oxidation: $\ce{2(MnO4^{2-} -> MnO4^{-} + 1e^{-})}$ = $\ce{2MnO4^{2-} -> 2MnO4^{-} + 2e^{-} }$

Combine both, and the final balanced equation is:

$\ce{3MnO4^{2-} + 2H2O -> MnO2 + 2MnO4^{-} + 4OH-}$