Stability of MnO4- in basic vs acidic conditions
The obviously feasible and spontaneous disproportionation reaction can be explained by considering the standard electrode potentials (standard reduction potential) involved (quoted as half–cell reductions, as is the convention).


$\displaystyle \ce{MnO4- (aq) + e- -&gt; MnO4^2- (aq)} \quad\quad (E^0 = +0.56~\mathrm{V})$
$\displaystyle \ce{MnO4^2- (aq) + 4H+(aq) + 2e-  -&gt; MnO2(s) + 2H2O(l)} \quad\quad  (E^0 = +1.70~\mathrm{V}, \text{in acid solution})$


2. has the more positive potential, so this will be the reduction half–cell reaction.

1. has the less positive potential, so this will be (reversed) the oxidation half–cell reaction.

$$E^0_\text{reaction} = E^0_\text{reduction} – E^0_\text{oxidation} = (+1.70~\mathrm{V}) - (+0.56~\mathrm{V}) = +1.14~\mathrm{V} \gg 0~\mathrm{V}$$

Therefore, the reaction is very feasible.

Incidentally: Given the two half–cell reactions, you get the complete balanced equation by adding 2. plus 2 x 1. reversed.

The greater stability of the manganate(VI) ion in alkali can also be explained by considering the electrode potential for 2. in an alkaline media.


$\displaystyle \ce{MnO4^2- (aq) + 2H2O(l) + 2e-  -&gt; MnO2(s) + 4OH- (aq)}\quad \quad  (E^0 = +0.59~\mathrm{V}, \text{in alkaline solution})$


so, re–calculating gives

$$E^0_\text{reaction} = E^0_\text{reduction} - E^0_\text{oxidation} = (+0.59~\mathrm{V}) - (+0.56~\mathrm{V}) = +0.03~\mathrm{V} &gt; 0~\mathrm{V}$$

Therefore just feasible! but on the basis of an equilibrium argument, here, the far lower $E^0_\text{reaction}$, suggests the $\ce{MnO4^2-}$ ion is far more likely to exist, i.e. more stable, in a very high pH solution and in practice it is stable for a few hours in alkali.