Please note that in my previous posts I showed the disproportionation reactions producing Mn2+. The extent of the disproportionation depends of the severity of the conditions and although the equations previously posted are not incorrect, it is generally accepted that these reactions produce the Mn(IV) state in the form of MnO2 along with the Mn(VII) in the form of MnO4(-). This is corrected below:
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The half equation for oxidation of Mn2+ in alkaline conditions is:
Mn2+ + 4H2O --> MnO4(2-) + 8H+ + 4e
The half equation for oxidation of Mn2+ in acid conditions is:
Mn2+ + 4H2O --> MnO4(-) + 5e + 8H+
so your argument could be equally applied to both reactions.
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These are redox half-equations and so only apply in the presence of an oxidising agent - which one, doesn't matter providing that it's strong enough.
However, the Mn(VI) state is not stable under acid conditions because it CAN disproportionate (notice that this is NOT a half-equation - it is a reaction):
3MnO4(2-) + 4H+ --> MnO2 + 2MnO4(-) + 2H2O
The Mn(VI) state cannot do this under alkaline conditions, and the alkaline disproportionation reaction (theoretical) would be:
3MnO4(2-) + 2H2O --> MnO2 + 2MnO4(-) + 4OH-
This is now in direct competition with the reverse reaction (as all reactions are) which we know DOES occur, as the MnO4(-) ion is a powerful oxidising agent and can drive the reaction backwards.
However, what we are really comparing is the Gibbs free energy of the two disproportionation equations. In the case of acid conditions:
3MnO4(2-) + 4H+ --> MnO2 + 2MnO4(-) + 2H2O
The Gibbs Free energy makes the forward reaction feasible. In the case of alkaline conditions, however, Gibbs Free energy makes the backward reaction equally feasible:
3MnO4(2-) + 2H2O <-- MnO2 + 2MnO4(-) + 4OH-
As stated in a previous post, this can be predicted with electrode potentials. If you break down each disporportionation into two half equations and compare the redox potentials using E = E(red) - E(ox).
OK, let's do it!
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Acid conditions:
3MnO4(2-) + 4H+ --> MnO2 + 2MnO4(-) + 2H2O
This disproportionation can be though of as two half equations:
MnO4(2-) + 4H+ + 2e --> MnO2 + 2H2O ........... Eº = +2.26V
MnO4(2-) --> MnO4(-) + 1e ............................Eº = +0.56V
Calculating Eº = E(red) - E(ox) = +2.26V - +0.56V = + 1.70V
This is a large positive value (remember that any Eº value greater than 0.3v means tha the reaction is spontaneous as shown) so the forward reaction proceeds.
Alkaline conditions:
3MnO4(2-) + 2H2O --> MnO2 + 2MnO4(-) + 4OH-
This disproportionation can be though of as two half equations:
MnO4(2-) + 2H2O + 2e --> MnO2 + 4OH- ........... Eº = +0.67V
MnO4(2-) --> MnO4(-) + 1e..............................Eº = +0.56V
Calculating Eº = E(red) - E(ox) = 0.67 - 0.56 = 0.11V
Although this is a positive value, it is very small (under 0.3V) indicating that an equilibrium will be established (under standard conditions) and that the reaction will proceed in the forward direction by only an immeasurable amount, and as the equation is in equilibrium, by Le Chatelier, addition of base will drive the equilibrium in the reverse direction.