Why does Manganese (VI) disproportionate in acid conditions but not in basic conditions

August 3rd, 2007

Manganese (VI) is usually in the form of the MnO₄^2- ion. This ion disproportionates in acid solution but not in base. What is going on here?

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.

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-

This can be predicted by reference to the electrode potential of the half equations involved. 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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Acidic conditions:

3MnO₄(2-) + 4H+ –> MnO₂ + 2MnO₄(-) + 2H₂O

This disproportionation can be though of as two half equations:
1. MnO₄(2-) + 4H+ + 2e –> MnO₂ + 2H₂O ……….. Eº = +2.26V ( the Manganate ion is reduced)
2. MnO₄(2-) –> MnO₄(-) + 1e ……………………….Eº = +0.56V ( the Manganate ion is oxidised

Calculating Eº = E(reduced state) – E(oxidised state) = +2.26V – +0.56V = + 1.70V
This is a large positive value (remember that any Eº value greater than 0.3v means that 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 (reduction)
MnO₄(2-) –> MnO₄(-) + 1e…………………………Eº = +0.56V (oxidation)

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.