Explain the difference in stability of permanganate ions in acidic/ alkaline solutions?

Question

The question is such:

Use the data to comment on the stability of $\ce{MnO4-}$ in acidic and basic solution. Suggest a reason for this difference.

$$\begin{align} &\text{acidic solution:}&\ \ce{MnO4- + 4H+ + 3e- &-> MnO2 + 2H2O}\qquad &E^\circ&=+1.70\ \mathrm V\\[6pt] &\text{basic solution}:&\ \ce{MnO4- + 2H2O + 3e- &-> MnO2 + 4OH-}\qquad &E^\circ&=+0.59\ \mathrm V \end{align}$$

I know that the permanganate ion is more stable in alkaline solution than in acidic solution, as the more positive electrode potential in acidic solution means more negative Gibbs free energy, therefore the reaction is more feasible in acidic solution, therefore less stable...

However, is there a chemical reason for this?

Answer 1

Actually, yes.

As we know, $\ce{H2O = H+ +OH-}$ . Chemical equation can be threated as algebraic equations as long as the final result has chemical meaning (i.e. no negative coeffecients ... unless you understand their meaning), thus we can add $\ce{4H+}$ to both sides of the second equation. This gives us equation

$\ce{MnO4- +4H+ + 2H2O +3 e- \rightarrow MnO2 + 4H+ +4OH-}$

Obviously, we can collapse four pairs of $\ce{H+}$ and $\ce{OH-}$ into $\ce{4H2O}$

$\ce{MnO4- +4H+ + 2H2O +3 e- \rightarrow MnO2 + 4H2O}$

and now we can drop two $\ce{H2O}$ from both sides

$\ce{MnO4- +4H+ +3 e- \rightarrow MnO2 + 2H2O}$

which is exactly as your first equation. So, in fact it is exactly the same reaction. However, WTF they are written differently and the corresponding potentials are different? This is because standard potentials are implied to be measured at concentration of all ions and molecules (except water) in corresponding half-reaction 1 mol/l. However, what happend when one concentration deviates from 1 mol/l ? The corresponding potential deviates from its value as well, in accordance with Le Chatelier's principle.

When we set up concentration of $\ce{OH-}$ at 1 mol/l, we set $\ce{H+}$ at roughly $10^{-14}$, since $\ce{[H+][OH^{-}] \approx 10^{-14}}$ (water autoprotolysis constant). According to Le Chatelier's principle the equilibrium should move to the left, which is observable as a change in electrode potential.

The quantitative expression of this kind of dependence is known as Nernst equation.