Question #44129

In an acidic aqueous medium, the comproportionation of $\ce{MnO4^-}$ and $\ce{Mn^{2+}}$ yields $\ce{Mn^{3+}}$, which is highly unstable and undergoes rapid disproportionation into $\ce{Mn^{2+}}$ and $\ce{MnO2(s)}$.

Given standard reduction potentials at 298 K: $E^\circ(\ce{Mn^{3+}/Mn^{2+}}) = 1.51\text{ V}$ $E^\circ(\ce{MnO2(s)/Mn^{3+}}) = 0.95\text{ V}$

To stabilize the $\ce{Mn^{3+}}$ oxidation state, a neutral ligand $L$ is added, forming octahedral complexes $\ce{MnL6^{3+}}$ and $\ce{MnL6^{2+}}$ with overall formation constants $\beta_{III}$ and $\beta_{II}$, respectively. The disproportionation reaction in the presence of $L$ is:

$2\ce{MnL6^{3+}} + 2\ce{H2O(l)} \rightleftharpoons \ce{MnL6^{2+}} + \ce{MnO2(s)} + 4\ce{H^+} + 6L$

Assume standard conditions for all species ($[\ce{MnL6^{3+}}] = [\ce{MnL6^{2+}}] = [L] = 1\text{ M}$) except for $\ce{H^+}$. If the solution is buffered at $\text{pH} = 2.0$, what is the minimum value of $\log_{10}\left(\frac{\beta_{III}^2}{\beta_{II}}\right)$ required to thermodynamically prevent the disproportionation of $\ce{MnL6^{3+}}$?

(Take $\frac{2.303 RT}{F} = 0.056\text{ V}$ at 298 K)