Question #44861

Manganese(III) ions readily disproportionate in aqueous acidic solutions. However, the presence of fluoride ions strongly stabilizes the +3 oxidation state via the formation of the $\ce{[MnF6]^{3-}}$ complex. Given the following standard reduction potentials at $298\text{ K}$:

$E^\circ(\ce{Mn^{3+}/Mn^{2+}}) = 1.50\text{ V}$ $E^\circ(\ce{MnO2/Mn^{3+}}) = 0.95\text{ V}$

The overall formation constant ($\beta_6$) for $\ce{[MnF6]^{3-}}$ is $10^{15}$. A solution is prepared containing $0.1\text{ M } \ce{[MnF6]^{3-}}$, $0.01\text{ M } \ce{Mn^{2+}}$, and $0.1\text{ M } \ce{F^-}$ in a buffer of $\text{pH} = 0$ with excess solid $\ce{MnO2}$.

Calculate the cell potential ($E_{\text{cell}}$) for the disproportionation of $\ce{[MnF6]^{3-}}$ under these conditions. (Assume $\frac{2.303RT}{F} = 0.06\text{ V}$ at $298\text{ K}$ and that $\ce{Mn^{2+}}$ does not form significant fluoride complexes).