Question #44502

An electrochemical cell is constructed using Platinum electrodes in an aqueous buffer solution at $298\text{ K}$ and $\text{pH} = 4.0$. The solution contains $2.0\text{ M } \ce{CH3COOK}$ and $10^{-3}\text{ M } \ce{Mn(CH3COO)2}$. The applied voltage is gradually increased from $0\text{ V}$.

Given the following standard reduction potentials at $298\text{ K}$: $E^\circ(\ce{C2H6(g)}, \ce{CO2(g)} / \ce{CH3COO^-}) = +2.15\text{ V}$ $E^\circ(\ce{O2(g)} / \ce{H2O(l)}) = +1.23\text{ V}$ $E^\circ(\ce{MnO2(s)} / \ce{Mn^{2+}}, \ce{H^+}) = +1.23\text{ V}$

Assume the overpotential for Kolbe electrolysis on Pt is $0.10\text{ V}$, for $\ce{O2}$ evolution is $0.60\text{ V}$, and for $\ce{MnO2}$ deposition is negligible. Identify the first major product formed at the anode and the minimum anodic potential (vs SHE) required for its formation. (Take $\frac{2.303 RT}{F} = 0.06\text{ V}$ and assume partial pressures of all gases to be $1\text{ bar}$)