Question #44766

An experimental all-iron secondary cell utilizes the \ce{Fe^{3+}/Fe^{2+}} and \ce{Fe^{2+}/Fe} redox couples. To suppress self-discharge at the anode, the bulk electrolyte is buffered at pH = 4.0. At this pH, any \ce{Fe^{3+}} that crosses over to the anode precipitates as \ce{Fe(OH)3(s)}.

During a high-rate discharge, a localized parasitic reaction drops the pH at the anode surface to 1.0. The \ce{Fe(OH)3(s)} dissolves, reaching a local steady-state \ce{Fe^{3+}} concentration of $10^{-2}$ M.

The self-corrosion reaction at the anode is: $\ce{Fe(s) + 2Fe^{3+}(aq) -> 3Fe^{2+}(aq)}$

Given: $E^\circ(\ce{Fe^{3+}/Fe^{2+}}) = +0.77$ V $E^\circ(\ce{Fe^{2+}/Fe}) = -0.44$ V $K_{sp}(\ce{Fe(OH)3}) = 10^{-38}$ The local $[\ce{Fe^{2+}}]$ at the anode remains constant at $10^{-1}$ M. Use $\frac{2.303 RT}{F} = 0.06$ V and $F = 96500$ C/mol.

Calculate the change in the Gibbs free energy of corrosion, $\Delta \Delta G_{corr} = \Delta G_{corr, pH=1} - \Delta G_{corr, pH=4}$, due to this localized pH shift.