Question #44769

During the rapid exothermic discharge of a secondary lead-acid battery, the local temperature at the cathode increases from $300\text{ K}$ to $350\text{ K}$.

The cathode reduction half-reaction is: $\ce{PbO2(s) + 4H+(aq) + 2e- <=> Pb^{2+}(aq) + 2H2O(l)}$

The generated $\ce{Pb^{2+}}$ immediately establishes equilibrium with the electrolyte: $\ce{PbSO4(s) <=> Pb^{2+}(aq) + SO4^{2-}(aq)}$

Given:

1. $[\ce{H+}] = 1\text{ M}$ and $[\ce{SO4^{2-}}] = 1\text{ M}$ (assume concentrations remain constant).
2. $K_{sp}(\ce{PbSO4})$ at $300\text{ K} = 10^{-8}$
3. $K_{sp}(\ce{PbSO4})$ at $350\text{ K} = 10^{-5}$
4. $\frac{2.303 RT}{F} = 0.06\text{ V}$ at $300\text{ K}$.

Assuming the standard reduction potential ($E^\circ$) for the $\ce{PbO2/Pb^{2+}}$ couple is independent of temperature, calculate the change in the reduction potential of the cathode ($\Delta E_{\text{cathode}}$) due to this temperature increase.