Question #44765

A lead storage battery operates at $298 \text{ K}$ with the following overall discharge reaction:

$Pb(s) + PbO_2(s) + 4H^+(aq) + 2SO_4^{2-}(aq) \rightarrow 2PbSO_4(s) + 2H_2O(l)$

The standard cell potential is $E^\circ_{cell} = 2.05 \text{ V}$. Initially, the bulk electrolyte concentrations are $[H^+] = 1.0 \text{ M}$ and $[SO_4^{2-}] = 1.0 \text{ M}$. During a rapid discharge at a constant current of $I = 2.0 \text{ A}$, severe concentration polarization occurs, causing the local surface concentrations at the electrodes to drop to $[H^+]_{surf} = 0.1 \text{ M}$ and $[SO_4^{2-}]_{surf} = 0.01 \text{ M}$.

Given the activation overpotential $\eta_{act} = 0.10 \text{ V}$ and the internal resistance of the cell $R_{int} = 0.05 \ \Omega$, calculate the actual terminal voltage ($V_{cell}$) of the battery during this discharge.

(Assume $\frac{2.303 RT}{F} = 0.06 \text{ V}$ at $298 \text{ K}$)