For a cell involving one electron, $E_{cell}^{\circ }=0.59\text{V}$ at $298\text{K}$. The equilibrium constant for the cell reaction is: (Given that $\frac{2.303RT}{F}=0.059\text{V}$ at $T=298\text{K}$)

Solution & Explanation

\textbf{Answer:} (C) $1.0\times 10^{10}$ \textbf{Solution:} At equilibrium, $\Delta G^{\circ }=-nF{E}_{cell}^{\circ }=-RT\ln K$, which rearranges to $\log K=\frac{n{E}_{cell}^{\circ }}{0.059}$. With $n=1$ and $E_{cell}^{\circ }=0.59\text{V}$: $\log K=\frac{1\times 0.59}{0.059}=10$, so $K=1.0\times 10^{10}$. Mapped: source numeric key "3" $=1.0\times 10^{10}$ → option C.