Q.Using $E^{\circ}_{cell} = 1.10\ \text{V}$ for the Daniell cell ($n=2$) at $298\ \text{K}$, calculate the equilibrium constant $K_c$ for the reaction $\text{Zn}(s) + \text{Cu}^{2+}(aq) \rightleftharpoons \text{Zn}^{2+}(aq) + \text{Cu}(s)$.

[!TLDR] $\log K = \dfrac{nE^{\circ}_{cell}}{0.059} = \dfrac{2\times1.10}{0.059} \approx 37.3$. [!ANSWER] $K_c \approx 1.9\times10^{37}$ — an astronomicall …

At equilibrium, $\Delta G^{\circ} = -RT\ln K = -nFE^{\circ}_{cell}$, which combines (converting $\ln$ to $\log_{10}$ and evaluating $2.303RT/F$ at $298\ \text{K}$) into the working relation $\log K = \dfrac{nE^{\circ}_{cell}}{0.059}$. Substituting $n=2$ and $E^{\circ}_{cell}=1.10\ \text{V}$: $\log K = \dfrac{2\times1.10}{0.059} = \dfrac{2.20}{0.059} \approx 37.29$. Taking the antilog, $K_c = 10^{37.29} \approx 1.9\times10^{37}$. A value this large means the equilibrium lies overwhelmingly on the product side — at equilibrium, essentially all the $\text{Cu}^{2+}$ originally presen …

Use $\log K = nE^{\circ}_{cell}/0.059$ (from combining $\Delta G^{\circ}=-nFE^{\circ}_{cell}$ with $\Delta G^{\c …

Reporting only the exponent ("$K \approx 37$") instead of $10^{37}$ — a very common slip that changes the …