At $298 \text{ K}$ the equilibrium constant for the reaction, $\text{M}(\text{s})+2\text{Ag}^+(\text{aq}) \rightleftharpoons \text{M}^{2+}(\text{aq})+2\text{Ag}(\text{s})$ is $10^{15}$. What is the $E_{\text{cell}}^{\circ}$ (in $\text{V}$) for this reaction? ($\frac{2.303 \text{ RT}}{F} = 0.06 \text{ V}$)
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The relation $E_{\text{cell}}^{\circ} = \frac{2.303 \text{ RT}}{nF} \log K$ is a key equation in electrochemistry. In many exam problems, the factor $\frac{2.303 \text{ RT}}{F}$ is approximated as $0.0592 \text{ V}$ or $0.06 \text{ V}$ at $298 \text{ K}$ for simpler calculation.