The equilibrium constant for the reaction $M(s) + 2Ag^{+}(aq) \rightarrow M^{2+}(aq) + 2Ag(s)$ is $10^{15}$. What is the Gibbs energy change ($\Delta_r G^{\ominus}$ in KJ mol$^{-1}$) for this reaction?
($\frac{RT}{F} \times 2.303 = 0.06V$; F = 96500 C mol$^{-1}$)
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Remember: $\Delta G^\ominus = -nFE^\ominus_{cell}$.
Also $E^\ominus_{cell} = \frac{0.0591}{n} \log K_c$ (at 298K).
Combining these gives the same result.