Given the equilibrium constant KC of the reaction: {Cu(s) + 2Ag^+(aq) Cu^{2+}(aq) + 2Ag(s)} is 10 1015, calculate the…

Strategy: Use the Nernst-derived relation $log K = (n \cdot E^\circ) / 0.059$ at equilibrium. First, balan\ce the reaction to find the number of electrons transferred ($n$).

Step 1: Balan\ce reaction and find $n$ Reaction: $\ce{Cu(s) + 2Ag^+(aq) \rightarrow Cu^{2+}(aq) + 2Ag(s)}$

• Cu is oxidized: $\ce{Cu \rightarrow Cu^{2+} + 2e^-}$
• Ag is redu\ced: $\ce{2Ag^+ + 2e^- \rightarrow 2Ag}$ Total $n = 2$.

Step 2: Calculate $E^\circ$ $K = 10 \times 10^{15} = 10^{16}$. $log K = 16$ $16 = \frac{2 \times E^\circ}{0.059}$ $E^\circ = \frac{16 \times 0.059}{2} = 8 \times 0.059 = 0.472 \text{ V}$

Comparing with options, $0.4736 \text{ V}$ is the closest (often using slightly different constants like $0.0592$).

$\boxed{\text{Answer: (c) 0.4736 V}}$