Question

For the cell reaction, \[ \text{Cu(s)} + 2\text{Ag}^+(aq) \rightarrow \text{Cu}^{2+}(aq) + 2\text{Ag(s)}, \quad E^\circ_{cell} = 0.46\,V \] The equilibrium constant of the reaction is:

• A. \(3.92 \times 10^{12}\)
• B. \(3.92 \times 10^{15}\)
• C. \(8.92 \times 10^{17}\)
• D. \(8.92 \times 10^{10}\)

Hint:

Higher \(E^\circ\) $\Rightarrow$ larger \(K\).
Use \( \log K = \frac{nE^\circ}{0.0591} \) directly at \(298\,K\) for fast solving.

Answer 1

The Correct Option is B

Concept: The relationship between Gibbs free energy, cell potential, and equilibrium constant is: \[ \Delta G^\circ = -nFE^\circ \quad \text{and} \quad \Delta G^\circ = -RT \ln K \] Equating: \[ \ln K = \frac{nFE^\circ}{RT} \] At \(298\,K\), this simplifies to: \[ \log K = \frac{nE^\circ}{0.0591} \]

Step 1: Number of electrons transferred: \[ n = 2 \]

Step 2: Calculate \(\log K\): \[ \log K = \frac{2 \times 0.46}{0.0591} \approx \frac{0.92}{0.0591} \approx 15.57 \]

Step 3: Calculate \(K\): \[ K = 10^{15.57} \approx 3.92 \times 10^{15} \] Final: \[ {3.92 \times 10^{15}} \]