Equilibrium constant is \(2 \times 10^{37}\) at 298 K for \(Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)\). Predict the spontaneity of cell and calculate \(E^\circ_{cell}\). [log 2 = 0.3010]
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Spontaneous, 1.103 V
E°cell and the equilibrium constant are related by
\( \displaystyle E^\circ_{cell} = \frac{2.303RT}{nF}\log K_c = \frac{0.059}{n}\log K_c \) at 298 K
For the Daniell cell n = 2 and Kc = 2 × 1037:
\( \log K_c = \log 2 + 37 = 0.3010 + 37 = 37.301 \)
\( \displaystyle E^\circ_{cell} = \frac{0.0591}{2} \times 37.301 \approx 1.10\ \text{V} \)
E°cell is positive, so ΔG° = −nFE°cell is negative and the cell reaction is spontaneous. A very large Kc (> 1) points the same way.