Question:
The following reaction takes place in a cell at 298 K: $2M^{3+}(aq) + 2I^{-}(aq) \rightarrow 2M^{2+}(aq) + I_2(s)$. What is the value of $\log K_c$ for this reaction? (Given: $E_{cell}^\circ = 0.235 V, F = 96500 C mol^{-1}, R = 8.3 J mol^{-1} K^{-1}$)
The following reaction takes place in a cell at 298 K: $2M^{3+}(aq) + 2I^{-}(aq) \rightarrow 2M^{2+}(aq) + I_2(s)$. What is the value of $\log K_c$ for this reaction? (Given: $E_{cell}^\circ = 0.235 V, F = 96500 C mol^{-1}, R = 8.3 J mol^{-1} K^{-1}$)
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$\log K_c = \frac{n E_{cell}^\circ}{0.0591}$ at 298 K.
Updated On: Jun 10, 2026
- 7.04
- 7.96
- 9.04
- 6.55
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The Correct Option is B
Solution and Explanation
Step 1: Concept
$E_{cell}^\circ = \frac{0.0591}{n} \log K_c$ or $\Delta G^\circ = -nFE_{cell}^\circ = -RT \ln K_c$. For $n=2$ electrons transferred: $2 \times 96500 \times 0.235 = 8.3 \times 298 \times \ln K_c$.
Step 2: Analysis
$45355 = 2473.4 \times 2.303 \log K_c$. $45355 = 5696.2 \log K_c \implies \log K_c = 45355 / 5696.2 \approx 7.96$.
Step 3: Conclusion
$\log K_c \approx 7.96$.
Final Answer: (B)
$E_{cell}^\circ = \frac{0.0591}{n} \log K_c$ or $\Delta G^\circ = -nFE_{cell}^\circ = -RT \ln K_c$. For $n=2$ electrons transferred: $2 \times 96500 \times 0.235 = 8.3 \times 298 \times \ln K_c$.
Step 2: Analysis
$45355 = 2473.4 \times 2.303 \log K_c$. $45355 = 5696.2 \log K_c \implies \log K_c = 45355 / 5696.2 \approx 7.96$.
Step 3: Conclusion
$\log K_c \approx 7.96$.
Final Answer: (B)
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