Question:
At 298 K, $E^{\circ}$ value of the cell involving $2 Fe^{3+}(aq) + 2 I^-(aq) \rightarrow 2 Fe^{2+}(aq) + I_2(s)$ is X V. The X (in V) and $\log K_c$ for the reaction are respectively: ($E^{\circ}_{Fe^{3+}/Fe^{2+}} = 0.77$ V, $E^{\circ}_{I_2/I^-} = 0.54$ V)
At 298 K, $E^{\circ}$ value of the cell involving $2 Fe^{3+}(aq) + 2 I^-(aq) \rightarrow 2 Fe^{2+}(aq) + I_2(s)$ is X V. The X (in V) and $\log K_c$ for the reaction are respectively: ($E^{\circ}_{Fe^{3+}/Fe^{2+}} = 0.77$ V, $E^{\circ}_{I_2/I^-} = 0.54$ V)
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$\log K_c = \frac{n E^{\circ}}{0.0591}$ at 298 K!
Updated On: Jun 6, 2026
- 0.23, 8.79
- -0.23, 9.79
- 0.23, 7.79
- -0.23, 6.79
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The Correct Option is C
Solution and Explanation
Step 1: Concept
$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$; $\log K_c = \frac{n E^{\circ}}{0.0591}$.
Step 2: Meaning
Cathode: $Fe^{3+} \rightarrow Fe^{2+}$, Anode: $2I^- \rightarrow I_2$.
Step 3: Analysis
$E^{\circ}_{cell} = 0.77 - 0.54 = 0.23$ V. $n = 2$. $\log K_c = \frac{2 \times 0.23}{0.0591} \approx 0.46 / 0.0591 \approx 7.79$.
Step 4: Conclusion
X = 0.23 V, $\log K_c = 7.79$.
Final Answer: (C)
$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$; $\log K_c = \frac{n E^{\circ}}{0.0591}$.
Step 2: Meaning
Cathode: $Fe^{3+} \rightarrow Fe^{2+}$, Anode: $2I^- \rightarrow I_2$.
Step 3: Analysis
$E^{\circ}_{cell} = 0.77 - 0.54 = 0.23$ V. $n = 2$. $\log K_c = \frac{2 \times 0.23}{0.0591} \approx 0.46 / 0.0591 \approx 7.79$.
Step 4: Conclusion
X = 0.23 V, $\log K_c = 7.79$.
Final Answer: (C)
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