Question:
The correct value for log\(_{10}\) \(K_c\) (where \(K_c\) is the equilibrium constant) for the equilibrium reaction at 298 K \[ \text{2Fe}^{3+} + 3\text{I}^- \rightleftharpoons 2\text{Fe}^{2+} + \text{I}_3^- \quad (\text{E}^\circ_{\text{cell}} = x \, \text{V}) \]
The correct value for log\(_{10}\) \(K_c\) (where \(K_c\) is the equilibrium constant) for the equilibrium reaction at 298 K \[ \text{2Fe}^{3+} + 3\text{I}^- \rightleftharpoons 2\text{Fe}^{2+} + \text{I}_3^- \quad (\text{E}^\circ_{\text{cell}} = x \, \text{V}) \]
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In the Nernst equation, the number of electrons transferred and the standard cell potential are key factors in determining the equilibrium constant.
Updated On: Mar 18, 2026
- \( \frac{0.059x}{3} \)
- \( \frac{0.029x}{3} \)
- \( \frac{2x}{0.059} \)
- \( \frac{0.059}{x} \)
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The Correct Option is C
Solution and Explanation
The Nernst equation is used to calculate the equilibrium constant \( K_c \) from the cell potential \( E^\circ_{\text{cell}} \):
\[
E^\circ_{\text{cell}} = \frac{0.059}{n} \log_{10} K_c
\]
Where:
- \( n \) is the number of moles of electrons transferred in the reaction (which is 2 for this reaction),
- \( E^\circ_{\text{cell}} \) is the standard cell potential,
- \( K_c \) is the equilibrium constant.
Rearranging the equation to solve for \( \log_{10} K_c \):
\[ \log_{10} K_c = \frac{E^\circ_{\text{cell}} \cdot n}{0.059} \] Substituting \( E^\circ_{\text{cell}} = x \) and \( n = 2 \), we get: \[ \log_{10} K_c = \frac{2x}{0.059} \] Thus, the correct answer is (3) \( \frac{2x}{0.059} \).
- \( E^\circ_{\text{cell}} \) is the standard cell potential,
- \( K_c \) is the equilibrium constant.
Rearranging the equation to solve for \( \log_{10} K_c \):
\[ \log_{10} K_c = \frac{E^\circ_{\text{cell}} \cdot n}{0.059} \] Substituting \( E^\circ_{\text{cell}} = x \) and \( n = 2 \), we get: \[ \log_{10} K_c = \frac{2x}{0.059} \] Thus, the correct answer is (3) \( \frac{2x}{0.059} \).
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