AIIMS2018Chemistry-Electrochemistry

AIIMS 2018 Chemistry Nernst Equation MCQ Question

Type: MCQ-numerical-Hard-Class 12

Cell equation:

A + 2B²⁺ → A²⁺ + 2B

A²⁺ + 2e⁻ → A E°=+0.34 V

And log₁₀K=15.6 at 300 K for cell reactions. Find E° for B⁺ + e⁻ → B. Given [2.303RTF=0.059]\left[\frac{2.303RT}{F} = 0.059\right] at 300 K300\text{ K}

A

0.80

B

1.26

C

−0.54

D

+0.94

Correct Answer

Option A

Detailed Explanation

To find the standard reduction potential EE^\circ for the reaction B++eBB^+ + e^- \rightarrow B, we can use the relationship between the equilibrium constant KK and the standard cell potential, given by the Nernst equation. The overall cell reaction can be expressed in terms of KK as E=0.059nlog10KE^\circ = \frac{0.059}{n} \log_{10} K. With log10K=15.6\log_{10} K = 15.6 and n=2n = 2 (since 2 electrons are transferred in the overall reaction), we calculate EE^\circ for B++eBB^+ + e^- \rightarrow B to be approximately +0.80 V, aligning with option A.

Options B, C, and D are incorrect because they do not match the calculated potential based on the provided equilibrium constant and the stoichiometry of the reactions involved. Specifically, option B (1.26 V) and option D (+0.94 V) suggest higher potentials than what is derived from the equilibrium constant, while option C (−0.54 V) incorrectly indicates a negative potential, which is inconsistent with the positive KK value.

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