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Chemistry-Nernst Equation

NEET Chemistry Nernst Equation MCQ Question

Type: MCQ-numerical-Easy-Class 12

Using the Nernst equation, calculate the cell potential (E) at 298 K for the reaction Zn(s)+Cu2+(aq)→Zn2+(aq)+Cu(s)Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s), given Ecell∘=1.10 VE^\circ_{cell} = 1.10 \text{ V}, [Cu2+]=0.01 M[Cu^{2+}] = 0.01 \text{ M}, and [Zn2+]=1.0 M[Zn^{2+}] = 1.0 \text{ M}. Use R=8.314 J mol−1 K−1R = 8.314 \text{ J mol}^{-1} \text{ K}^{-1} and F=96500 C mol−1F = 96500 \text{ C mol}^{-1}.

A

09 V

B

05 V

C

15 V

D

0.95 V

Correct Answer

Option A

Detailed Explanation

Using the Nernst equation, E=Ecell∘−RTnFln⁡QE = E^\circ_{cell} - \frac{RT}{nF} \ln Q. Substituting the values, n=2n = 2, Q=[Zn2+][Cu2+]=1.00.01=100Q = \frac{[Zn^{2+}]}{[Cu^{2+}]} = \frac{1.0}{0.01} = 100, we get E=1.10−8.314×2982×96500ln⁡(100)≈1.09 VE = 1.10 - \frac{8.314 \times 298}{2 \times 96500} \ln(100) \approx 1.09 \text{ V}.

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