NEET Chemistry Nernst Equation MCQ Question with Solution

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

A

09 V

B

05 V

C

15 V

D

0.95 V

Correct Answer

Option A

Detailed Explanation

Using the Nernst equation, $E = E^\circ_{cell} - \frac{RT}{nF} \ln Q$ . Substituting the values, $n = 2$ , $Q = \frac{[Zn^{2+}]}{[Cu^{2+}]} = \frac{1.0}{0.01} = 100$ , we get $E = 1.10 - \frac{8.314 \times 298}{2 \times 96500} \ln(100) \approx 1.09 \text{ V}$ .

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NEET Chemistry Nernst Equation MCQ Question

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