AIIMS 2019 Chemistry Nernst Equation MCQ Question

To calculate the emf of the cell, we use the Nernst equation: $E = E^\circ - \frac{RT}{nF} \ln \frac{[M^{2+}]_{anode}}{[M^{2+}]_{cathode}}$. Given $E^\circ_{cell} = 4 \, \text{V}$, $[M^{2+}]_{anode} = 0.0001 \, \text{M}$, and $[M^{2+}]_{cathode} = 0.01 \, \text{M}$, we find $E = 4 - 0.06 \ln \frac{0.0001}{0.01} = 4 - 0.06 \ln 0.01 = 4 - 0.06 \times (-4.605) \approx 3.94 \, \text{V}$, confirming option A as correct.

Options B, C, and D are incorrect as they do not accurately reflect the calculated emf based on the Nernst equation, either by miscalculating the logarithmic term or the overall voltage adjustment.