NEET Chemistry Nernst Equation MCQ Question

Using the Nernst equation $E_{\text{cell}} = E^o_{\text{cell}} - \frac{RT}{nF} \ln Q$, where $Q = \frac{[Ni^{2+}]}{[Ag^+]^2}$, we calculate $E_{\text{cell}} = 1.05 - \frac{0.059}{2} \ln \frac{0.100}{(0.010)^2}$, resulting in a potential of 0.91 V.