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(A) From the Nernst equation:$E = E^{\circ} - \frac{0.059}{n} \log \left( \frac{[Zn^{2+}]}{[Cu^{2+}]} \right)$Here,$n = 2$ and $E^{\circ}_{cell} = E^{

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$C_1 = 0.1 \ M, C_2 = 0.01 \ M$

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(A) From the Nernst equation:$E = E^{\circ} - \frac{0.059}{n} \log \left( \frac{[Zn^{2+}]}{[Cu^{2+}]} \right)$Here,$n = 2$ and $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = 0.34 \ V - (-0.76 \ V) = 1.1 \ V$.Substituting the values:$E = 1.1 - \frac{0.059}{2} \log \left( \frac{C_2}{C_1} \right)$To make $E$ maximum,the term $\frac{0.059}{2} \log \left( \frac{C_2}{C_1} \right)$ must be as small as possible (most negative).This occurs when $\log \left( \frac{C_2}{C_1} \right)$ is the most negative value.For option $A$: $\log \left( \frac{0.01}{0.1} \right) = \log(0.1) = -1$.For option $B$: $\log \left( \frac{0.1}{0.01} \right) = \log(10) = 1$.For option $C$: $\log \left( \frac{0.2}{0.1} \right) = \log(2) \approx 0.301$.For option $D$: $\log \left( \frac{0.1}{0.2} \right) = \log(0.5) \approx -0.301$.The value is minimum for option $A$,hence $E$ is maximum.

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