According to the Nernst equation for the Daniell cell: $E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{2.303RT}{nF} \log \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]}$ For the first case ($E_1$), $[\text{Zn}^{2+}] = 0.01 \text{ M}$ and $[\text{Cu}^{2+}] = 1.0 \text{ M}$, $n=2$. $E_1 = E^\circ_{\text{cell}} - \frac{0.059}{2} \log \left(\frac{0.01}{1.0}\right) = E^\circ_{\text{cell}} - \frac{0.059}{2} (-2) = E^\circ_{\text{cell}} + 0.059 \text{ V}$ For the second case ($E_2$), $[\text{Zn}^{2+}] = 1.0 \text{ M}$ and $[\text{Cu}^{2+}] = 0.01 \text{ M}$. $E_2 = E^\circ_{\text{cell}} - \frac{0.059}{2} \log \left(\frac{1.0}{0.01}\right) = E^\circ_{\text{cell}} - \frac{0.059}{2} (2) = E^\circ_{\text{cell}} - 0.059 \text{ V}$ Comparing the two equations, it is clear that $E_1 > E_2$.