The cell reaction for a Daniell cell is: $Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)$

According to the Nernst equation at $298\text{ K}$: $E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{2} \log \frac{[Zn^{2+}]}{[Cu^{2+}]}$

For the first case ($E_1$): $[Zn^{2+}] = 0.01\text{ M} = 10^{-2}\text{ M}$ and $[Cu^{2+}] = 1.0\text{ M}$ $E_1 = E^{\circ}_{cell} - \frac{0.0591}{2} \log \left(\frac{10^{-2}}{1}\right) = E^{\circ}_{cell} - \frac{0.0591}{2} (-2) = E^{\circ}_{cell} + 0.0591\text{ V}$

For the second case ($E_2$): $[Zn^{2+}] = 1.0\text{ M}$ and $[Cu^{2+}] = 0.01\text{ M} = 10^{-2}\text{ M}$ $E_2 = E^{\circ}_{cell} - \frac{0.0591}{2} \log \left(\frac{1}{10^{-2}}\right) = E^{\circ}_{cell} - \frac{0.0591}{2} (2) = E^{\circ}_{cell} - 0.0591\text{ V}$

Comparing $E_1$ and $E_2$, we get $E_1 > E_2$.