To solve this problem, we will use the Nernst equation to calculate the emf of the cell under different conditions.$\\$The Nernst equation for a cell is given by:$\\ E = E^\circ - \frac{RT}{nF} \ln Q \\$Where:$\\$• $E$ is the cell potential (emf).$\\$• $E^\circ$ is the standard cell potential.$\\$• $R$ is the universal gas constant.$\\$• $T$ is the temperature in Kelvin.$\\$• $n$ is the number of moles of electrons transferred in the reaction.$\\$• $F$ is Faraday's constant.$\\$• $Q$ is the reaction quotient.$\\$For the given cell reaction:$\\$Zn$+$Cu$^{2+} \rightarrow$Zn$^{2+} +$Cu$\\$The standard cell potential $E^\circ$ is calculated as:$\newline E^\circ = E^\circ_{cathode} - E^\circ_{anode} \\$Given that $\frac{RT}{F} = 0.059,$ we can write the Nernst equation as:$\\ E = E^\circ - \frac{0.059}{n} \log Q \\$For the first condition:$\\$• Concentration of ZnSO$_4 = 0.1$ M$\\$• Concentration of CuSO$_4 = 1.0$ M$\\$The reaction quotient $Q_1$ is:$\\ Q_1 = \frac{[{Zn}^{2+}]}{[{Cu}^{2+}]} = \frac{0.1}{1.0} = 0.1 \\ E_1 = E^\circ - \frac{0.059}{2} \log 0.1 \\$For the second condition:$\\$• Concentration of ZnSO$_4 = 1.0$ M$\\$• Concentration of CuSO$_4 = 0.01$ M$\\$The reaction quotient $Q_2$ is:$\\ Q_2 = \frac{[{Zn}^{2+}]}{[{Cu}^{2+}]} = \frac{1.0}{0.01} = 100 \\ E_2 = E^\circ - \frac{0.059}{2} \log 100 \\$Now, calculate the difference between $E_1$ and $E_2: \\ E_1 = E^\circ - \frac{0.059}{2} \log 0.1 \\ E_2 = E^\circ - \frac{0.059}{2} \log 100 \\$Since $\log 0.1 = -1$ and $\log 100 = 2,$ we have:$\\ E_1 = E^\circ + \frac{0.059}{2} \cdot 1 \\ E_2 = E^\circ - \frac{0.059}{2} \cdot 2 \\ E_1 = E^\circ + 0.0295 \\ E_2 = E^\circ - 0.059 \\$Comparing $E_1$ and $E_2: \\ E_1 = E^\circ + 0.0295 \\ E_2 = E^\circ - 0.059 \\$Thus, $E_1 > E_2. \\$Therefore, the correct relationship is: Option 2: $E_1 > E_2.$