To find the emf of the cell, we will use the Nernst equation.$\\$The Nernst equation is given by:$\\ E_{cell} = E^\circ_{cell} - \frac{0.059}{n} \log Q \\$where:$\\$• $E^\circ_{cell} = 1.05$ V (standard cell potential)$\\$• $n = 2$ (number of electrons transferred)$\\$• $Q$ is the reaction quotient$\\$The cell reaction is:$\\$Ni(s) + 2Ag$^+(0.001 \, \mathrm{M}) \rightarrow$Ni$^{2+} (0.001 \, \mathrm{M}) + 2Ag(s) \\$The reaction quotient $Q$ is given by:$\\ Q = \frac{[{Ni}^{2+}]}{[{Ag}^+]^2} \\$Substitute the concentrations:$\\ Q = \frac{0.001}{(0.001)^2} = \frac{0.001}{0.000001} = 1000 \\$Now, substitute the values into the Nernst equation:$\\ E_{cell} = 1.05 - \frac{0.059}{2} \log 1000 \\$Calculate $\log 1000: \\ \log 1000 = 3 \\$Substitute back into the equation:$\\ E_{cell} = 1.05 - \frac{0.059}{2} \times 3 \\ E_{cell} = 1.05 - 0.0885 \\ E_{cell} = 0.9615 \, \mathrm{V} \\$Therefore, the emf of the cell is $0.9615 \, \mathrm{V}. \\$This corresponds to Option 3.