Calculate the emf of the following cell: 3 Ni ( s )+2 Ag^+(0.01 M ) → Ni^2+(0.1 M )+2 Ag ( s ) Given that E_cell^°=1.05 V, log 10=1

\(\displaystyle \mathrm{E}^{\circ}{ }_{\text {cell }}=1.05 \mathrm{~V}\) \[\begin{aligned} & \mathrm{E}_{\text {cell }}=1.05-\frac{0.059}{2} \log \frac{[\mathrm{Ni}]^{2+}}{[\mathrm{Ag}+]^{2}} \\ & =1.05-0.059 / 2 \log \frac{(0.1)}{(0.01)^{2}} \\ & =1.05-0.0825 \\ & =0.9615 \mathrm{~V}(\text { Deduct } 1 / 2 \text { mark for incorrect unit or no unit }) \end{aligned} \]