For the given cell reaction: $Ni(s) + 2Ag^+(aq) \rightarrow Ni^{2+}(aq) + 2Ag(s)$ The number of electrons transferred, $n = 2$.

According to the Nernst equation at $298\text{ K}$: $E_{cell} = E^{\circ}_{cell} - \frac{0.059}{n} \log \frac{[Ni^{2+}]}{[Ag^+]^2}$

Substituting the given values: $E^{\circ}_{cell} = 1.05\text{ V}$ $[Ni^{2+}] = 0.001\text{ M} = 10^{-3}\text{ M}$ $[Ag^+] = 0.001\text{ M} = 10^{-3}\text{ M}$ $E_{cell} = 1.05 - \frac{0.059}{2} \log \frac{10^{-3}}{(10^{-3})^2}$ $E_{cell} = 1.05 - 0.0295 \times \log \left(\frac{10^{-3}}{10^{-6}}\right)$ $E_{cell} = 1.05 - 0.0295 \times \log (10^3)$ $E_{cell} = 1.05 - 0.0295 \times 3$ $E_{cell} = 1.05 - 0.0885$ $E_{cell} = 0.9615\text{ V}$