Derive the Nernst equation for the following galvanic cell: $Ni_{(s)}|Ni_{(aq)}^{2+}\parallel Ag_{(aq)}^{+}|Ag_{(s)}$

(N/A) The cell reaction is as follows:
Anode (Oxidation): $Ni_{(s)} \rightarrow Ni_{(aq)}^{2+} + 2e^{-}$
Cathode (Reduction): $2Ag_{(aq)}^{+} + 2e^{-} \rightarrow 2Ag_{(s)}$
Overall Redox reaction: $Ni_{(s)} + 2Ag_{(aq)}^{+} \rightarrow Ni_{(aq)}^{2+} + 2Ag_{(s)}$
Here,$n = 2$ (number of electrons transferred).
The Nernst equation is given by: $E_{cell} = E_{cell}^{\theta} - \frac{RT}{nF} \ln Q$
Substituting the values: $E_{cell} = E_{cell}^{\theta} - \frac{RT}{2F} \ln \frac{[Ni_{(aq)}^{2+}]}{[Ag_{(aq)}^{+}]^{2}}$
At $298 \ K$,converting to $\log_{10}$: $E_{cell} = E_{cell}^{\theta} - \frac{0.0591}{2} \log_{10} \frac{[Ni_{(aq)}^{2+}]}{[Ag_{(aq)}^{+}]^{2}}$