Find the emf of the cell in which the following reaction takes...

Given reaction: $Ni ( s ) + 2{Ag}^{+} ( 0 . 001M ) \to {Ni}^{2 +} ( 0 . 001M ) + 2Ag ( s )$

Applying Nernst equation we have:
${E}_{\left(cell\right)} = {E}_{\left(cell\right)}^{0} - \dfrac{2 . 303RT}{nF}log\dfrac{\left[{Ni}^{2 +}\right] [ Ag ]}{{\left[{Ag}^{+}\right]}^{2} [ Ni ]}$
Active mass of solid is taken to be unity so \left[Ni\left(s\right)\right] = \left[Ag\left(s\right)\right] = 1$<br />${E}_{\left(cell\right)} = {E}_{\left(cell\right)}^{0} - \dfrac{0.059}{n}log\dfrac{\left[N{i}^{2 +}\right]}{{\left[A{g}^{+}\right]}^{2}}$<br />$= 1.05 - \dfrac{0.0591}{2}log\dfrac{\left(0.001\right)}{{\left(0.001\right)}^{2}}$<br />$= 1.05 - 0.0295\log (1\times {10}^{3} )$<br />$= 1.05 - 0.0295 \times 3$<br />$= 0.9615V$</p><p>Therefore, the emf of the cell is$0 . 9615V$. </p><p>Note: Question is modified little bit for academic accuracy.$E{°}_{cell} = 1 . 05VinplaceofE{°}_{cell} = 10 . 5V$ as given in NEET 2022 paper.