The Nernst equation corresponding to the cell reaction given below, is

$ \text{Ni}(\text{s})+2\text{Ag}^{+}(\text{aq})\rightarrow\text{Ni}^{2+}(\text{aq})+2\text{Ag}(\text{s}) $

$E_{\text{cell}}=E_{\text{cell}}^{\circ}-\frac{RT}{2F}\ln\frac{[\text{Ni}^{2+}]}{[\text{Ag}^{+}]^2}$

The correct answer is Option (1) → $E_{\text{cell}}=E_{\text{cell}}^{\circ}-\frac{RT}{2F}\ln\frac{[\text{Ni}^{2+}]}{[\text{Ag}^{+}]^2}$

For any electrode reaction,

 $ \text{M}^{n+}(\text{aq})+n\text{e}^{-}\rightarrow\text{M}(\text{s}) $

According to Nernst equation,

The electrode potential at any concentration measured with respect to standard hydrogen electrode can be represented by:

 $ E_{(\text{M}^{n+}/\text{M})}=E_{(\text{M}^{n+}/\text{M})}^{\circ}-\frac{RT}{nF}\ln\frac{[\text{M}]}{[\text{M}^{n+}]} $

but concentration of solid M is taken as unity 

The cell reaction is:

$ \text{Ni}(\text{s})+2\text{Ag}^{+}(\text{aq})\rightarrow\text{Ni}^{2+}(\text{aq})+2\text{Ag}(\text{s}) $

The Nernst equation can be written as

$ E_{(\text{cell})}=E_{(\text{cell})}^{\circ}-\frac{RT}{2F}\ln\frac{[\text{Ni}^{2+}]}{[\text{Ag}^{+}]^2} $