Dissociation constant for $Ag{\left(N{H}_3\right)}_2^{+}$ into $A{g}^{+}$ and $N{H}_3$ is $6\times {10}^{-14}$.Calculate Eº for the half reaction,

$Ag{\left(N{H}_3\right)}_2^{+}+e^{-}\to Ag+2N{H}_3$

Given,   $A{g}^{+}+e^{-}\to Ag$  has  $E^o=0.799V$

$[\frac{2.303RT}{F}=0.059,\log 2=0.3,\log 3=0.48]$

Use  $Ag\to A{g}^{+}+e;E_{OP}^0=-0.799V$

$\frac{Ag{\left(N{H}_3\right)}_2^{+}+e\to Ag+2N{H}_3}{Ag{\left(N{H}_3\right)}_2^{+}\rightleftharpoons A{g}^{+}+2N{H}_3}$

And  $E_{cell}=E_{cell}^0+\frac{0.059}{1}{\log }_{10}\frac{\left[Ag{\left(N{H}_3\right)}_2^{+}\right]}{\left[A{g}^{+}\right]{\left[N{H}_3\right]}^2}$

and  $E_{cell}=0$

and  $E_{cell}^0=E_{O{P}_{Ag/A{g}^{+}}}^0+E_{R{P}_{Ag{\left(N{H}_3\right)}_2^{+}/Ag}}^0$

$$\begin{gathered} \begin{array}{l}{E}_{cell}^0=\frac{0.059}{1}{\log }_{10}{K}_c\\ =\frac{0.059}{1}{\log }_{10}6\times {10}^{-14}=-0.780,V \end{array} \\ {E}_{A{g}_{Ag{\left(N{H}_3\right)}_2^{+}/Ag}}^0=-0.780+0.799=+0.019V \end{gathered}$$