If $E^{\circ }\left({\mathrm{Ag}}^{+}\mid \mathrm{Ag}\right)=0.80\mathrm{V}$ and $E^{\circ }\left({\mathrm{CN}}^{-},\mathrm{Ag}(\mathrm{CN}{)}_{ȷ}^{-}\mid \mathrm{Ag}\right)=-0.38\mathrm{V}$  then the standard equilibrium constant of the reaction $\mathrm{Ag}(\mathrm{CN}{)}_2^{-}\rightleftharpoons {\mathrm{Ag}}^{+}+2{\mathrm{CN}}^{-}\text{at }298\mathrm{K}$ will be

$\mathrm{Ag}(\mathrm{CN}{)}_2^{-}+{\mathrm{e}}^{-}\to \mathrm{Ag}+2{\mathrm{CN}}^{-} E^{\circ }=-0.38\mathrm{V}$

$\text{ Add }\frac{\mathrm{Ag}\to {\mathrm{Ag}}^{+}+{\mathrm{e}}^{-}}{\mathrm{Ag}(\mathrm{CN}{)}_2^{-}\to {\mathrm{Ag}}^{+}+2{\mathrm{CN}}^{-}}\begin{array}{r}{E}^{\circ }=-0.8\mathrm{V}\\ E_{\text{cell }}^{\circ }=-1.18\mathrm{V}\end{array}$

Now $\mathrm{\Delta }{G}^{\circ }=-nF{E}_{\text{cell }}^{\circ }=-RT\ln K^{\circ }=-2.303RT\log K^{\circ }$

Hence $\log K^{\circ }=\frac{n{E}_{\text{cell }}^{\circ }}{(RT/F)}=\frac{\left(1\right)(-1.18\mathrm{V})}{\left(0.059\mathrm{V}\right)}=-20$ This gives $K^{\circ }=1.0\times {10}^{-20}$