$E^{\circ }(\mathrm{Ag}{}^{+}/\mathrm{Ag})=0.80\mathrm{V}$ and $E^{\circ }$ for $[\mathrm{Ag}(\mathrm{NH}{}_3){}_2]{}^{+}+\mathrm{e}{}^{-}\to \mathrm{Ag}+2\mathrm{NH}{}_3$ is $0.37\mathrm{V}$ at 298 K. The value of ${\log }_{10}{K}_f$ (one decimal) for the formation of $[\mathrm{Ag}(\mathrm{NH}{}_3){}_2]{}^{+}$ from $\mathrm{Ag}{}^{+}$ and $\mathrm{NH}{}_3$ is. Take $\frac{2.303RT}{F}=0.0591\mathrm{V}$.

Subtract the two half reactions: $\mathrm{Ag}{}^{+}+\mathrm{e}{}^{-}\to \mathrm{Ag}$ ($0.80\mathrm{V}$) minus $[\mathrm{Ag}(\mathrm{NH}{}_3){}_2]{}^{+}+\mathrm{e}{}^{-}\to \mathrm{Ag}+2\mathrm{NH}{}_3$ ($0.37\mathrm{V}$) gives $\mathrm{Ag}{}^{+}+2\mathrm{NH}{}_3\to [\mathrm{Ag}(\mathrm{NH}{}_3){}_2]{}^{+}$, whose equilibrium constant is $K_f$. Treating this as a cell with $\mathrm{Ag}{}^{+}/\mathrm{Ag}$ as cathode and the ammine couple as anode, $E^{\circ }=0.80-0.37=0.43\mathrm{V}$, $n=1$. $\log K_f=0.43/0.0591=7.28\approx 7.3$. Complexation lowers the reduction potential because free $\mathrm{Ag}{}^{+}$ is removed.