For a Ag-Zn button cell, net reaction is 
$$\begin{gathered} \mathrm{Zn}\left(\mathrm{s}\right)+{\mathrm{Ag}}_2\mathrm{O}\left(\mathrm{s}\right)⟶\mathrm{ZnO}\left(\mathrm{s}\right)+2\mathrm{Ag}\left(\mathrm{s}\right) \\ {\mathrm{\Delta G}}_{\mathrm{f}}^0\left({\mathrm{Ag}}_2\mathrm{O}\right)=-11.21{\mathrm{kJmol}}^{-1} \\ {\mathrm{\Delta G}}_{\mathrm{f}}^0\left(\mathrm{ZnO}\right)=-318.3\mathrm{kJ}{\mathrm{mol}}^{-1} \\ \mathrm{Then}, {\mathrm{E}}_{\mathrm{cell} } \mathrm{of}\mathrm{the}\mathrm{button}\mathrm{cell}\mathrm{is} \end{gathered}$$

The equations taking palce in button cell are 

$$\begin{gathered} Anode:\hspace{0.17em}Zn \hspace{0.17em}+\hspace{0.17em}\frac{1}{2}{O}_2\to ZnO\hspace{0.17em}+\hspace{0.17em}2e^{- };∆G=-318.3 kJ mo{l}^{-1} \\ Cathode \hspace{0.17em}A{g}_{2}O +\hspace{0.17em}2e^{-}\to 2Ag +\hspace{0.17em}\frac{1}{2}{O}_2; ∆G= + 11.21 kJ\hspace{0.17em}mo{l}^{-1} \\ -------------------------------- \\ The overall reaction:\hspace{0.17em}Zn +\hspace{0.17em}A{g}_{2}O\hspace{0.17em} \to ZnO\hspace{0.17em}+\hspace{0.17em}2Ag n= 2 \\ -------------------------------- \end{gathered}$$

$$\begin{gathered} ∆G_{reaction}^o=∆G_{\left(ZnO\right)}^o-∆G_{\left(A{g}_{2}O\right)}^o \\ =-318.3 +\hspace{0.17em}11.21 \\ =-307 kJ \\ △G^o =- nF{E}_{cell}^o \\ {E}_{cell}^o=\frac{-△G^o}{nF}=-\frac{-307\hspace{0.17em}\times 1000 }{2\times 96500}=+1.591V \end{gathered}$$