The zinc/silver oxide cell is used in electric watches. The reaction is as following,

${\mathrm{Zn}}^{2+}+2{\mathrm{e}}^{-}\to \mathrm{Zn};\mathrm{E}°=-0.760 \mathrm{V}$

${\mathrm{Ag}}_2\mathrm{O}+{\mathrm{H}}_2\mathrm{O}+2{\mathrm{e}}^{-}\to 2\mathrm{Ag}+2{\mathrm{OH}}^{-};\mathrm{E}°=0.344 \mathrm{V}$

$\mathrm{If} \mathrm{F} \mathrm{is} 96,500{\mathrm{Cmol}}^{-1},\mathrm{\Delta G}° \mathrm{of} \mathrm{the} \mathrm{cell} \mathrm{will} \mathrm{be}$

${\mathrm{\Delta G}}^0=-{\mathrm{nFE}}^0$

From the reaction, n = 2

$\begin{array}{l}{\mathrm{\Delta G}}^0=-2\left(96500\right)\left({\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}\right).........\left(1\right)\\ {\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}={\mathrm{E}}_{\mathrm{cathode}}^{\mathrm{o}}-{\mathrm{E}}_{\mathrm{anode}}^{\mathrm{o}}\\ {\mathrm{Zn}}^{+2}+2{\mathrm{e}}^{-}\stackrel{ }{\to }\mathrm{Zn},{\mathrm{E}}^0=-0.76\mathrm{V}\\ {\mathrm{Ag}}_2\mathrm{O}+{\mathrm{H}}_2\mathrm{O}+2\overline{\mathrm{e}}\stackrel{ }{\to }2\mathrm{Ag}+2{\mathrm{OH}}^{-},{\mathrm{E}}^0=+0.344\mathrm{V}\end{array}$

SRP value of ${\mathrm{Ag}}_2\mathrm{O}$  half cell is high 
$\therefore$ it acts like cathode
Thus …   $\mathrm{Anode}$ ;$\mathit{ }Zn\to {\mathrm{Zn}}^{+2} +2\stackrel{-}{\mathrm{e}};$$\mathrm{Cathode}; {\mathrm{Ag}}_2\mathrm{O} +{\mathrm{H}}_2\mathrm{O} +2\stackrel{-}{\mathrm{e}}\underset{ }{\to }2\mathrm{Ag} +2{\mathrm{OH}}^{-}$

$\begin{array}{l}\mathrm{Zn}+{\mathrm{Ag}}_2\mathrm{O}+{\mathrm{H}}_2\mathrm{O}\stackrel{ }{\to }{\mathrm{Zn}}^{+2}+2\mathrm{Ag}+2{\mathrm{OH}}^{-}\\ {\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}=+0.344-\left(-0.76\right)\end{array}$

= + 1.104 V
Substituting  the value in ---(1)
${\mathrm{\Delta G}}^0=-2\left(96.5\right)\left(1.104\right)\mathrm{kJ}$

= –213.072 kJ