$ \mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^0-\frac{0.059}{2} \log \frac{\mathrm{P}_{\mathrm{H}_2}}{\left[\mathrm{H}^{+}\right]^2} $
$ =0-0.059 \times 3=-0.177 \text { volts. }=-17.7 \times 10^{-2} \mathrm{~V} .$
$Zn ^{2+}+2 e ^{-} \rightarrow Zn ; E ^{\circ}=-0.760 \,V$
$Ag _{2} O + H _{2} O +2 e ^{-} \rightarrow 2 Ag +2 OH ^{-} ; E ^{\circ}=0.344 \,V$
જો $F$ $96,500 C mol ^{-1}$ હોય, તો કોષનો $\Delta G ^{\circ}$ શોધો. ($kJ mol ^{-1}$ માં)
(આપેલ છે: $\mathrm{E}_{\mathrm{Sn}^{2+} / \mathrm{Sn}}^{0}=-0.14 \mathrm{\;V}$ $\left.\mathrm{E}_{\mathrm{Pb}^{+2}/{\mathrm{Pb}}}^{0}=-0.13 \;\mathrm{V}, \frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.06\right)$