Chemistry - Electrochemistry Question with Solution | TestHub

$\frac{{\displaystyle \stackrel{2H^{+}+2O{H}^{-}\rightleftharpoons 2H_{2}O \Delta G_1°=-RT\ln {\left(\frac{1}{{10}^{-14}}\right)}^2}{\frac{1}{2}{O}_2+H_{2}O+2e^{-}\to 2O{H}^{-} \Delta G_2°=-2F\times 0.4 }}}{{\displaystyle 2H^{+}+\frac{1}{2}{O}_2+2e^{-}⟶H_{2}O \Delta G_3°=-2FE° }}$

$⟹E^{\circ }=1.23V$.