How many Coulombs of electricity are required for the oxidation of one mole of water to dioxygen?

$$\begin{gathered} {\mathrm{H}}_2\mathrm{O}\to 2{\mathrm{H}}^{+\hspace{0.17em}}+\hspace{0.17em}\mathrm{O}\hspace{0.17em}+\hspace{0.17em}2{\mathrm{e}}^{-} \\ \mathrm{n}-\mathrm{factor} \mathrm{for} \mathrm{Oxygen} = 2 \end{gathered}$$

Therefore, 2 moles of electricity is required for the oxidation of 1 mole of water.
 Hence, charge required =$2\times 96500\mathrm{c} = 1.93\times {10}^5\mathrm{c}$