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

$$\begin{gathered} \begin{array}{l}{\mathrm{H}}_2\mathrm{O}\to 2{\mathrm{H}}^{+}+1/2{\mathrm{O}}_2+2{\mathrm{e}}^{-}\\ 1\mathrm{mol}2\mathrm{mol}\end{array} \\ (\mathrm{In} \mathrm{this} \mathrm{case}, \mathrm{the} \mathrm{oxygen}\text{'}\mathrm{s} \mathrm{oxidation} \mathrm{number} \mathrm{changes} \mathrm{from} -2 \mathrm{in} {\mathrm{H}}_2\mathrm{O} \\ \mathrm{to} 0 \mathrm{in} {\mathrm{O}}_2, \mathrm{resulting} \mathrm{in} \mathrm{the} \mathrm{loss} \mathrm{of} 2 \mathrm{moles} \mathrm{of} \mathrm{electrons} \mathrm{during} \mathrm{the} \\ \mathrm{oxidation} \mathrm{of} 1 \mathrm{mol} \mathrm{of} \mathrm{water}.) \mathrm{Q}=\mathrm{nF} \mathrm{provides} \mathrm{the} \mathrm{charge} \mathrm{Q} \mathrm{on} \mathrm{n} \mathrm{mol} \\ \mathrm{of} \mathrm{electrons}. \mathrm{Consequently}, \mathrm{the} \mathrm{amount} \mathrm{of} \mathrm{electricity} \mathrm{needed} \mathrm{to} \mathrm{oxidize} \\ 1 \mathrm{mol} \mathrm{of} {\mathrm{H}}_2\mathrm{O} \text{to }{\mathrm{O}}_2, \\ \mathrm{Q}=2\mathrm{mol}\times 96500{\text{C mol}}^{-1}=193000\mathrm{C}=1.93\times {10}^5\mathrm{C} \end{gathered}$$

Therefore, the correct option is C