Question

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

• A. $9.65\times 10^{4}C$
• B. $1.93\times 10^{4}C$
• C. $1.93\times 10^{5}C$
• D. $19.3\times 10^{5}C$

Answer 1

Answer: (C)

The electrode reaction for 1 mole $H_{2}O$ is

$H_{2}O\to H_2+\frac{1}{2}{O}_2,O^{2-}\to \frac{1}{2}{O}_2+2e^{-}$

$2H^{+}+2e^{-}\to H_2$

or $H_{2}O\to 2H^{+}+\frac{1}{2}{O}_2+2e^{-}$

Electricity required for oxidation of 1 mole of $H_{2}O=2F,1F=96500C$

$\therefore$ Total charge $=2\times 96500=1.93\times 10^{5}C$