ELECTROCHEMISTRY
19968
In the electrolysis of water, one Faraday of electrical energy would evolve
1 One mole of oxygen
2 One \(g\) atom of oxygen
3 \(8 \,g \) of oxygen
4 \(22.4\) litres of oxygen
Explanation:
(c)Hydrolysis of water : \(2{H_2}O\) \( \Leftrightarrow \) \(4{H^ + } + 4{e^ - } + {O_2}\)
\(4\,F \) charge will produce \(= 1\) mole \({O_2} = 32\;gm\;{O_2}\)
\(1\,F\) charge will produce \( = \frac{{32}}{4} = 8\;gm\;{O_2}\).