Question

In the electrolysis of acidified water:
(a) Name the gas collected at (i) anode, and (ii) cathode
(b) Why is the volume of gas collected at one electrode double than that at the other electrode?
​(c) What would happen if dilute sulphuric acid is not added to water?

Answer 1

(a) Electrolysis of acidified water:
$$\begin{gathered} \mathrm{At}\mathrm{cathode}:2{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}\to {\mathrm{H}}_2 \\ \mathrm{At}\mathrm{anode}:{\mathrm{OH}}^{-}\to \mathrm{OH}+{\mathrm{e}}^{-} \\ 4\mathrm{OH}\to 2{\mathrm{H}}_2\mathrm{O}+{\mathrm{O}}_2 \end{gathered}$$
Hence, the gas produced at cathode is hydrogen and the gas produced at anode is oxygen.

(b) A water molecule contains two atoms of hydrogen and one atom of oxygen. Therefore, during the electrolysis of water, the ratio of the amount of hydrogen gas and oxygen gas produced is 2:1. Hence, we can say that the volume of gas collected at one electrode is double than that at the other electrode

(c) As electrolysis of water is a slow process, to increase the rate of electrolysis we add dilute sulphuric acid.
Dilute sulphuric acid increases the rate of breakdown of ions.