Question

A group of students carried out electrolysis of acidified water in the laboratory. By the end of the experiment two gases were collected in the test tubes at both the electrodes.
(a) Name the gases collected at cathode and the anode respectively.
(b) The gas collected in one test tube is double the volume of gas collected in the other. Name the gas. Give reason for your answer.
(c) Write a balanced chemical equation to represent the electrolysis of water

Answer 1

Acidified water is the water which contains a small amount of acid dissolved in it. The electrolysis is the process of chemical decomposition produced by the passage of electric current through a solution which contains ions (electrolyte).

Complete step by step solution:
- As we know, electrolysis of acidified water is a way of splitting up or decomposition of the water using electrical energy. The process of electrolysis involves a flow of electrons in the external wires and electrodes and a flow of ions in the electrolyte.
(a) During electrolysis of acidified water the gases collected at cathode and the anode are hydrogen and oxygen gases respectively. The corresponding reaction is given below
$At\text{ }cathode:4{{H}_{2}}O+4{{e}^{-}}\to 2{{H}_{2}}+4O{{H}^{-}}$
$At\text{ }anode:2{{H}_{2}}O\to {{O}_{2}}+4{{H}^{+}}+4{{e}^{-}}$
(b) On electrolysis of acidified water the volume of hydrogen liberated at STP condition is 22.4L and the volume of oxygen liberated is 11.2 L. Thus the gas which is having double the volume of other gas is hydrogen. The reason is that, at cathode the ${{H}^{+}}$ ion takes two electrons to convert itself into hydrogen gas. Thus two moles of ${{H}^{+}}$ will give one mole ${{H}_{2}}$.
$2{{H}^{+}}_{\left( aq \right)}+2{{e}^{-}}\to {{H}_{2\left( g \right)}}$
In anode the $O{{H}^{-}}$ ion releases two electrons to convert into water and oxygen. Two mole of $O{{H}^{-}}$ gives 0.5 mole of oxygen. That is why the volume of one gas collected at one electrode is double of another.
(c) The balanced chemical equation to represent the electrolysis of water can be given as follows
$2{{H}_{2}}{{O}_{\left( l \right)}}\to 2{{H}_{2\left( g \right)}}+{{O}_{2\left( g \right)}}$

Note: Keep in mind that the gases which are formed at the electrodes can be identified by the following methods. The hydrogen gas which is evolved at the cathode will give pop sound and the oxygen which is formed at the anode rekindles a glowing splinter