Question

In the electrolysis of water,

a) Name the gas collected at the anode and cathode

b) Why is the volume of gas collected at one electrode double than the other?

c) What would happen if dilute ${\mathrm{H}}_2{\mathrm{SO}}_4$ is not added to water?

Answer 1

a.The gas formed at anode and cathode:

• The process of using electricity to split water into hydrogen and oxygen is known as electrolysis.
  $\underset{\mathrm{Water}}{2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}\to \underset{\mathrm{Hydrogen}}{2{\mathrm{H}}_2\left(\mathrm{g}\right)}+\underset{\mathrm{Oxygen}}{{\mathrm{O}}_2\left(\mathrm{g}\right)}$
• Hydrogen gas released at the cathode
• Oxygen gas is released at anode

b.The volume of gas collected:

• Here two moles of water undergo electrolysis and produce two moles of hydrogen and one mole of oxygen gas.
• Because the water molecule contains two hydrogens and one oxygen in its molecular formula.
• Therefore after electrolysis of water volume of one gas collected at one electrode is double (hydrogen gas) of the gas collected at the anode (oxygen gas).

c. If dilute ${\mathbf{H}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}$ is not added:

• Water is a bad conductor of electricity.
• So, when a direct current is passed through the water it won’t conduct.
• Therefore to make the electrolysis happen, a few drops of sulphuric acid $\left({\mathrm{H}}_2{\mathrm{SO}}_4\right)$ to the water in the electrolysis of water.