In the electrolysis of water

(a) Name the gases liberated at anode and cathode.

(b) Why is it that the volume of gas collected on one electrode is two times that on the other electrode?

(c) What would happen if dil. $H_2SO_4$ is not added to water?

(a) During the electrolysis of water:

• At the Anode (positive electrode): Oxygen gas ($O_2$) is liberated.
• At the Cathode (negative electrode): Hydrogen gas ($H_2$) is liberated.

(b) The balanced chemical equation for the electrolysis of water is:

$$2\text{H}_2\text{O}(l) \xrightarrow{\text{Electricity}} 2\text{H}_2(\text{g}) + \text{O}_2(\text{g})$$

According to stoichiometry, 2 molecules of water decompose to produce 2 molecules of hydrogen gas and 1 molecule of oxygen gas. Since the volume of a gas is directly proportional to the number of moles/molecules (at constant temperature and pressure), the volume of hydrogen gas collected at the cathode is exactly twice the volume of oxygen gas collected at the anode.

(c) Pure water is a very poor conductor of electricity because it does not dissociate significantly into ions. If dilute sulphuric acid ($H_2SO_4$) is not added, the electrical conductivity of water will be extremely low, and the electrolysis process will occur extremely slowly or not at all. Adding acid provides free ions ($H^+$ and $SO_4^{2-}$), making the water highly conducting.