Section C: Give a short answer to the question given below.
 
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. H₂SO₄ is not added to water?
 
a. Gases liberated at anode and cathode:
Hydrogen gas (\(H_2\)) is liberated at the .
Oxygen gas (\(O_2\)) is liberated at the .
 
b. Volume of gas collected at electrodes:

Due to the stoichiometry of the electrolysis reaction, the volume of gas collected at one electrode (cathode) is twice that at the other electrode (anode).
 
\(2H_2​O\)(l) → \(2H_2​\)(g) + \(O_2\)​(g)
This equation shows that molecules of water produce two molecules of hydrogen gas and one molecule of oxygen gas. Thus, the volume of hydrogen gas (\(H_2\)) produced at the cathode is twice that of oxygen gas (\(O_2\)) produced at the anode.
 
c. Importance of adding dilute \(H_2SO_4\) to water:
 
If dilute sulphuric acid (\(H_2SO_4\)) is not added to water, the electrolysis process would be significantly less efficient. Pure water is a poor conductor of electricity because it has a very low concentration of ions. Adding dilute \(H_2SO_4\) increases the solution's  concentration, enhancing its electrical conductivity.
 
The presence of \(H^+\) and \(SO_4^{2-}\) ions from the sulphuric acid allows the electrical current to pass more quickly through the water, facilitating the electrolysis process. Without the acid, the resistance of the water would be high, and the gas production rate would be very slow or negligible