Step 1: Understanding the Question:
The question concerns the electrolysis of a salt solution and the selective discharge of ions at inert (Pt) electrodes.
Step 2: Detailed Explanation:
In an aqueous solution of \(Na_2SO_4\), we have the following ions: \(Na^+\), \(SO_4^{2-}\), and \(H_2O\) molecules.
At the Cathode (reduction): There is a competition between \(Na^+\) ions and \(H_2O\) molecules. The reduction potential of water (\(-0.83 V\)) is much higher than that of Sodium ions (\(-2.71 V\)). Therefore, water is reduced preferentially:
\(2H_2O + 2e^- \rightarrow H_2(g) + 2OH^-\).
Thus, Hydrogen gas (\(H_2\)) is liberated at the cathode (Y).
At the Anode (oxidation): There is competition between \(SO_4^{2-}\) ions and \(H_2O\). Sulfate ions are extremely difficult to oxidize. Water molecules are oxidized more easily to yield Oxygen gas:
\(2H_2O \rightarrow O_2(g) + 4H^+ + 4e^-\).
Thus, Oxygen gas (\(O_2\)) is liberated at the anode (X).
Effectively, the electrolysis of aqueous sodium sulfate is simply the electrolysis of water, as the salt only serves to provide conductivity.
Step 3: Final Answer:
The gas liberated at the anode (X) is \(O_2\) and at the cathode (Y) is \(H_2\).