Question

Electrolysis of an aqueous solution of $Na_{2}SO_{4}$ between Pt electrodes liberate a gas X at anode and gas Y at cathode. X and Y respectively are}

• A. $H_{2}, O_{2}$
• B. $O_{2}, H_{2}$
• C. $SO_{2}, H_{2}$
• D. $H_{2}, SO_{2}$

Hint:

P-A-O: Positive Anode Oxidation. In aqueous salt solutions, water often "wins" and gives $O_{2}$ at the anode.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
In the electrolysis of aqueous $Na_2SO_4$, we must consider the discharge potential of the ions ($Na^+, SO_4^{2-}$) versus water ($H_2O$). Platinum (Pt) electrodes are inert and do not participate in the reaction.
Step 2: Key Formula or Approach:
Compare reduction/oxidation potentials. Water is oxidized more easily than $SO_4^{2-}$ and reduced more easily than $Na^+$.
Step 3: Detailed Explanation:

At Anode (Oxidation): Both $SO_4^{2-}$ and $H_2O$ are present. Water has a lower oxidation potential, so it oxidizes: $2H_2O \rightarrow O_2(g) + 4H^+ + 4e^-$. Thus, X is O₂.
At Cathode (Reduction): Both $Na^+$ and $H_2O$ are present. $H_2O$ is reduced more easily than $Na^+$: $2H_2O + 2e^- \rightarrow H_2(g) + 2OH^-$. Thus, Y is H₂.
Step 4: Final Answer:
X is Oxygen ($O_2$) and Y is Hydrogen ($H_2$).