Question

At 300 K, aqueous KCl and aqueous K$_2$SO$_4$ solutions were electrolyzed separately using Pt electrodes. The gases liberated at cathodes in these two electrolytic processes are respectively:

• A. \( Cl_2, O_2 \)
• B. \( O_2, O_2 \)
• C. \( H_2, O_2 \)
• D. \( H_2, H_2 \)

Hint:

- In the electrolysis of aqueous ionic solutions, the gases liberated depend on the standard reduction potentials of ions.
- For KCl solution, \( H_2 \) is evolved at the cathode and \( Cl_2 \) at the anode.
- For K$_2$SO$_4$ solution, \( H_2 \) is evolved at the cathode and \( O_2 \) at the anode.

Answer 1

The Correct Option is D

Step 1: Electrolysis of Aqueous KCl Solution
- Aqueous KCl contains \( K^+ \), \( Cl^- \), and \( H_2O \) molecules.
- At the cathode: Hydrogen gas (\( H_2 \)) is liberated by the reduction of water: \[ 2H_2O + 2e^- \rightarrow H_2 + 2OH^- \]
- At the anode: Chlorine gas (\( Cl_2 \)) is liberated by the oxidation of \( Cl^- \): \[ 2Cl^- \rightarrow Cl_2 + 2e^- \] Step 2: Electrolysis of Aqueous K$_2$SO$_4$ Solution
- Aqueous K$_2$SO$_4$ contains \( K^+ \), \( SO_4^{2-} \), and \( H_2O \) molecules.
- At the cathode: Hydrogen gas (\( H_2 \)) is liberated by the reduction of water: \[ 2H_2O + 2e^- \rightarrow H_2 + 2OH^- \]
- At the anode: Oxygen gas (\( O_2 \)) is liberated by the oxidation of water: \[ 2H_2O \rightarrow O_2 + 4H^+ + 4e^- \] Step 3: Conclusion
- At the cathode in both cases, hydrogen gas (\( H_2 \)) is liberated.
- Thus, the correct answer is Option (4) \( H_2, H_2 \).