Two statements are given below: Statement I: Molten NaCl is electrolysed using Pt electrodes. \( Cl_2 \) is liberated at the anode.
Statement II: Aqueous CuSO\(_4\) is electrolysed using Pt electrodes. \( O_2 \) is liberated at the cathode. The correct answer is:
Two statements are given below: Statement I: Molten NaCl is electrolysed using Pt electrodes. \( Cl_2 \) is liberated at the anode.
Statement II: Aqueous CuSO\(_4\) is electrolysed using Pt electrodes. \( O_2 \) is liberated at the cathode. The correct answer is:
Show Hint
- Both statement I and II are correct
- Both statement I and II are not correct
- Statement I is correct but statement II is not correct
Statement I is not correct but statement II is correct
The Correct Option is C
Solution and Explanation
Step 1: Understanding Electrolysis of Molten NaCl
- When molten sodium chloride (\( NaCl \)) is electrolysed using platinum electrodes, it undergoes the following reactions:
- At the cathode (\( - \)): \( Na^+ + e^- \rightarrow Na \) (Sodium metal is deposited)
- At the anode (\( + \)): \( 2Cl^- \rightarrow Cl_2 + 2e^- \) (Chlorine gas is liberated)
- Therefore, Statement I is correct since \( Cl_2 \) is indeed liberated at the anode.
Step 2: Understanding Electrolysis of Aqueous CuSO\(_4\)
- When aqueous copper sulfate (\( CuSO_4 \)) is electrolysed using platinum electrodes:
- At the cathode: \( Cu^{2+} + 2e^- \rightarrow Cu \) (Copper is deposited)
- At the anode: \( 2H_2O \rightarrow O_2 + 4H^+ + 4e^- \) (Oxygen gas is liberated)
- However, the oxygen gas is liberated at the anode, not at the cathode, making Statement II incorrect.
Step 3: Evaluating the Given Options
- Option (1): Incorrect, because Statement II is incorrect.
- Option (2): Incorrect, as Statement I is correct.
- Option (3): Correct, as Statement I is correct, and Statement II is incorrect.
- Option (4): Incorrect, as Statement I is correct.
Thus, the correct answer is
Option (3).
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