Question

When $CuSO_4 $ is electrolysed using platinum electrodes?

• A. Copper is liberated at cathode, sulphur at anode
• B. Copper is liberated at cathode, oxygen at anode
• C. Sulphur is liberated at cathode, oxygen at anode
• D. Oxygen is liberated at cathode, copper at anode

Answer 1

The Correct Option is B

To determine the outcome of the electrolysis of CuSO_4 (copper(II) sulfate) using platinum electrodes, let's analyze the reactions that occur at the cathode and the anode.

1. Cathode Reaction: In the electrolysis process, the cathode attracts cations (positive ions). For CuSO_4 in aqueous solution, the relevant cation is Cu^{2+} (copper ions).
   • At the cathode, Cu^{2+} ions gain electrons (reduction) to form copper metal. The reaction is: Cu^{2+} + 2e^- \rightarrow Cu(s).

   This means copper metal is deposited at the cathode.

2. Anode Reaction: At the anode, oxidation occurs, and anions are attracted to it. In the solution, sulfate ions (SO_4^{2-}) and hydroxide ions (OH^-) are present due to dissociation of water and copper sulfate.
   • The primary reaction at the anode in aqueous solutions often involves the oxidation of water rather than sulfate due to the higher oxidation potential required for sulfate ions. Thus, the reaction is: 2H_2O(l) \rightarrow O_2(g) + 4H^+ + 4e^-.

   As a result, oxygen gas is released at the anode.

Conclusively, in the electrolysis of CuSO_4 using platinum electrodes:

• Copper is liberated at the cathode.
• Oxygen is liberated at the anode.

Therefore, the correct answer is: Copper is liberated at cathode, oxygen at anode.