Correct Answer: Option B

Explanation

During the electrolysis of an aqueous copper sulfate (CuSO\(_4\)) solution using platinum electrodes, copper ions (Cu\(^{2+}\)) are reduced to solid copper (Cu) at the cathode, and water (H\(_2\)O) is oxidized to oxygen gas (O\(_2\)) and hydrogen ions (H\(^+\)) at the anode. 

Here's a more detailed explanation:

• Cathode (Reduction): Copper ions (Cu\(^{2+}\)) from the solution migrate towards the cathode and gain two electrons (2e-) to form solid copper, which deposits on the electrode. 

  • Reaction: Cu\(^{2+}\)(aq) + 2e- → Cu(s) 

• Anode (Oxidation): Water molecules (H\(_2\)O) are oxidized at the anode, losing electrons and forming oxygen gas (O\(_2\)) and hydrogen ions (H+). 

  • Reaction: 4OH\(^-\) + 4e\(^-\)(l) → 2H\(_2\)O(g) + O\(_2\) 

There is an explanation video available below.

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