Question

Electrolysis of aqueous copper (II) sulphate between Pt electrodes gives 'X' at anode and 'Y' at cathode. X and Y are respectively

• A. \(\text{Cu, O}_2\)
• B. \(\text{O}_2, \text{Cu}\)
• C. \(\text{SO}_2, \text{H}_2\)
• D. \(\text{O}_2, \text{H}_2\)

Hint:

Key Rule: At the anode, if the anion contains oxygen (like \(\text{SO}_4^{2-}, \text{NO}_3^-\)), it is generally not oxidised; instead, water is oxidised to \(\text{O}_2\). At the cathode, metals below Hydrogen in the activity series (Cu, Ag, Au) are deposited preferentially over \(\text{H}_2\).

Answer 1

The Correct Option is B

Step 1: Identify Species Present:
Electrolyte: \(\text{CuSO}_4(aq) \rightarrow \text{Cu}^{2+} + \text{SO}_4^{2-}\). Solvent: \(\text{H}_2\text{O} \rightleftharpoons \text{H}^+ + \text{OH}^-\). Electrodes: Platinum (Inert).
Step 2: Reaction at Cathode (Negative Electrode):
Cations migrate to cathode: \(\text{Cu}^{2+}\) and \(\text{H}^+\). Reduction potential of \(\text{Cu}^{2+}/ \text{Cu}\) (\(+0.34\text{V}\)) is higher than \(\text{H}^+ / \text{H}_2\) (\(0.0\text{V}\) or lower at pH 7). Reaction: \(\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}(s)\). So, Y is Copper (Cu).
Step 3: Reaction at Anode (Positive Electrode):
Anions migrate to anode: \(\text{SO}_4^{2-}\) and \(\text{H}_2\text{O}\) (or \(\text{OH}^-\)). Oxidation of water is energetically favored over oxidation of sulphate ions. Reaction: \(2\text{H}_2\text{O}(l) \rightarrow \text{O}_2(g) + 4\text{H}^+(aq) + 4e^-\). So, X is Oxygen (\(\text{O}_2\)). Final Answer:
X = \(\text{O}_2\), Y = Cu.