Question

Copper sulphate solution is electrolyzed using copper electrodes, the reaction taking place at anode is

• A. H$^+$ + e$^-$ $\longrightarrow$ H
• B. SO$_4^{2-}$ (aq) $\longrightarrow$ SO$_4$ + 2e$^-$
• C. Cu$^{2+}$ + 2e$^-$ $\longrightarrow$ Cu
• D. Cu (s) $\longrightarrow$ Cu$^{2+}$ (aq) + 2e$^-$

Answer 1

Answer: (D)

Cu (s) $\longrightarrow$ Cu$^{2+}$ (aq) + 2e$^-$

Answer 2

During the electrolysis of copper sulphate solution using copper electrodes, the following species are present: Cu²⁺, SO₄²⁻, and H₂O. The electrodes are active copper electrodes.

At the anode, oxidation occurs. The possible species to be oxidized are the copper electrode itself (Cu(s)), sulphate ions (SO₄²⁻), and water (H₂O).

1. Oxidation of copper electrode: Cu(s) $\longrightarrow$ Cu²⁺(aq) + 2e⁻ (E°_ox = -0.34 V)
2. Oxidation of water: 2H₂O(l) $\longrightarrow$ O₂(g) + 4H⁺(aq) + 4e⁻ (E°_ox = -1.23 V)

Sulphate ions are very difficult to oxidize and typically do not react.

Comparing the standard oxidation potentials, copper metal is much more easily oxidized (-0.34 V) than water (-1.23 V). Therefore, the copper anode itself dissolves.

The reaction taking place at the anode is:

Cu (s) $\longrightarrow$ Cu²⁺ (aq) + 2e⁻