Question

what occurs during reduction in the reaction for the electrolysis of aqueous cuso₄?
half-reaction | e° (v)
cu²⁺ + 2e⁻ → cu | +0.34
2h₂o + 2e⁻ → h₂ + 2oh⁻ | - 0.83
copper(ii) ions change to copper.
a base is formed.
hydrogen gas is produced.
copper changes to copper(ii) ions.

Explanation:

In electrolysis, reduction is the gain of electrons (or decrease in oxidation state). For the electrolysis of aqueous \(CuSO_4\), we compare the standard reduction potentials of the possible half - reactions. The half - reaction \(Cu^{2+}+2e^-\to Cu\) has a more positive \(E^{\circ}( + 0.34\ V)\) compared to \(2H_2O + 2e^-\to H_2+2OH^-\) (\(E^{\circ}=- 0.83\ V\)). A more positive reduction potential means the reaction is more likely to occur as a reduction (gain of electrons). So, \(Cu^{2+}\) ions will gain electrons and be reduced to \(Cu\) (copper metal).

• Option "A base is formed": The reduction of water to form \(H_2\) and \(OH^-\) is less favorable due to its lower \(E^{\circ}\), so base formation is not the main reduction process here.
• Option "Hydrogen gas is produced": The reduction of water to produce \(H_2\) is less favorable than the reduction of \(Cu^{2+}\) because of the lower \(E^{\circ}\) of the water - reduction half - reaction.
• Option "Copper changes to copper(II) ions": This is an oxidation process (loss of electrons), not reduction.

Answer:

Copper(II) ions change to copper.