Question

Given the balanced ionic equation representing a reaction: 2al(s) + 3cu2+(aq) → 2al3+(aq) + 3cu(s)which half-reaction represents the reduction that occurs?al → al3+ +3eal3+ +3e → alcu→ cu2+ +2ecu2+ + 2e → cu

Answer 1

Option D : $Cu^{2+}(aq)+2 e^{-}\rightarrow Cu(s)$

Reduction take place when oxidation state of atom of an element decrease. Here, addition of electron/s takes place. Opposite to that in oxidation, oxidation state increases and here, loss of electron/s take place.

The balanced chemical equation is as follows:

$2Al(s)+3Cu^{2+}(aq)\rightarrow 2Al^{3+}(aq)+3Cu(s)$

Here, oxidation state of Al changes from zero to +3 thus, it undergoes oxidation and oxidation state of Cu changes from +2 to zero thus, it undergoes reduction.

The half reactions will be:

Oxidation: $Al(s)\rightarrow Al^{3+}(aq)+3e^{-}$

Reduction: $Cu^{2+}(aq)+2 e^{-}\rightarrow Cu(s)$

Therefore, option D is correct.

Answer 2

Cu2+ + 2e- --->Cu is the answer