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

The ionic equation represents the electrolytes in the dissociated ion form in the aqueous solution. Half-reaction which represents the reduction is the increased number of electrons in the copper atom.

What is reduction?

Reduction in the chemical reaction is the increase in the number of electrons in the atoms, while oxidation is the decrease in the number of electrons from an atom.

The balanced chemical reaction is shown as:

$\rm 2Al + 3Cu^{2+} \rightarrow 2Al^{3+} + 3Cu$

Reduction occurs when the oxidation number of the atom decreases and electrons are added to it. In the reaction the oxidation number of copper changes from +2 to 0 hence, undergoes reduction.

The half-reaction of the equation is written as,

Oxidation: $\rm Al \rightarrow Al^{3+} + 3e^{-}$

Reduction: $\rm Cu^{2+} + 2e^{-} \rightarrow Cu$

Therefore, reduction takes place at option D. $\rm Cu^{2+} + 2e^{-} \rightarrow Cu.$

Learn more about reduction here:

https://brainly.com/question/13896138

#SPJ3