Which one represents the correct reduction reaction for the $\mathrm{Cu^{2+}|Cu}$ half-cell?

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

The correct answer is Option (1) → $\mathrm{Cu^{2+}(aq)+2e^{-}\rightarrow Cu(s)} $

Reduction is the gain of electrons. In a metal ion–metal half-cell, reduction refers to the conversion of a metal ion into solid metal by gaining electrons.

Explanation

Option 1: $\text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s)$

Here, the copper ion gains two electrons and gets converted into metallic copper. This is the correct reduction process for a $\text{Cu}^{2+} | \text{Cu}$ half-cell.

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

Here electrons are released instead of gained. This represents oxidation, not reduction.

Option 3: $\text{Cu}(s) \rightarrow \text{Cu}^{2+}(aq) + 2e^-$

This is also oxidation where solid copper loses electrons.

Option 4: $\text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}^+(s) + e^-$

This equation is incorrect and unbalanced. Also $\text{Cu}^+$ is not the final product in this half-cell.