To analyze the given half-reaction \( Cu^{2+}(aq) + 2e^{-} \longrightarrow Cu(s) \), let's go through the process step-by-step:
1. Identify the species involved:
- \( Cu^{2+}(aq) \) represents the copper ion in an aqueous solution with a charge of +2.
- \( 2e^{-} \) represents two electrons.
- \( Cu(s) \) represents solid copper.
2. Understand the half-reaction:
- The reaction shows that copper ions (\( Cu^{2+} \)) are gaining two electrons (\( 2e^{-} \)) to form solid copper (\( Cu \)).
3. Determine the type of process:
- Gaining electrons in a chemical reaction is called reduction. This can be remembered by the acronym "OIL RIG," which stands for "Oxidation Is Loss, Reduction Is Gain (of electrons)."
- Since the \( Cu^{2+} \) ion is gaining electrons, the process taking place is reduction.
Based on the understanding that gaining electrons is reduction, we can evaluate the given options:
1. Copper is being oxidized - This is incorrect because oxidation involves the loss of electrons.
2. Copper is being reduced - This is correct because the \( Cu^{2+} \) ion is gaining electrons.
3. Copper is losing electrons - This is incorrect because losing electrons would mean copper is being oxidized, not reduced.
4. Copper is a reducing agent - This is incorrect because a reducing agent donates electrons and gets oxidized, while here copper is gaining electrons.
Therefore, the best statement that describes what is taking place in this half-reaction is:
Copper is being reduced.
So, the correct choice is option 2.
