Which of the following involves the reduction of copper:

Options

A. `Cu(s)+(1)/(2)O_(2)(g)rarrCuO(s)`
B. `Cu^(2+)(aq)+2I^(-)(aq)rarr2Cul(aq)`
C. `CuCl_(2)(g)+2F^(-)(aq)rarrCuF_(2)+Cl_(2)(g)`
D. None of the above.

Explanation

To determine which of the given options involves the reduction of copper, we need to analyze each option and identify the changes in the oxidation state of copper.

### Step-by-Step Solution:

1. **Analyze Option 1: Cu + 1/2 O2 → CuO**
- In this reaction, copper (Cu) is in its elemental form with an oxidation state of 0.
- Oxygen (O2) has an oxidation state of 0, and in CuO, oxygen has an oxidation state of -2.
- Copper in CuO has an oxidation state of +2.
- Since the oxidation state of copper increases from 0 to +2, this represents an **oxidation** process, not reduction.

2. **Analyze Option 2: Cu²⁺ (aq) + 2I⁻ (aq) → 2CuI (aq)**
- Here, copper starts as Cu²⁺ with an oxidation state of +2.
- In the product CuI, the oxidation state of copper is +1 (since iodine is -1).
- The oxidation state of copper decreases from +2 to +1, indicating that copper is being **reduced**.

3. **Analyze Option 3: Cu²⁺ (aq) + 2Cl⁻ (aq) → CuCl₂ (aq)**
- In this reaction, copper starts as Cu²⁺ with an oxidation state of +2.
- In CuCl₂, copper remains at +2 since chlorine has an oxidation state of -1.
- There is no change in the oxidation state of copper, so this reaction does not involve reduction.

4. **Analyze Option 4: None of the above**
- Since we have identified that Option 2 involves the reduction of copper, this option is not correct.

### Conclusion:
The correct option that involves the reduction of copper is **Option 2**: Cu²⁺ (aq) + 2I⁻ (aq) → 2CuI (aq).