The oxidation state of copper changes when aqueous copper (II) ions react with:
(I)`NaOH(aq)` (II)`Fe(s)` (III)`KI(aq)`

Options

A. I, II, III
B. II only
C. II, III
D. I only

Answer

To determine the changes in the oxidation state of copper when aqueous copper (II) ions react with sodium hydroxide (NaOH), iron (Fe), and potassium iodide (KI), we can analyze each reaction step by step.

### Step 1: Reaction with Sodium Hydroxide (NaOH)
1. **Write the reaction**: When copper (II) ions react with sodium hydroxide, the reaction can be represented as:
\[ \text{Cu}^{2+} (aq) + 2 \text{NaOH} (aq) \rightarrow \text{Cu(OH)}_2 (s) + 2 \text{Na}^+ (aq) \]
2. **Determine oxidation states**:
- In \(\text{Cu}^{2+}\), the oxidation state of copper is +2.
- In \(\text{Cu(OH)}_2\), the oxidation state of copper remains +2.
3. **Conclusion**: The oxidation state of copper does not change in this reaction.

### Step 2: Reaction with Iron (Fe)
1. **Write the reaction**: When copper (II) ions react with solid iron, the reaction can be represented as:
\[ \text{Cu}^{2+} (aq) + \text{Fe} (s) \rightarrow \text{Cu} (s) + \text{Fe}^{2+} (aq) \]
2. **Determine oxidation states**:
- In \(\text{Cu}^{2+}\), the oxidation state of copper is +2.
- In solid copper (\(\text{Cu}\)), the oxidation state is 0.
3. **Conclusion**: The oxidation state of copper decreases from +2 to 0 in this reaction.

### Step 3: Reaction with Potassium Iodide (KI)
1. **Write the reaction**: When copper (II) ions react with potassium iodide, the reaction can be represented as:
\[ 2 \text{Cu}^{2+} (aq) + 4 \text{KI} (aq) \rightarrow 2 \text{CuI} (s) + \text{I}_2 (g) + 2 \text{K}^+ (aq) \]
2. **Determine oxidation states**:
- In \(\text{Cu}^{2+}\), the oxidation state of copper is +2.
- In \(\text{CuI}\), the oxidation state of copper is +1.
3. **Conclusion**: The oxidation state of copper decreases from +2 to +1 in this reaction.

### Final Conclusion
- The oxidation state of copper does not change when reacting with sodium hydroxide (remains +2).
- The oxidation state of copper decreases from +2 to 0 when reacting with iron.
- The oxidation state of copper decreases from +2 to +1 when reacting with potassium iodide.

Thus, the oxidation state of copper changes in the reactions with iron and potassium iodide.

### Summary of Results
- **NaOH**: Oxidation state remains +2 (no change).
- **Fe**: Oxidation state changes from +2 to 0 (decrease).
- **KI**: Oxidation state changes from +2 to +1 (decrease).