Which of the following ions will liberated iodine when trated with KI?

Options

A. `Cu^(2+)`
B. `Fe^(2+)`
C. `Pb^(2+)`
D. `Sn^(2+)`

Answer

To determine which ions will liberate iodine when treated with potassium iodide (KI), we need to analyze the oxidation states and the ability of the ions to act as reducing agents. Here’s a step-by-step solution:

### Step 1: Understand the Reaction
When KI is treated with a reducing agent, the iodide ion (I⁻) can be oxidized to iodine (I₂). This means that the iodide ion is losing electrons (oxidation), and the other species must be reduced.

**Hint:** Identify the oxidation states of iodine in KI (I⁻) and in I₂ (0).

### Step 2: Identify the Oxidation States of the Ions
- **Iodide ion (I⁻)**: Oxidation state is -1.
- **Iodine (I₂)**: Oxidation state is 0.

### Step 3: Determine the Reducing Agents
A reducing agent is a species that donates electrons and gets oxidized in the process. We need to check the ions provided in the options to see if they can be reduced.

1. **Cu²⁺ (Copper II ion)**:
- Common oxidation states are +1 and +2.
- Can be reduced to Cu⁺ by gaining an electron.
- Therefore, Cu²⁺ can be reduced, and I⁻ will oxidize to I₂.

2. **Fe²⁺ (Iron II ion)**:
- Common oxidation states are +2 and +3.
- Cannot be reduced further (it cannot go to a lower oxidation state).
- Therefore, Fe²⁺ will not liberate iodine.

3. **Pb²⁺ (Lead II ion)**:
- Common oxidation states are +2 and +4.
- Similar to Fe²⁺, it cannot be reduced further.
- Therefore, Pb²⁺ will not liberate iodine.

4. **Sn²⁺ (Tin II ion)**:
- Common oxidation states are +2 and +4.
- Like the previous two, it cannot be reduced further.
- Therefore, Sn²⁺ will not liberate iodine.

### Step 4: Conclusion
Based on the analysis, only Cu²⁺ can act as a reducing agent and will liberate iodine when treated with KI.

**Final Answer:** The ion that will liberate iodine when treated with KI is **Cu²⁺**.

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