Potassium iodide act as reducing agent when treated with

Options

A. Acidified `K_(2)Cr_(2)O_(7)` solution
B. An acidified `KMnO_(4)` solution
C. A `CuSO_(4)` solution
D. A lead acetate solution

Answer

To determine when potassium iodide (KI) acts as a reducing agent, we need to analyze its behavior in different reactions. A reducing agent is a substance that donates electrons to another substance and gets oxidized in the process.

Let's evaluate each option step by step:

### Step 1: Evaluate Reaction with Acidified Potassium Dichromate (K2Cr2O7)
1. **Write the reaction**:
\[ K_2Cr_2O_7 + H_2SO_4 + KI \rightarrow Cr_2(SO_4)_3 + I_2 + K_2SO_4 \]
2. **Identify oxidation states**:
- In K2Cr2O7, chromium (Cr) is in the +6 oxidation state.
- In the product Cr2(SO4)3, chromium is in the +3 oxidation state.
- Iodine (I) in KI is in the -1 oxidation state and in I2 it is 0.
3. **Determine changes**:
- Chromium is reduced from +6 to +3 (gains electrons).
- Iodine is oxidized from -1 to 0 (loses electrons).
4. **Conclusion**: KI acts as a reducing agent because it gets oxidized while reducing Cr.

### Step 2: Evaluate Reaction with Acidified KMnO4
1. **Write the reaction**:
\[ KMnO_4 + H_2SO_4 + KI \rightarrow MnO_2 + I_2 + KHSO_4 \]
2. **Identify oxidation states**:
- In KMnO4, manganese (Mn) is in the +7 oxidation state.
- In MnO2, manganese is in the +4 oxidation state.
- Iodine in KI is -1 and in I2 is 0.
3. **Determine changes**:
- Manganese is reduced from +7 to +4 (gains electrons).
- Iodine is oxidized from -1 to 0 (loses electrons).
4. **Conclusion**: KI acts as a reducing agent because it gets oxidized while reducing Mn.

### Step 3: Evaluate Reaction with Copper Sulfate (CuSO4)
1. **Write the reaction**:
\[ CuSO_4 + 2KI \rightarrow CuI + I_2 + K_2SO_4 \]
2. **Identify oxidation states**:
- In CuSO4, copper (Cu) is in the +2 oxidation state.
- In CuI, copper is in the +1 oxidation state.
- Iodine in KI is -1 and in I2 is 0.
3. **Determine changes**:
- Copper is reduced from +2 to +1 (gains electrons).
- Iodine is oxidized from -1 to 0 (loses electrons).
4. **Conclusion**: KI acts as a reducing agent because it gets oxidized while reducing Cu.

### Step 4: Evaluate Reaction with Lead Acetate (Pb(C2H3O2)2)
1. **Write the reaction**:
\[ Pb(C_2H_3O_2)_2 + 2KI \rightarrow PbI_2 + 2K(C_2H_3O_2) \]
2. **Identify oxidation states**:
- Lead (Pb) is in the +2 oxidation state in both reactants and products.
- Iodine in KI is -1 and in PbI2 is also -1.
3. **Determine changes**:
- There is no change in oxidation states for lead or iodine.
4. **Conclusion**: KI does not act as a reducing agent because it does not undergo oxidation.

### Final Conclusion
Potassium iodide acts as a reducing agent when treated with:
- Acidified potassium dichromate (A)
- Acidified KMnO4 (B)
- Copper sulfate (C)

**Correct options**: A, B, and C.

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