What would be the normality of a 0.1 M `K_(2)Cr_(2)O_(7)` solution used as a oxidizing agent of Pb^(2+)?

Options

A. 0.1N
B. 0.6N
C. 0.4N
D. 0.2N

Answer

To find the normality of a 0.1 M \( K_2Cr_2O_7 \) solution when used as an oxidizing agent for \( Pb^{2+} \), we need to follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction**:
The potassium dichromate (\( K_2Cr_2O_7 \)) acts as an oxidizing agent and will reduce while oxidizing lead ions (\( Pb^{2+} \)). The chromium in \( K_2Cr_2O_7 \) is reduced from +6 oxidation state to +3 oxidation state.

2. **Determine the Change in Oxidation States**:
- In \( K_2Cr_2O_7 \), chromium (Cr) is in the +6 oxidation state.
- In the reduced form (Cr\(^{3+}\)), chromium is in the +3 oxidation state.
- The change in oxidation state for one chromium atom is \( +6 \) to \( +3 \), which corresponds to a gain of 3 electrons.

3. **Calculate the Total Electrons Transferred**:
- Since there are 2 chromium atoms in \( K_2Cr_2O_7 \), the total number of electrons transferred when both chromium atoms are reduced is:
\[ 2 \text{ Cr} \times 3 \text{ electrons/Cr} = 6 \text{ electrons} \]

4. **Relate Normality to Molarity**:
- Normality (N) is related to molarity (M) and the number of equivalents (n) by the formula:
\[ N = M \times n \]
- Here, \( n \) is the number of electrons transferred per formula unit of \( K_2Cr_2O_7 \) in the reaction, which we found to be 6.

5. **Calculate the Normality**:
- Given that the molarity of the \( K_2Cr_2O_7 \) solution is 0.1 M, we can substitute into the formula:
\[ N = 0.1 \, \text{M} \times 6 = 0.6 \, \text{N} \]

6. **Conclusion**:
- Therefore, the normality of the 0.1 M \( K_2Cr_2O_7 \) solution is \( 0.6 \, \text{N} \).

### Final Answer:
The normality of the 0.1 M \( K_2Cr_2O_7 \) solution is **0.6 N**.