When ferrous oxalate is titrated against `K_(2)Cr_(2)O_(7)`, milli equivalents of `Fe^(2+),C_(2)O_(4)^(2-)andCr_(2)O_(7)` in the redox reaction are X,Y and Z respectively , then :-

Options

A. `X=Y`
B. `X+Y=Z`
C. `X+2Y=Z`
D. `2X+6Y=6Z`

Answer

To solve the problem, we need to analyze the redox reaction that occurs when ferrous oxalate (FeC2O4) is titrated against potassium dichromate (K2Cr2O7) in an acidic medium. We will determine the milli equivalents of Fe²⁺, C2O4²⁻, and Cr2O7²⁻ involved in the reaction.

### Step-by-step Solution:

1. **Write the Balanced Redox Reaction:**
The reaction between ferrous oxalate and potassium dichromate in an acidic medium can be represented as follows:
\[ 6 \text{Fe}^{2+} + \text{Cr}_2\text{O}_7^{2-} + 14 \text{H}^+ \rightarrow 6 \text{Fe}^{3+} + 2 \text{Cr}^{3+} + 7 \text{H}_2\text{O} + 6 \text{CO}_2 \]
Here, ferrous ions (Fe²⁺) are oxidized to ferric ions (Fe³⁺), and dichromate ions (Cr2O7²⁻) are reduced to chromium ions (Cr³⁺).

2. **Identify the Changes in Oxidation States:**
- Fe²⁺ is oxidized to Fe³⁺ (oxidation state changes from +2 to +3).
- Cr₂O₇²⁻ is reduced to Cr³⁺ (oxidation state changes from +6 to +3).
- C₂O₄²⁻ (oxalate) is also involved in the reaction, producing CO₂.

3. **Calculate the Milli Equivalents:**
- For Fe²⁺: Each Fe²⁺ gives 1 equivalent of electrons when it is oxidized to Fe³⁺. Since 6 moles of Fe²⁺ are involved, the milli equivalents (X) of Fe²⁺ will be:
\[ X = 6 \text{ moles of Fe}^{2+} = 6 \text{ milli equivalents} \]

- For C₂O₄²⁻: Each C₂O₄²⁻ can donate 2 electrons when it is oxidized to CO₂. Since 6 moles of C₂O₄²⁻ are involved (from the stoichiometry of the reaction), the milli equivalents (Y) of C₂O₄²⁻ will also be:
\[ Y = 6 \text{ moles of C}_2\text{O}_4^{2-} = 6 \text{ milli equivalents} \]

- For Cr₂O₇²⁻: Each Cr₂O₇²⁻ ion accepts 6 electrons when it is reduced to Cr³⁺. Since only 1 mole of Cr₂O₇²⁻ is involved, the milli equivalents (Z) of Cr₂O₇²⁻ will be:
\[ Z = 1 \text{ mole of Cr}_2\text{O}_7^{2-} = 6 \text{ milli equivalents} \]

4. **Establish Relationships:**
From the calculations:
- X (Fe²⁺) = 6
- Y (C₂O₄²⁻) = 6
- Z (Cr₂O₇²⁻) = 6

Thus, we can conclude that:
\[ X = Y = Z \]

### Conclusion:
The relationship between the milli equivalents of Fe²⁺, C₂O₄²⁻, and Cr₂O₇²⁻ is:
\[ X = Y = Z \]