2 moles of `FeSO_4` in acid medium are oxidized by x moles of `KMnO_4` , whereas 2 moles of `FeC_2O_4` in acid medium are oxidized by y moles of `KMnO_4` . The ratio of x and y is

Options

A. `1/3`
B. `1/2`
C. `1/4`
D. `1/5`

Answer

To solve the problem, we need to determine the ratio of moles of KMnO4 (x and y) required to oxidize 2 moles of FeSO4 and 2 moles of FeC2O4 in an acidic medium.

### Step-by-Step Solution:

1. **Oxidation of FeSO4:**
- In FeSO4, iron (Fe) is in the +2 oxidation state (Fe²⁺) and is oxidized to +3 oxidation state (Fe³⁺).
- The change in oxidation state for Fe is from +2 to +3, which corresponds to the loss of 1 electron.
- Therefore, the n-factor for FeSO4 is 1 (since 1 electron is lost per Fe atom).

2. **Determine the n-factor for KMnO4:**
- In acidic medium, KMnO4 is reduced from +7 (in MnO4⁻) to +2 (in Mn²⁺).
- The change in oxidation state for Mn is from +7 to +2, which corresponds to the gain of 5 electrons.
- Therefore, the n-factor for KMnO4 is 5.

3. **Calculate the equivalences for FeSO4:**
- For 2 moles of FeSO4, the total number of equivalents is:
\[ \text{Equivalents of FeSO4} = \text{moles} \times \text{n-factor} = 2 \times 1 = 2 \]

4. **Set up the equivalence equation:**
- The equivalence of KMnO4 used (in terms of x moles) is:
\[ \text{Equivalents of KMnO4} = x \times 5 \]
- Setting the equivalences equal gives:
\[ 2 = 5x \implies x = \frac{2}{5} \]

5. **Oxidation of FeC2O4:**
- In FeC2O4, Fe is again in the +2 oxidation state and is oxidized to +3 (1 electron lost).
- The carbon in C2O4²⁻ is in the +3 oxidation state and is oxidized to +4 (1 electron lost per carbon atom).
- Since there are 2 carbon atoms, the total change for carbon is 2 electrons.
- Therefore, the total n-factor for FeC2O4 is:
\[ \text{n-factor} = 1 (for Fe) + 2 (for 2 C) = 3 \]

6. **Calculate the equivalences for FeC2O4:**
- For 2 moles of FeC2O4, the total number of equivalents is:
\[ \text{Equivalents of FeC2O4} = 2 \times 3 = 6 \]

7. **Set up the equivalence equation for FeC2O4:**
- The equivalence of KMnO4 used (in terms of y moles) is:
\[ \text{Equivalents of KMnO4} = y \times 5 \]
- Setting the equivalences equal gives:
\[ 6 = 5y \implies y = \frac{6}{5} \]

8. **Calculate the ratio of x to y:**
- Now, we can find the ratio:
\[ \frac{x}{y} = \frac{\frac{2}{5}}{\frac{6}{5}} = \frac{2}{6} = \frac{1}{3} \]

### Final Answer:
The ratio of x to y is \( \frac{1}{3} \).