A certain amount of reducing agent reduces x mole of `KMnO_(4)` & y mole of `K_(2)Cr_(2)O_(7)` in difference experiments in acidic medium. If the change in oxidation state of reducing agent is same in both experiments then `x:y` is

Options

A. `5:3`
B. `3:5`
C. `5:6`
D. `6:5`

Explanation

To solve the problem, we need to determine the ratio \( x:y \) based on the equivalence of the reducing agent when reducing \( KMnO_4 \) and \( K_2Cr_2O_7 \) in acidic medium.

### Step-by-Step Solution:

1. **Identify the Oxidation States**:
- In \( KMnO_4 \), manganese (Mn) has an oxidation state of +7. When it is reduced to \( Mn^{2+} \), the change in oxidation state is from +7 to +2. This means that each mole of \( KMnO_4 \) requires 5 moles of electrons for reduction.
- In \( K_2Cr_2O_7 \), chromium (Cr) has an oxidation state of +6. When it is reduced to \( Cr^{3+} \), the change in oxidation state is from +6 to +3. This means that each mole of \( K_2Cr_2O_7 \) requires 6 moles of electrons for reduction.

2. **Calculate the Equivalents**:
- For \( x \) moles of \( KMnO_4 \), the total equivalents of reducing agent used is:
\[ \text{Equivalents from } KMnO_4 = 5x \]
- For \( y \) moles of \( K_2Cr_2O_7 \), the total equivalents of reducing agent used is:
\[ \text{Equivalents from } K_2Cr_2O_7 = 6y \]

3. **Set Up the Equation**:
- Since the change in oxidation state of the reducing agent is the same in both experiments, the equivalents of reducing agent must be equal:
\[ 5x = 6y \]

4. **Solve for the Ratio \( x:y \)**:
- Rearranging the equation gives:
\[ \frac{x}{y} = \frac{6}{5} \]
- Therefore, the ratio \( x:y \) is:
\[ x:y = 6:5 \]

5. **Conclusion**:
- The final answer is that the ratio \( x:y \) is \( 6:5 \).