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 analyze the reduction of potassium permanganate (KMnO4) and potassium dichromate (K2Cr2O7) by a reducing agent in acidic medium. We are given that the change in oxidation state of the reducing agent is the same in both experiments.

### Step-by-Step Solution:

1. **Understanding the Reaction**:
- In acidic medium, KMnO4 is reduced from Mn(VII) to Mn(II), which involves a change in oxidation state of 5.
- K2Cr2O7 is reduced from Cr(VI) to Cr(III), which involves a change in oxidation state of 3.

2. **Equivalence of Reducing Agents**:
- The number of equivalents of KMnO4 reduced can be calculated as:
\[ \text{Equivalents of KMnO4} = 5x \]
where \(x\) is the number of moles of KMnO4.
- The number of equivalents of K2Cr2O7 reduced can be calculated as:
\[ \text{Equivalents of K2Cr2O7} = 6y \]
where \(y\) is the number of moles of K2Cr2O7.

3. **Setting Up the Equation**:
- Since the change in oxidation state of the reducing agent is the same in both experiments, we can equate the equivalents:
\[ 5x = 6y \]

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

### Final Answer:
The ratio \(x:y\) is \(6:5\).