If equal volumes of `0.1 M KMnO_(4)` and `0.1 M K_(2)Cr_(2)O_(7)` solutions are allowed to oxidise `Fe^(2+)` to `Fe^(3+)` in acidic medium, then `Fe^(2+)` oxidised will be:

Options

A. more by `KMnO_(4)`
B. more by `K_(2)CrO_(7)`
C. equal in both cases
D. can't be determined

Explanation

To solve the problem, we need to determine how much \( \text{Fe}^{2+} \) will be oxidized when equal volumes of \( 0.1 \, M \, \text{KMnO}_4 \) and \( 0.1 \, M \, \text{K}_2\text{Cr}_2\text{O}_7 \) are used in acidic medium.

### Step-by-Step Solution:

1. **Identify the Reducing Agents:**
- The reducing agent here is \( \text{Fe}^{2+} \), which will be oxidized to \( \text{Fe}^{3+} \).

2. **Determine the N-Factors:**
- For \( \text{KMnO}_4 \) in acidic medium, the n-factor is 5 (as \( \text{Mn}^{7+} \) is reduced to \( \text{Mn}^{2+} \)).
- For \( \text{K}_2\text{Cr}_2\text{O}_7 \) in acidic medium, the n-factor is 6 (as \( \text{Cr}^{6+} \) is reduced to \( \text{Cr}^{3+} \)).

3. **Calculate the Equivalents of Each Oxidizing Agent:**
- Let the volume of each solution used be \( V \) liters (since equal volumes are used).
- For \( \text{KMnO}_4 \):
\[ \text{Equivalents of KMnO}_4 = \text{Molarity} \times \text{n-factor} \times \text{Volume} = 0.1 \, M \times 5 \times V = 0.5V \]
- For \( \text{K}_2\text{Cr}_2\text{O}_7 \):
\[ \text{Equivalents of K}_2\text{Cr}_2\text{O}_7 = \text{Molarity} \times \text{n-factor} \times \text{Volume} = 0.1 \, M \times 6 \times V = 0.6V \]

4. **Compare the Equivalents:**
- From the calculations:
- \( \text{KMnO}_4 \) provides \( 0.5V \) equivalents.
- \( \text{K}_2\text{Cr}_2\text{O}_7 \) provides \( 0.6V \) equivalents.
- Since \( 0.6V > 0.5V \), \( \text{K}_2\text{Cr}_2\text{O}_7 \) will oxidize more \( \text{Fe}^{2+} \) than \( \text{KMnO}_4 \).

5. **Conclusion:**
- The amount of \( \text{Fe}^{2+} \) oxidized will be more by \( \text{K}_2\text{Cr}_2\text{O}_7 \). Therefore, the answer is that \( \text{Fe}^{2+} \) oxidized will be more by \( \text{K}_2\text{Cr}_2\text{O}_7 \).

### Final Answer:
The \( \text{Fe}^{2+} \) oxidized will be more by \( \text{K}_2\text{Cr}_2\text{O}_7 \).

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