If equal volumes of 1 M `KMnO_(4)` and 1M `K_(2)Cr_(2)O_(7)` solutions are allowed to oxidise Fe(II) to Fe(III) in acidic medium, then Fe(II) oxidised will be

Options

A. More with `KMnO_(4)`
B. More with `K_(2)Cr_(2)O_(7)`
C. Equal with both oxidising agent
D. Cannot be determined

Answer

To solve the problem of how much Fe(II) is oxidized to Fe(III) when equal volumes of 1 M KMnO₄ and 1 M K₂Cr₂O₇ are used in acidic medium, we need to analyze the oxidation states and stoichiometry of the reactions involved.

### Step-by-Step Solution:

1. **Identify the Oxidation States:**
- In KMnO₄, manganese (Mn) has an oxidation state of +7.
- In K₂Cr₂O₇, chromium (Cr) has an oxidation state of +6.

2. **Write the Reduction Half-Reactions:**
- The reduction of KMnO₄ in acidic medium:
\[ \text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \]
- The reduction of K₂Cr₂O₇ in acidic medium:
\[ \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{e}^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \]

3. **Determine the Moles of Electrons Transferred:**
- From the half-reaction of KMnO₄, 1 mole of KMnO₄ can oxidize 5 moles of Fe(II).
- From the half-reaction of K₂Cr₂O₇, 1 mole of K₂Cr₂O₇ can oxidize 6 moles of Fe(II).

4. **Calculate the Amount of Fe(II) Oxidized:**
- If we take equal volumes of 1 M solutions of KMnO₄ and K₂Cr₂O₇, let's assume we take 1 L of each solution.
- This means we have 1 mole of KMnO₄ and 1 mole of K₂Cr₂O₇.
- KMnO₄ will oxidize:
\[ 1 \text{ mole KMnO}_4 \times 5 \text{ moles Fe}^{2+}/\text{mole KMnO}_4 = 5 \text{ moles Fe}^{2+} \]
- K₂Cr₂O₇ will oxidize:
\[ 1 \text{ mole K}_2\text{Cr}_2\text{O}_7 \times 6 \text{ moles Fe}^{2+}/\text{mole K}_2\text{Cr}_2\text{O}_7 = 6 \text{ moles Fe}^{2+} \]

5. **Conclusion:**
- Since K₂Cr₂O₇ oxidizes more Fe(II) than KMnO₄, the answer to the question is that more Fe(II) will be oxidized by K₂Cr₂O₇.

### Final Answer:
Fe(II) oxidized will be more with K₂Cr₂O₇.