Equal volumes of 1M `KMnO_4` and 1M `K_2Cr_2O_7` solution are allowed to oxidise `Fe^(2+)` ions to `Fe^(3+)` ions in acidic medium. The number of moles of `Fe^(2+)` ions oxidised in the two cases are in the ratio:

Options

A. `1:1`
B. `3:1`
C. `5:6`
D. `6:5`

Explanation

To solve the problem, we need to determine the number of moles of \( Fe^{2+} \) ions oxidized by equal volumes of 1M \( KMnO_4 \) and 1M \( K_2Cr_2O_7 \) in an acidic medium.

### Step-by-Step Solution:

1. **Identify the Redox Reactions**:
- The oxidation of \( Fe^{2+} \) to \( Fe^{3+} \) can be represented as:
\[ Fe^{2+} \rightarrow Fe^{3+} + e^- \]
- For \( KMnO_4 \) in acidic medium, the reduction reaction is:
\[ MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O \]
- For \( K_2Cr_2O_7 \) in acidic medium, the reduction reaction is:
\[ Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O \]

2. **Determine the Moles of Electrons Required**:
- To oxidize 1 mole of \( Fe^{2+} \), 1 mole of electrons is required.
- From the \( KMnO_4 \) reaction, 1 mole of \( MnO_4^- \) can provide 5 moles of electrons. Therefore, 1 mole of \( KMnO_4 \) can oxidize:
\[ 5 \text{ moles of } Fe^{2+} \]
- From the \( K_2Cr_2O_7 \) reaction, 1 mole of \( Cr_2O_7^{2-} \) can provide 6 moles of electrons. Therefore, 1 mole of \( K_2Cr_2O_7 \) can oxidize:
\[ 6 \text{ moles of } Fe^{2+} \]

3. **Calculate the Ratio of Moles of \( Fe^{2+} \) Oxidized**:
- The ratio of moles of \( Fe^{2+} \) oxidized by \( KMnO_4 \) to that oxidized by \( K_2Cr_2O_7 \) is:
\[ \text{Ratio} = \frac{5 \text{ moles of } Fe^{2+}}{6 \text{ moles of } Fe^{2+}} = \frac{5}{6} \]

4. **Final Answer**:
- The number of moles of \( Fe^{2+} \) ions oxidized in the two cases are in the ratio:
\[ 5 : 6 \]