The number of moles of `KMnO_(4)` required to oxidise `1 mol` of `Fe(C_(2)O_(4))` in acidic medium is

Options

A. `0.6`
B. `1.67`
C. `0.2`
D. `0.4`

Explanation

To determine the number of moles of KMnO₄ required to oxidize 1 mole of Fe(C₂O₄) in acidic medium, we can follow these steps:

### Step 1: Understand the Oxidation and Reduction Processes
- KMnO₄ acts as an oxidizing agent and will be reduced in the process.
- Fe(C₂O₄) consists of ferrous ions (Fe²⁺) and oxalate ions (C₂O₄²⁻), both of which will be oxidized.

### Step 2: Write the Half-Reactions
1. **Reduction Half-Reaction for KMnO₄:**
- In acidic medium, KMnO₄ is reduced from Mn(VII) to Mn(II).
- The half-reaction can be written as:
\[ \text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \]
- This shows that 1 mole of KMnO₄ consumes 5 electrons.

2. **Oxidation Half-Reaction for Fe(C₂O₄):**
- The ferrous ion (Fe²⁺) is oxidized to ferric ion (Fe³⁺) and the oxalate ion (C₂O₄²⁻) is oxidized to carbon dioxide (CO₂).
- The half-reactions can be written as:
- For Fe²⁺:
\[ \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + \text{e}^- \]
- For C₂O₄²⁻:
\[ \text{C}_2\text{O}_4^{2-} \rightarrow 2\text{CO}_2 + 2\text{e}^- \]

### Step 3: Combine the Half-Reactions
- The total number of electrons released by the oxidation of 1 mole of Fe(C₂O₄) is:
- 1 electron from Fe²⁺ and 2 electrons from C₂O₄²⁻, totaling 3 electrons.
- To balance the electrons, we need to multiply the oxidation half-reaction by 5 (to get 15 electrons) and the reduction half-reaction by 3 (to consume 15 electrons).

### Step 4: Write the Balanced Equation
- After balancing, we have:
\[ 5\text{Fe}^{2+} + 5\text{C}_2\text{O}_4^{2-} + 3\text{MnO}_4^- + 15\text{H}^+ \rightarrow 5\text{Fe}^{3+} + 10\text{CO}_2 + 3\text{Mn}^{2+} + 15\text{H}_2\text{O} \]
- This shows that 5 moles of Fe(C₂O₄) require 3 moles of KMnO₄.

### Step 5: Calculate the Moles of KMnO₄ Required
- From the balanced equation, we see that:
\[ 5 \text{ moles of Fe(C}_2\text{O}_4) \text{ require } 3 \text{ moles of KMnO}_4 \]
- Therefore, for 1 mole of Fe(C₂O₄):
\[ \text{Moles of KMnO}_4 = \frac{3}{5} = 0.6 \text{ moles} \]

### Conclusion
The number of moles of KMnO₄ required to oxidize 1 mole of Fe(C₂O₄) in acidic medium is **0.6 moles**.

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