Calculate mol of (a) `MnO_4^–` and (b) `Cr_2O_7^(2–)` to oxidise 1 mol of `FeC_2O_4` (ferrous oxalate) in acidic medium.

Answer

To calculate the moles of \( \text{MnO}_4^- \) and \( \text{Cr}_2\text{O}_7^{2-} \) required to oxidize 1 mole of \( \text{FeC}_2\text{O}_4 \) (ferrous oxalate) in acidic medium, we need to follow these steps:

### Step 1: Determine the number of electrons required for oxidation of \( \text{FeC}_2\text{O}_4 \)

1. **Oxidation of \( \text{Fe}^{2+} \)**:
- \( \text{Fe}^{2+} \) needs 1 electron to be oxidized to \( \text{Fe}^{3+} \).

2. **Oxidation of \( \text{C}_2\text{O}_4^{2-} \)** (oxalate ion):
- \( \text{C}_2\text{O}_4^{2-} \) needs 2 electrons to be oxidized to \( \text{CO}_2 \).

3. **Total electrons required**:
- For 1 mole of \( \text{FeC}_2\text{O}_4 \):
- \( 1 \, \text{electron (for Fe)} + 2 \, \text{electrons (for C}_2\text{O}_4) = 3 \, \text{electrons} \).

### Step 2: Calculate moles of \( \text{MnO}_4^- \)

1. **Electrons provided by \( \text{MnO}_4^- \)**:
- In acidic medium, \( \text{MnO}_4^- \) is reduced to \( \text{Mn}^{2+} \) and it donates 5 electrons.

2. **Moles of \( \text{MnO}_4^- \) needed**:
- To find the moles of \( \text{MnO}_4^- \) required to provide 3 electrons:
\[ \text{Moles of } \text{MnO}_4^- = \frac{\text{Electrons required}}{\text{Electrons provided by } \text{MnO}_4^-} = \frac{3}{5} = 0.6 \, \text{moles} \]

### Step 3: Calculate moles of \( \text{Cr}_2\text{O}_7^{2-} \)

1. **Electrons provided by \( \text{Cr}_2\text{O}_7^{2-} \)**:
- In acidic medium, \( \text{Cr}_2\text{O}_7^{2-} \) is reduced to \( \text{Cr}^{3+} \) and it donates 6 electrons.

2. **Moles of \( \text{Cr}_2\text{O}_7^{2-} \) needed**:
- To find the moles of \( \text{Cr}_2\text{O}_7^{2-} \) required to provide 3 electrons:
\[ \text{Moles of } \text{Cr}_2\text{O}_7^{2-} = \frac{3}{6} = 0.5 \, \text{moles} \]

### Final Answer:
- Moles of \( \text{MnO}_4^- \) required: **0.6 moles**
- Moles of \( \text{Cr}_2\text{O}_7^{2-} \) required: **0.5 moles**

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