Moles of KMnO4 Required to Oxidise Fe Mixture in Acidic Medium

To find the number of moles of $\text{KMnO}_4$ required to oxidise the given mixture in an acidic medium, we can calculate the total equivalents of reducing agents using the $n$-factor concept.

1. Calculation of $n$-factor for each component:

• $\text{FeC}_2\text{O}_4$ (Ferrous oxalate):

  • $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-$ (change in oxidation state = $1$)
  • $\text{C}_2\text{O}_4^{2-} \rightarrow 2\text{CO}_2 + 2e^-$ (change in oxidation state = $2$)
  • $n\text{-factor}_1 = 1 + 2 = 3$

• $\text{Fe}_2(\text{C}_2\text{O}_4)_3$ (Ferric oxalate):

  • $\text{Fe}^{3+}$ is already in its maximum stable oxidation state (+3), so it undergoes no further oxidation.
  • $3\text{C}_2\text{O}_4^{2-} \rightarrow 6\text{CO}_2 + 6e^-$ (change in oxidation state = $3 \times 2 = 6$)
  • $n\text{-factor}_2 = 6$

• $\text{FeSO}_4$ (Ferrous sulphate):

  • $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-$ (change in oxidation state = $1$)
  • $\text{SO}_4^{2-}$ is not oxidised.
  • $n\text{-factor}_3 = 1$

• $\text{Fe}_2(\text{SO}_4)_3$ (Ferric sulphate):

  • Both $\text{Fe}^{3+}$ and $\text{SO}_4^{2-}$ cannot be oxidised further by $\text{KMnO}_4$.
  • $n\text{-factor}_4 = 0$

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2. Calculation of total equivalents of the reducing mixture:

Since the mixture contains $1\text{ mole}$ of each substance:

$\text{Total Equivalents} = \sum (\text{Moles}_i \times n\text{-factor}_i)$

$\text{Total Equivalents} = (1 \times 3) + (1 \times 6) + (1 \times 1) + (1 \times 0) = 3 + 6 + 1 + 0 = 10$

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3. Calculation of moles of $\text{KMnO}_4$ required:

In an acidic medium, $\text{KMnO}_4$ is reduced according to the half-reaction:

$\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$

Thus, the $n$-factor of $\text{KMnO}_4$ in acidic medium is $5$.

By the principle of equivalence:

$\text{Equivalents of KMnO}_4 = \text{Total Equivalents of reducing mixture}$

$\text{Moles of KMnO}_4 \times n\text{-factor of KMnO}_4 = 10$

$\text{Moles of KMnO}_4 \times 5 = 10$

$\text{Moles of KMnO}_4 = \frac{10}{5} = 2$

Thus, the required number of moles of $\text{KMnO}_4$ is 2.