Question

In order to oxidise a mixture of 1 mole each of FeC₂O₄, Fe₂(C₂O₄)₃, FeSO₄ and Fe₂(SO₄)₃ in acidic medium, the number of moles of KMnO₄ required is

• A. 3
• B. 2
• C. 5
• D. 7

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
In acidic medium, $KMnO_4$ acts as an oxidizing agent where $Mn^{7+}$ is reduced to $Mn^{2+}$, meaning its n-factor is 5. We must calculate the total moles of electrons lost by the reducing agents and equate them using the principle of equivalence.

Step 2: Key Formula or Approach:
1. Total equivalents of $KMnO_4$ = Total equivalents of reducing agents. 2. Equivalents = Moles $\times$ n-factor. 3. n-factor for $KMnO_4$ (acidic) = 5.

Step 3: Detailed Explanation:
1. $FeC_2O_4$: $Fe^{2+} \to Fe^{3+}$ (1e) and $C_2O_4^{2-} \to 2CO_2$ (2e). Total n-factor = 3. Equivalents = $1 \times 3 = 3$. 2. $Fe_2(C_2O_4)_3$: $Fe$ is already $3+$. Only $3 \times C_2O_4^{2-}$ oxidizes. Total n-factor = $3 \times 2 = 6$. Equivalents = $1 \times 6 = 6$. 3. $FeSO_4$: $Fe^{2+} \to Fe^{3+}$ (1e). Total n-factor = 1. Equivalents = $1 \times 1 = 1$. 4. $Fe_2(SO_4)_3$: $Fe$ is $3+$ and $SO_4^{2-}$ cannot be further oxidized. Equivalents = 0. 5. Total Equivalents = $3 + 6 + 1 = 10$. 6. Moles of $KMnO_4$ = $\frac{\text{Total Equivalents}}{\text{n-factor of } KMnO_4} = \frac{10}{5} = 2$.

Step 4: Final Answer:
The number of moles of $KMnO_4$ required is 2.