Chemistry Q56 – JEE-MAIN 2026

Home Chemistry Redox ReactionsQuestion #553

In order to oxidise a mixture of 1 mole each of

F e C_{2} O_{4}

F e_{2} (C_{2} O_{4})_{3}

F e S O_{4}

and

F e_{2} (S O_{4})_{3}

in acidic medium, the number of moles of

K M n O_{4}

required is

Clue 1 of 3: Redox Reactions

Identify all species that can be oxidized in the given mixture and determine the change in their oxidation states.

🧠 Full Solution Path

Step 1: Determine n-factors for each species✦ Active

In acidic medium, $K M n O_{4}$ acts as an oxidizing agent, where $M n$ changes its oxidation state from $+ 7$ to $+ 2$ . Thus, the n-factor for $K M n O_{4}$ is 5.

For the reducing agents:

F e C_{2} O_{4} : F e^{2 +} \to F e^{3 +} (n = 1); C_{2} O_{4}^{2 -} \to 2 C O_{2} (n = 2) . Total n-factor = 1 + 2 = 3.

F e_{2} (C_{2} O_{4})_{3} : F e^{3 +} is already oxidized; 3 C_{2} O_{4}^{2 -} \to 6 C O_{2} (n = 3 \times 2 = 6) . Total n-factor = 6.

F e S O_{4} : F e^{2 +} \to F e^{3 +} (n = 1) . Total n-factor = 1.

F e_{2} (S O_{4})_{3} : F e^{3 +} is already oxidized . Total n-factor = 0.

Step 2: Calculate total equivalents of reducing agents○ Expand

Since 1 mole of each compound is present, the total equivalents of reducing agents is the sum of their individual n-factors:

Total equivalents = (1 \times 3) + (1 \times 6) + (1 \times 1) + (1 \times 0) = 3 + 6 + 1 + 0 = 10.

💡 Teacher's Secret Hint

Remember to consider all oxidizable components within each compound.

Step 3: Calculate moles of KMnO4 required○ Expand

According to the equivalence principle, the equivalents of the oxidizing agent must equal the equivalents of the reducing agents.

Moles of K M n O_{4} \times n-factor of K M n O_{4} = Total equivalents of reducing agents

Moles of K M n O_{4} \times 5 = 10

Moles of K M n O_{4} = \frac{10}{5} = 2.

💡 Teacher's Secret Hint

Ensure the correct n-factor for $K M n O_{4}$ in acidic medium is used.

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Chemistry Question 56 – JEE-MAIN 2026

Clue 1 of 3: Redox Reactions

🧠 Full Solution Path