Faraday Charge to Produce 1 Mole Fe from Fe3O4

How Much Faraday Charge is Required to Produce 1 Mole of Fe from Fe₃O₄?

Published by Abhishek Sengar | CHEMCA India

In Electrochemistry, calculating the required Faraday charge for simple ions like Cu²⁺ or Al³⁺ is straightforward. But what happens when you encounter a mixed oxide like Fe₃O₄ that has a fractional oxidation state?

Let's break down this classic JEE Main integer-type question step-by-step.

Video Tutorial: Solving the Fe₃O₄ Puzzle

Watch Abhishek Sengar sir from CHEMCA demonstrate the mathematical workflow to effortlessly calculate the oxidation state, set up the reduction equation, and arrive at the final integer answer.

Step-by-Step Problem Breakdown

Problem Statement:

How much Faraday charge is required to produce exactly 1 mole of Fe from Fe₃O₄?

1. Find the Average Oxidation Number:
   Let the oxidation state of Iron (Fe) be x. Oxygen is usually -2.
   3x + 4(-2) = 0
   3x - 8 = 0 → 3x = +8
   x = +8/3
   This fractional oxidation state is normal for mixed oxides!
2. Write the Total Charge Equation:
   If the average charge on 1 Fe atom is +8/3, then for 3 Fe atoms, the total positive charge is:
   3 × (8/3) = +8.
3. Set up the Reduction Reaction:
   To completely reduce those 3 Iron atoms down to neutral solid Iron (Fe⁰), we must supply exactly 8 electrons to cancel out the +8 charge.
   
   3 Fe^(8/3)+ + 8 e^- → 3 Fe(s)

Applying Faraday's Law:
1 Mole of electrons is equal to 1 Faraday (1F) of charge.
Our equation shows that producing 3 moles of Fe requires 8 moles of electrons (8F).

Fig: Using Stoichiometry to isolate the charge required for exactly 1 Mole of product.

4. Calculate for 1 Mole:
   If 3 Moles of Fe = 8F
   Then 1 Mole of Fe = 8/3 F
   8/3 ≈ 2.66 F

Since JEE Main asked for an integer answer, we round 2.66 to the nearest whole number.
Final Answer: 3

Practice Questions for JEE & NEET

Ensure you grasp the logic of mixed oxides and reduction stoichiometry with these conceptual checks.

Question 1: Why is the oxidation state of Fe in Fe₃O₄ a fraction (+8/3)? Is it possible to have a fraction of an electron lost?

Answer: No, the fraction is just a mathematical average.

Reasoning:

It is impossible to lose a fraction of an electron! Fe₃O₄ (Magnetite) is not a simple compound; it is a Mixed Oxide composed of an equimolar mixture of FeO and Fe₂O₃.

• In FeO, the Fe is in a +2 state (1 atom).
• In Fe₂O₃, the Fe is in a +3 state (2 atoms).

The total charge is (1 \times +2) + (2 \times +3) = +8 distributed over 3 atoms. Therefore, the average is +8/3.

Question 2: How many Faradays of charge are required to completely reduce 1 mole of Dichromate ions (Cr₂O₇^2-) to Cr³⁺ in an acidic medium?

Answer: 6 Faradays

Reasoning:

• Step 1: Find the oxidation state of Cr in Cr₂O₇^2-.
  2x + 7(-2) = -2 \rightarrow 2x = +12 \rightarrow x = +6.
• Step 2: Determine the change per atom. Each Cr atom goes from +6 down to +3, meaning it gains 3 electrons.
• Step 3: Calculate for 1 mole of the ion. There are exactly two Cr atoms in every 1 mole of Cr₂O₇^2-.
  Total electrons needed = 2 atoms × 3 electrons/atom = 6 moles of electrons.
• 6 moles of electrons = 6 Faradays.

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