n-factor of ferric oxalate acting like reducing agent in acidic medium is

Analysis of the Problem

1. Key Idea: The n-factor is the total number of moles of electrons exchanged per mole of the compound in a redox reaction.
2. Oxidation Process: In acidic medium, oxalate ions (C₂O₄²⁻) act as a reducing agent and oxidize to carbon dioxide (CO₂). Simultaneously, Fe³⁺ can act as an oxidizing agent.

Here, we are concerned with ferric oxalate (Fe₂(C₂O₄)₃) acting as a reducing agent. Only C₂O₄²⁻ (oxalate) is reduced, not Fe³⁺.

Reduction Half-Reaction of Oxalate (C₂O₄²⁻)

Oxalate ions oxidize as follows:

C₂O₄²⁻ → 2CO₂ + 2e⁻
 

For 1 mole of C₂O₄²⁻, 2 electrons are exchanged.

Decomposition of Ferric Oxalate (Fe₂(C₂O₄)₃)

The compound dissociates as:

Fe₂(C₂O₄)₃ → 2Fe³⁺ + 3C₂O₄²⁻
 

From the dissociation:

• 3 moles of oxalate ions (C₂O₄²⁻) are present per mole of ferric oxalate.

Total Electrons Exchanged

For 3 moles of oxalate ions, the total number of electrons exchanged is:

3 × 2 = 6 electrons
 

Thus, the n-factor of ferric oxalate acting as a reducing agent is 6.

The correct answer is: d) 6

Solution Recap

1. Dissociation of ferric oxalate yields 3 moles of oxalate ions per mole of Fe₂(C₂O₄)₃.
2. Each oxalate ion exchanges 2 electrons during oxidation.
3. Total electrons exchanged = 3 × 2 = 6.
4. n-factor = 6.

This makes d) 6 the correct choice.