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Redox titration problems can be solved two ways: write the balanced equation and use mole ratios, or use n-factors and equivalents and skip the full equation. Many students prefer the second route for JEE because it is faster, but it only works if the n-factor is right. This post explains what the n-factor is, how to find it, and works three problems both ways so you can check one method against the other.
Is the equivalent concept on the syllabus?
The JEE (Advanced) 2026 brochure lists calculations "involving common oxidation-reduction, neutralisation, and displacement reactions" and concentration in terms of molarity, molality and normality. The JEE (Main) 2026 syllabus lists balancing of redox reactions, and its practical unit includes oxalic acid against KMnO₄ and Mohr's salt against KMnO₄; it does not name normality. The mole method works for every exam, so learn equivalents as a shortcut, not a replacement. Check the syllabus for your exam year.
What is the n-factor?
For a redox reaction, the n-factor of a substance is the number of electrons one formula unit gains or loses. For an acid or base, it is the number of H⁺ or OH⁻ one formula unit gives or takes up in the reaction.
Then:
equivalents = n-factor × moles
normality = n-factor × molarity
at the end point, equivalents of oxidant = equivalents of reductant
The NCERT Class 12 chemistry lab manual, Unit 6, uses exactly this for its redox titrations, written as a₁M₁V₁ = a₂M₂V₂. It defines a₁ and a₂ as the number of electrons lost or gained per formula unit in the balanced half reaction. So a is the n-factor, and aM is the normality.
How do you find n-factors?
Write the half reaction or track oxidation numbers, and multiply by the number of atoms that change.
KMnO₄ in acid: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O. Mn goes from +7 to +2, so n = 5.
KMnO₄ in neutral or faintly alkaline solution: the product is MnO₂, as NCERT's d-block chapter shows, and the Redox Reactions chapter balances it as MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻, so n = 3.
K₂Cr₂O₇ in acid: Cr₂O₇²⁻ → 2Cr³⁺. Each Cr goes from +6 to +3, and there are two, so n = 6.
Oxalic acid or oxalate as a reductant: C₂O₄²⁻ → 2CO₂. Each C goes from +3 to +4, two carbons, so n = 2. The lab manual uses a₁ = 2.
Mohr's salt, FeSO₄·(NH₄)₂SO₄·6H₂O: Fe²⁺ → Fe³⁺, n = 1. The manual uses a₁ = 1.
FeC₂O₄: both parts are oxidised: Fe²⁺ loses 1 and C₂O₄²⁻ loses 2, so n = 3.
Worked example 1: standardising KMnO₄
This mirrors Experiment 6.1 in the manual, with a sample burette reading. 10.0 mL of 0.100 M oxalic acid needs 20.0 mL of KMnO₄ in acid. Find the molarity of KMnO₄.
By equivalents: 2 × 0.100 × 10.0 = 5 × M × 20.0, so M = 2.00 ÷ 100 = 0.0200 M.
By moles: the balanced ionic equation in the manual is 2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O. Oxalate = 1.00 mmol, so MnO₄⁻ = 1.00 × 2/5 = 0.400 mmol in 20.0 mL = 0.0200 M. Same answer.
Worked example 2: Mohr's salt
The manual gives the molar mass of Mohr's salt as 392 g mol⁻¹. Dissolve 3.92 g in 100 mL to get 0.100 M. What volume of the 0.0200 M KMnO₄ oxidises 10.0 mL of it in acid?
By equivalents: 1 × 0.100 × 10.0 = 5 × 0.0200 × V, so V = 1.00 ÷ 0.100 = 10.0 mL.
By moles: 5Fe²⁺ react with each MnO₄⁻, so 1.00 mmol Fe²⁺ needs 0.200 mmol MnO₄⁻, which is 10.0 mL of 0.0200 M.
Worked example 3: when both ions react
How many moles of KMnO₄ in acid oxidise 1 mol of FeC₂O₄ completely?
Equivalents of FeC₂O₄ = 3 × 1 = 3. Moles of KMnO₄ = 3 ÷ 5 = 0.6 mol.
A common wrong answer is 0.2 mol, from counting only the iron. Track every atom whose oxidation number changes.
Why does the lab manual avoid HCl?
The manual explains that dilute sulphuric acid is used because nitric acid is itself an oxidising agent, and HCl reacts with KMnO₄ to give chlorine. In equivalent terms, HCl would consume some permanganate, so the burette reading would overstate the reductant.
Common mistakes
Using n = 3 for dichromate by forgetting there are two Cr atoms.
Using n = 5 for KMnO₄ in neutral solution.
Mixing normality on one side with molarity on the other.
Missing a second reducing ion, as in FeC₂O₄.
For balancing the equations behind these numbers, see our guide to balancing redox reactions, and for reactions where one element is both oxidised and reduced, our post on disproportionation reactions. If you get the right method but a wrong number, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning.
A practice routine
Find the n-factor of K₂Cr₂O₇, H₂C₂O₄, FeSO₄, FeC₂O₄ and KMnO₄ in both media.
Solve each worked example again with the other method from the one you prefer.
How many mL of 0.0200 M K₂Cr₂O₇ oxidise 10.0 mL of 0.100 M Fe²⁺ in acid? You should get 8.33 mL.
Write one line on every titration problem: "n-factors: oxidant ___, reductant ___, because ___".
Before your next practical, work out the expected burette reading from the concentrations given, so a wrong end point stands out.
