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Balancing Redox Reactions by the Oxidation Number Method (Step by Step)

Chemistry · Redox Equilibrium · NEET

In the oxidation number method you balance a redox reaction by making the total electrons LOST equal to the total electrons GAINED. First find the oxidation number of every atom, see which ones go up (oxidised) and which go down (reduced), then cross-multiply so the rise equals the fall, and finally balance charge with H+ or OH- and balance atoms with H2O. Memory hook: "What goes UP must come DOWN by the same amount." This shows up in NEET as "find the coefficients a, b, c" questions, so it is a direct-marks topic.
Oxidation Number Method: match the rise and fallMnO4-Mn 2+Mn: +7 to +2 = gains 5 e- (reduced)C2O4 2-CO2C: +3 to +4 (x2 C) = loses 2 e- each (oxidised)Cross-multiply2 : 5, then 16 H+
The electrons gained by manganese (falls +7 to +2, gains 5) must equal the electrons lost by carbon (rises +3 to +4, 2 carbons lose 2 each). Cross-multiplying gives the 2 : 5 ratio and 16 H+ balance charge.

Your doubts, answered

What are the exact 5 steps of the oxidation number method?

NCERT gives 5 steps. Step 1: write the correct formula of every reactant and product. Step 2: assign oxidation numbers to all atoms and mark the ones that change. Step 3: find the increase and decrease in oxidation number per atom (and per molecule/ion), then multiply each side by a small number so the total increase equals the total decrease. Step 4: if the reaction is in water, add H+ (acidic medium) or OH- (basic medium) to one side so the total charge is equal on both sides. Step 5: add H2O molecules to balance hydrogen, then check that oxygen is also balanced. If O matches, the equation is fully balanced.

Why do we cross-multiply the change in oxidation number?

Electrons are never created or destroyed. The atom that is oxidised loses exactly as many electrons as the atom that is reduced gains. The change in oxidation number tells you how many electrons each atom moves. If one atom rises by 5 and another falls by 2, you multiply the first by 2 and the second by 5, so both moves become 10. This makes electrons lost = electrons gained, which is the whole rule.

When do I add H+ and when do I add OH-?

Look at the medium given in the question. If the reaction is in acidic solution, add H+ ions. If it is in basic (alkaline) solution, add OH- ions. You add them to the side that needs charge fixing so that the total charge is the same on both sides. Never add both in the same equation. In NEET, most MnO4- and Cr2O7 2- questions are acidic, so you will usually add H+.

How is this different from the half-reaction method?

Both give the same balanced equation. In the oxidation number method you work with the WHOLE equation at once and match the rise and fall of oxidation numbers. In the half-reaction (ion-electron) method you split the reaction into two separate half-reactions, balance each with electrons, then add them. Oxidation number method is faster for quick MCQ coefficient questions; half-reaction is cleaner for messy ionic equations.

After matching electrons, how do I finish balancing H and O?

First balance charge with H+ or OH-. Then add H2O to the side short of hydrogen to balance H atoms. Do NOT touch O directly. If your H+/OH- and H2O steps are correct, the oxygen atoms will automatically be equal on both sides. Always count O at the end as your final check; if O does not match, an earlier step is wrong.

What is the n-factor and how does it give the coefficients?

The n-factor is the total change in oxidation number for that species (electrons lost or gained per formula unit). For MnO4- going Mn +7 to +2, n = 5. For oxalate C2O4 2- with each C going +3 to +4 (2 carbons), n = 2. The mole ratio in the balanced equation is the INVERSE of the n-factors: MnO4- : oxalate = 2 : 5. That is why cross-multiplying works.

⚠️ The NEET trap
✗ Adding H+ ions in a basic-medium reaction, or forgetting to make the oxidation-number rise equal the fall before writing coefficients.
✓ For MnO4- + C2O4 2- + H+ in acidic medium, Mn falls by 5 and each oxalate rises by 2, so the ratio is 2 MnO4- : 5 oxalate, and balancing charge needs 16 H+. The coefficients are 2, 5, 16.
🧠 First balance the electrons (up = down), THEN balance charge with H+/OH-, THEN balance H with water. Doing these in the wrong order gives a wrong coefficient and a lost mark.

Real NEET questions

NEET 2018

For the redox reaction MnO4- + C2O4^2- + H+ -> Mn^2+ + CO2 + H2O, the correct coefficients of the reactants (MnO4-, C2O4^2-, H+ respectively) for the balanced equation are:

A · 2, 16, 5
B · 2, 5, 16 ✓
C · 16, 5, 2
D · 5, 16, 2
Solution: Mn goes from +7 in MnO4- to +2 in Mn^2+, so it gains 5 electrons (n-factor = 5). Each C goes from +3 in C2O4^2- to +4 in CO2, so each oxalate loses 2 electrons (n-factor = 2). To make electrons lost equal to electrons gained, take 2 MnO4- (gain 10 e-) and 5 C2O4^2- (lose 10 e-). Balancing charge in acidic medium needs 16 H+: 2MnO4- + 5C2O4^2- + 16H+ -> 2Mn^2+ + 10CO2 + 8H2O. So the reactant coefficients are 2, 5, 16, which is option (b).
NEET 2023 Phase 1

On balancing the redox reaction a Cr2O7^2- + b SO3^2- + c H+ -> 2Cr^3+ + b SO4^2- + (c/2) H2O, the coefficients a, b and c are, respectively:

A · 1, 3, 8 ✓
B · 3, 8, 1
C · 1, 8, 3
D · 8, 1, 3
Solution: Cr is reduced from +6 to +3; 2 Cr atoms gain 6 electrons total, so Cr2O7^2- has n-factor 6. S is oxidised from +4 in SO3^2- to +6 in SO4^2-, losing 2 electrons each. To match 6 electrons gained, you need 3 sulphite ions (3 x 2 = 6 electrons lost). Balancing charge in acidic medium gives: Cr2O7^2- + 3SO3^2- + 8H+ -> 2Cr^3+ + 3SO4^2- + 4H2O. So a = 1, b = 3, c = 8, which is option (a).

Solved Redox Equilibrium NEET PYQs

Try the real previous-year questions from this chapter — each with the answer and a full solution.

See all 22 Redox Equilibrium NEET PYQs ›
Next concept: Balancing Redox Equations by the Half-Reaction Method (Acidic and Basic Medium) →Keep learning — 2 minFeeling ready? Solve the Redox Equilibrium NEET PYQs ›Or practice on your phone — get the free MedicNEET app ›

Frequently asked

Does the oxidation number method work for reactions not in water?

Yes. Steps 1 to 3 (matching the rise and fall of oxidation number) work for any redox reaction. Steps 4 and 5 (adding H+/OH- and H2O) are only needed when the reaction happens in aqueous solution. For a dry reaction like C + 2H2SO4 -> CO2 + 2SO2 + 2H2O you only balance atoms after matching the electron change.

How do I know which medium a NEET question uses if it is not stated?

The species in the equation tell you. If you see H+ among the reactants or products, it is acidic. If you see OH-, it is basic. If H2O appears with H+, treat it as acidic. In NEET, permanganate and dichromate coefficient questions are almost always acidic, so you add H+.

Is the oxidation number method the same as the ion-electron method?

No. The ion-electron (half-reaction) method is a separate technique where you split the reaction into two halves. The oxidation number method balances the whole equation together using oxidation-number changes. Both give the same final answer, so NEET may ask either, but coefficient MCQs are fastest with the oxidation number method.

Why is balancing redox equations important for NEET?

NEET repeatedly asks for the coefficients of a balanced redox equation (2018 and 2023 are examples). These are quick, formula-based marks. It also builds the base for redox titration and equivalence-point numericals, where the mole ratio you get from balancing decides the final concentration.

Related concepts:
Balancing Redox by Half-Reaction Method (Acidic and Basic)How to Find Oxidation Number (Worked Examples)Rules for Assigning Oxidation NumbersKMnO4 as an Oxidising AgentRedox Titrations and Equivalence-Point Calculations
Practise Chemistry for NEET:
Chemistry MCQs (free, chapter-wise)Redox Equilibrium PYQs