On this page
Redox balancing is one of the few chemistry skills that pays off in every paper: titration numericals, electrochemistry, the d-block chapter and organic oxidations all depend on it. Most students know the steps but still make errors in the same three places. This post gives you one reliable method, two fully worked examples (one in acid, one in base) and the traps to watch for.
Where does redox balancing appear in the syllabus?
The JEE (Main) 2026 syllabus and the NEET (UG) 2026 syllabus both include, under "Redox reactions and electrochemistry", oxidation number, rules for assigning it and balancing of redox reactions. Both practical units also list the titrations of oxalic acid against KMnO₄ and Mohr's salt against KMnO₄, which are redox reactions. The method below follows the NCERT Class 11 chapter on redox reactions. Check the current year's syllabus for your exam.
How do you assign oxidation numbers quickly?
Use these rules in order:
An element on its own has oxidation number 0.
A single-atom ion has its charge as its oxidation number.
Fluorine is always −1. Oxygen is usually −2, but −1 in peroxides such as H₂O₂ and positive in OF₂.
Hydrogen is +1 with non-metals and −1 in metal hydrides such as NaH.
The oxidation numbers in a species add up to its overall charge.
Example: chromium in Cr₂O₇²⁻. Seven oxygens give −14. The total must be −2, so 2x − 14 = −2 and x = +6.
Fractional answers are averages. In the tetrathionate ion S₄O₆²⁻ the rules give 2.5, and the NCERT chapter points out that the real values for the four sulphur atoms are +5, 0, 0 and +5. When an answer looks strange, look at the structure.
Why do students find redox hard?
Garnett and Treagust (1992) interviewed 32 students in their final year of high-school chemistry after they had been taught electrochemistry. Many had trouble even identifying which equations were redox reactions, and several misconceptions came from applying the definitions of oxidation and reduction where they did not fit. A safe habit is to rely on one test, a change in oxidation number, every time.
The half-reaction method in seven steps
Write the skeleton ionic equation.
Assign oxidation numbers and split into an oxidation half and a reduction half.
Balance every atom except O and H.
Balance O by adding H₂O to the side short of oxygen.
Balance H by adding H⁺ to the side short of hydrogen.
Balance charge by adding electrons, then multiply the halves so the electrons are equal.
Add the halves, cancel what appears on both sides, and check atoms and charge.
For basic solution, finish step 5, then add as many OH⁻ to both sides as there are H⁺, and combine H⁺ and OH⁻ into water.
Worked example 1: permanganate and oxalate in acid
This is the reaction in the oxalic acid titration.
Skeleton: MnO₄⁻ + C₂O₄²⁻ → Mn²⁺ + CO₂
Manganese goes from +7 to +2 (reduction). Carbon goes from +3 to +4 (oxidation).
Oxidation half: C₂O₄²⁻ → 2CO₂ + 2e⁻
Reduction half: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O. Check the charge: left is −1 + 8 − 5 = +2, right is +2.
Multiply the oxidation half by 5 and the reduction half by 2, so both involve 10 electrons.
Overall: 2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O
Final check: oxygen is 8 + 20 = 28 on the left and 20 + 8 = 28 on the right. Charge is −2 − 10 + 16 = +4 on the left and +4 on the right.
Using it: 25.0 mL of 0.0200 M KMnO₄ contains 0.0250 × 0.0200 = 5.00 × 10⁻⁴ mol of MnO₄⁻. The equation says 5 mol of oxalate per 2 mol of permanganate, so the oxalate is 5/2 × 5.00 × 10⁻⁴ = 1.25 × 10⁻³ mol. If that was in 20.0 mL of solution, its concentration is 0.0625 M.
Worked example 2: permanganate and sulphite in base
Skeleton: MnO₄⁻ + SO₃²⁻ → MnO₂ + SO₄²⁻
Manganese goes from +7 to +4 (gains 3 electrons). Sulphur goes from +4 to +6 (loses 2).
Reduction half, balanced in acid: MnO₄⁻ + 4H⁺ + 3e⁻ → MnO₂ + 2H₂O. Add 4OH⁻ to both sides and simplify: MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻
Oxidation half, the same way: SO₃²⁻ + 2OH⁻ → SO₄²⁻ + H₂O + 2e⁻
Multiply the reduction half by 2 and the oxidation half by 3 (6 electrons each), add, and cancel water and hydroxide.
Overall: 2MnO₄⁻ + 3SO₃²⁻ + H₂O → 2MnO₂ + 3SO₄²⁻ + 2OH⁻
Check: oxygen is 8 + 9 + 1 = 18 on each side; charge is −8 on each side.
Notice that permanganate takes 5 electrons in acid but 3 when it forms MnO₂. That is why the "n-factor" of KMnO₄ depends on the medium, and why you should always balance before you calculate.
The three mistakes that cost the most marks
Adding H⁺ in a basic solution. If the question says alkaline, the final equation must not contain H⁺.
Forgetting to multiply the whole half-reaction. When you multiply by 5, the water and H⁺ get multiplied too.
Assuming the electrons from oxidation numbers alone. In C₂O₄²⁻, each carbon loses one electron, so each oxalate ion loses two. Count per formula unit, not per atom.
If the same step keeps going wrong, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning, so you can compare your half-reactions line by line. Balanced equations feed straight into cell potentials, covered in our guide to electrochemistry and the Nernst equation.
A practice routine for redox
Day 1: Twenty oxidation-number questions, including peroxides, hydrides and one structure-based case.
Day 2: Eight equations in acid, using all seven steps written out.
Day 3: Eight equations in base, plus two disproportionation reactions such as chlorine in cold alkali.
Day 4: Four titration numericals that need a balanced equation first.
A week later: Redo every equation you got wrong without looking at your old working.
Start today by balancing dichromate with Fe²⁺ in acid, then check it against the worked solution in your NCERT textbook.
