ChemistryCBSE Class 11 · Redox Reactions

Oxidation Number and Balancing Redox Equations Class 11

Updated 17 October 2026 · Fact-checked

The oxidation number is the charge an atom would carry if all bonding electrons were assigned to the more electronegative atom. The sum over a species equals its charge. To balance a redox equation, the total increase in oxidation number of the oxidised atoms must equal the total decrease of the reduced atoms (oxidation number method), or the two half-reactions must be balanced for atoms and charge and then added with equal electrons (ion-electron method). In acidic medium balance O with H₂O and H with H⁺; in basic medium add OH⁻ to cancel H⁺.

Understand Oxidation Number and Balancing Redox Equations

Rules for oxidation numbers. An atom in its elemental form is 0. A monatomic ion has its charge. Fluorine is −1 in compounds. Oxygen is −2 except in peroxides (−1), in superoxides (−1/2) and in OF₂ (+2). Hydrogen is +1 except in metal hydrides (−1). Alkali metals are +1 and alkaline earth metals +2. The sum equals zero in a neutral species and the ionic charge in an ion.

Fractional values. The calculated number can be a fraction because it is an average over atoms of the same element that are in different environments. In $S_{4} O_{6}^{2 -}$ the average is $+ 2.5$ although the real states are 0 and +5. In $F e_{3} O_{4}$ the average is $+ 8/3$ .

Types of redox reaction. Combination, decomposition (when an element is a reactant or product), displacement, and disproportionation, where one element in one oxidation state is both oxidised and reduced, as in $2 H_{2} O_{2} \to 2 H_{2} O + O_{2}$ (oxygen from −1 to −2 and to 0). In a comproportionation two states of an element converge to a single intermediate state.

Oxidation number method. (1) Write the skeleton equation. (2) Assign oxidation numbers and find the atoms that change. (3) Find the increase and decrease per formula unit (multiply by the number of atoms changing). (4) Multiply the formulas so that the total increase equals the total decrease. (5) Balance the remaining atoms, adding H₂O and H⁺ (acidic) or H₂O and OH⁻ (basic), and verify atoms and charge.

Ion-electron (half-reaction) method. (1) Split into oxidation and reduction half-reactions. (2) Balance every atom except H and O. (3) Balance O with H₂O. (4) Balance H with H⁺. (5) Balance charge with electrons. (6) In basic medium add one OH⁻ for every H⁺ on both sides, combining H⁺ + OH⁻ into H₂O. (7) Multiply to equalise the electrons, add and cancel.

Oxidation Number and Balancing Redox Equations: formulas and key facts

Sum rule: $\sum (oxidation numbers) = charge on the species$; Zero for a neutral molecule.
Balance of electrons: $(increase per atom \times n)_{o x} = (decrease per atom \times n)_{r e d}$; n is the number of atoms of that element changing.
Reduction half-reaction, permanganate in acid: $M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O$; A standard half-reaction in acidic medium.
Reduction half-reaction, permanganate in base: $M n O_{4}^{-} + 2 H_{2} O + 3 e^{-} \to M n O_{2} + 4 O H^{-}$; Neutral or weakly basic solution gives MnO₂.
Dichromate in acid: $C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 C r^{3 +} + 7 H_{2} O$; Six electrons per dichromate.

How to solve Oxidation Number and Balancing Redox Equations questions

Identify the atoms that change oxidation number, balance the electron exchange, and then balance everything else by H₂O and H⁺ or OH⁻. Verify with an atom count and a charge count at the end.

1Assign oxidation numbers to every element and note which change.
2For the oxidation number method: compute the net change per formula unit and cross-multiply.
3For the half-reaction method: write and balance each half-reaction for atoms (other than H and O), O, H and charge.
4In basic medium convert H⁺ into H₂O by adding OH⁻ to both sides.
5Add the equations after equalising the electrons and cancel the common species.
6Check atoms of every element and the net charge on both sides.

Oxidation Number and Balancing Redox Equations in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

An iron nail in a copper salt solution

Suppose an iron nail of mass 5.6 g reacts completely in an excess of copper(II) sulphate solution: Fe + Cu²⁺ → Fe²⁺ + Cu (Fe = 56, Cu = 63.5 g mol⁻¹).

Find the mass of copper deposited and the number of moles of electrons transferred.

Try it first, then show the solutionSolution

Moles of Fe = 5.6/56 = 0.10 mol, and one mole of Fe gives one mole of Cu.
Mass of Cu = 0.10 × 63.5 = 6.35 g. Each Fe loses 2 e⁻, so 0.20 mol of electrons are transferred.

Answer: 6.35 g of Cu; 0.20 mol of electrons.

Memory hook: Count electrons per atom, then multiply by moles.

Oxygen bubbles from an antiseptic

Suppose 3.4 g of pure H₂O₂ (34 g mol⁻¹) decomposes completely: 2H₂O₂ → 2H₂O + O₂.

Find the mass of oxygen released and name the type of reaction.

Try it first, then show the solutionSolution

Moles of H₂O₂ = 3.4/34 = 0.10 mol; moles of O₂ = 0.05 mol.
Mass = 0.05 × 32 = 1.6 g. Oxygen goes from −1 to −2 and to 0, so it is a disproportionation.

Answer: 1.6 g of O₂; disproportionation.

Memory hook: The same element rises and falls: that is disproportionation.

Tricks to remember Oxidation Number and Balancing Redox Equations

Short memory aids. Say each one aloud once, then use it on a question the same day.

OIL RIG
Oxidation Is Loss, Reduction Is Gain of electrons.
Use it when: Naming half-reactions.
Average, not real
A fractional oxidation number is an average over atoms in different environments.
Use it when: S₄O₆²⁻, Fe₃O₄.
Acid: H₂O then H⁺, base: add OH⁻
Balance O with H₂O, H with H⁺; in base add the same number of OH⁻ to both sides.
Use it when: Ion-electron method.
Final check
Atoms and charge on both sides before you stop.
Use it when: Every balancing question.

Oxidation Number and Balancing Redox Equations: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Oxidation numbers including a fractional value · 2 marks · Very short answerBoard level

Find (a) the oxidation number of iron in K₄[Fe(CN)₆] (cyanide is −1) and (b) the average oxidation number of sulphur in the tetrathionate ion, S₄O₆²⁻.

Try it first, then show the solutionSolution

(a) 4(+1) + x + 6(−1) = 0, so x = +2. (1)
(b) 4x + 6(−2) = −2, so 4x = +10 and x = +2.5 (an average over four sulphur atoms). (1)

Answer: (a) +2; (b) +2.5.

Example 2: Disproportionation of hydrogen peroxide · 3 marks · Short answerHarder board level

Show that 2H₂O₂ → 2H₂O + O₂ is a disproportionation reaction, identifying the species oxidised, the species reduced and the change in oxidation number.

Try it first, then show the solutionSolution

In H₂O₂ oxygen is −1 (a peroxide); in H₂O it is −2 and in O₂ it is 0. (1)
Oxygen goes from −1 to −2 (decrease of 1): reduction, so H₂O₂ acts as an oxidising agent. (1)
Oxygen goes from −1 to 0 (increase of 1): oxidation, so H₂O₂ also acts as a reducing agent. One element is oxidised and reduced in the same reaction: disproportionation. (1)

Answer: Oxygen is reduced (−1 to −2) and oxidised (−1 to 0) in the same reaction.

Example 3: Oxidation number method in acidic medium · 3 marks · Short answerHarder board level

Balance by the oxidation number method in acidic medium: Cr₂O₇²⁻ + SO₃²⁻ → Cr³⁺ + SO₄²⁻.

Try it first, then show the solutionSolution

Cr: +6 to +3, decrease 3 × 2 atoms = 6 per Cr₂O₇²⁻. S: +4 to +6, increase 2 per SO₃²⁻. LCM gives 1 Cr₂O₇²⁻ : 3 SO₃²⁻. (1)
Cr₂O₇²⁻ + 3SO₃²⁻ → 2Cr³⁺ + 3SO₄²⁻. The charge on the left is −8, on the right is 0, so add 8H⁺ to the left. (1)
O: left 7 + 9 = 16, right 12, so add 4H₂O on the right; H is then 8 = 8. Final: Cr₂O₇²⁻ + 3SO₃²⁻ + 8H⁺ → 2Cr³⁺ + 3SO₄²⁻ + 4H₂O. (1)

Answer: Cr₂O₇²⁻ + 3SO₃²⁻ + 8H⁺ → 2Cr³⁺ + 3SO₄²⁻ + 4H₂O.

Example 4: Ion-electron method in basic medium · 4 marks · Long answerHarder board level

Balance by the half-reaction method in basic medium: MnO₄⁻ + I⁻ → MnO₂ + I₂.

Try it first, then show the solutionSolution

Oxidation: 2I⁻ → I₂ + 2e⁻. (1)
Reduction: MnO₄⁻ + 4H⁺ + 3e⁻ → MnO₂ + 2H₂O; adding 4OH⁻ to both sides gives MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻. (1)
Multiply the oxidation by 3 and the reduction by 2 to equalise the electrons at 6. (1)
Add: 2MnO₄⁻ + 4H₂O + 6I⁻ → 2MnO₂ + 8OH⁻ + 3I₂. Check: charge −8 = −8, O 12 = 12, H 8 = 8. (1)

Answer: 2MnO₄⁻ + 6I⁻ + 4H₂O → 2MnO₂ + 3I₂ + 8OH⁻.

Example 5: Disproportionation of white phosphorus in alkali · 4 marks · Long answerHOTS / competency

Balance in basic medium: P₄ + OH⁻ → PH₃ + H₂PO₂⁻. Then (a) state the change in oxidation number for each product, and (b) find the moles of OH⁻ consumed and of PH₃ formed when 1.24 g of P₄ (P = 31) reacts.

Try it first, then show the solutionSolution

Phosphorus: 0 to −3 in PH₃ (gain of 3) and 0 to +1 in H₂PO₂⁻ (loss of 1). One P gaining 3 balances three P losing 1 each, so the ratio PH₃ : H₂PO₂⁻ = 1 : 3. (1)
P₄ + 3OH⁻ → PH₃ + 3H₂PO₂⁻; then add 3H₂O on the left to balance O and H: P₄ + 3OH⁻ + 3H₂O → PH₃ + 3H₂PO₂⁻. (1)
Check: P 4 = 4; O 6 = 6; H 9 = 9; charge −3 = −3. (1)
(b) Moles of P₄ = 1.24/124 = 0.010 mol, so OH⁻ consumed = 0.030 mol and PH₃ formed = 0.010 mol. (1)

Answer: P₄ + 3OH⁻ + 3H₂O → PH₃ + 3H₂PO₂⁻; 0.030 mol OH⁻ and 0.010 mol PH₃.

Where students lose marks in Oxidation Number and Balancing Redox Equations

The slips examiners see most often, and what to write instead.

Mistake: Assigning −2 to oxygen in a peroxide such as H₂O₂ or BaO₂.
Using the usual value without checking the structure.
Do this: In peroxides oxygen is −1; in OF₂ it is +2.
Usually costs: The oxidation number.
Mistake: Forgetting to multiply the change per atom by the number of atoms in the formula (for example Cr in Cr₂O₇²⁻).
Using the change for one atom only.
Do this: Total change = change per atom × atoms changing.
Usually costs: The balancing.
Mistake: Writing H⁺ in a basic-medium equation.
Skipping the OH⁻ conversion step.
Do this: Add OH⁻ to both sides to turn each H⁺ into H₂O and cancel water.
Usually costs: The final equation.
Mistake: Not checking charge at the end.
Balancing atoms only.
Do this: The net charge must be equal on both sides.
Usually costs: Final marks.

Exam tips

Write the oxidation number above each atom that changes and mark the change as a number.
Name the oxidising agent and reducing agent as the species, not the atom.
State whether the medium is acidic or basic before you balance.
End the answer with an atom count and a charge count.
Explain a fractional oxidation number as an average.

Practice questions on Oxidation Number and Balancing Redox Equations

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The oxidation number of nitrogen in hydroxylamine, NH₂OH, is
(a) −3 (b) −1 (c) +1 (d) +3
Hint
N + 3(+1) + (−2) = 0.
Answer
(b) −1.
Q2 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The oxidation number of sulphur in Na₂S₄O₆ can be a fraction.
Reason (R): The oxidation number obtained by calculation is an average over all atoms of that element in the species.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Average over inequivalent sulphur atoms.
Answer
(a).
Q3 · 2 marks · Very short answerBoard level
In the reaction 2Al + Fe₂O₃ → Al₂O₃ + 2Fe, identify the oxidising agent and the reducing agent and write the change in oxidation number of each.
Hint
Al is a metal in the elemental state.
Answer
Al goes from 0 to +3 (oxidised), so Al is the reducing agent. Fe goes from +3 to 0 (reduced), so Fe₂O₃ is the oxidising agent.
Q4 · 3 marks · Short answerHarder board level
Balance by the half-reaction method in acidic medium: H₂O₂ + I⁻ → I₂ + H₂O. Write both half-reactions.
Hint
Reduction: H₂O₂ + 2H⁺ + 2e⁻ → 2H₂O.
Answer
Oxidation: 2I⁻ → I₂ + 2e⁻. Reduction: H₂O₂ + 2H⁺ + 2e⁻ → 2H₂O. Adding: H₂O₂ + 2I⁻ + 2H⁺ → I₂ + 2H₂O.
Q5 · 4 marks · Long answerHOTS / competency
Balance by the oxidation number method in acidic medium: As₂S₃ + NO₃⁻ → H₃AsO₄ + SO₄²⁻ + NO. State the total number of electrons lost per formula unit of As₂S₃ and the number of NO₃⁻ that react with 3 formula units.
Hint
As goes +3 to +5 and S goes −2 to +6; N goes +5 to +2.
Answer
Per As₂S₃: As 2 × 2 = 4 and S 3 × 8 = 24, so 28 electrons lost. N gains 3, so 3As₂S₃ react with 28NO₃⁻. 3As₂S₃ + 28NO₃⁻ + 4H₂O + 10H⁺ → 6H₃AsO₄ + 9SO₄²⁻ + 28NO. Check: H 18 = 18, O 88 = 88, charge −18 = −18.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Oxidation number of O in H₂O₂?Show answer
−1.
2.What is a disproportionation reaction?Show answer
One element is oxidised and reduced in the same reaction.
3.How do you balance O in the half-reaction method?Show answer
With H₂O.
4.What is added to convert acidic to basic equations?Show answer
OH⁻ on both sides.
5.Oxidation number of H in NaH?Show answer
−1.
6.Electrons gained by Cr₂O₇²⁻ in acid?Show answer
Six.

Oxidation Number and Balancing Redox Equations: frequently asked questions

Can the oxidation number be zero in a compound?

Yes, for an atom in a particular environment, for example carbon in CH₂Cl₂, even though no atom is in the elemental state.

Why do both balancing methods give the same equation?

Both conserve atoms and charge; they differ only in the route.

Is a decomposition reaction always a redox reaction?

No. It is redox only when an element appears in the products in a changed oxidation state, such as the free element.

Is comproportionation the opposite of disproportionation?

Yes: two oxidation states of an element form a single intermediate one.

Which method is better for ionic equations?

The ion-electron method, because it works directly with the ions and charges.