NEET UG Chemistry · Chapter 7

Redox Reactions

Oxidation number, types of redox reactions, balancing by oxidation-number and half-reaction methods.

NCERT Class 11NEET: 1 Q per paper 12 min read

1. Chapter at a Glance

In a redox reaction one substance loses electrons (oxidation) while another gains them (reduction). Burning, rusting, respiration, photosynthesis, bleaching and batteries are all redox processes. This chapter teaches how to find oxidation numbers, how to classify redox reactions, and how to balance them. It also connects to titrations and to electrochemistry. NEET usually asks one question from here, most often on oxidation numbers or balancing.

What NEET asks most
  • Oxidation number of an element in a compound or ion (including fractional values).
  • Identifying oxidising and reducing agents.
  • Types: combination, decomposition, displacement, disproportionation.
  • Balancing by the ion–electron (half-reaction) method in acid or base.
  • Simple redox titration calculations (KMnO₄, K₂Cr₂O₇).

2. Oxidation and Reduction

IdeaOxidationReduction
ClassicalAddition of oxygen / removal of hydrogenRemoval of oxygen / addition of hydrogen
ElectronicLoss of electronsGain of electrons
Oxidation numberIncrease in ONDecrease in ON
The agent that causes itOxidising agent (itself reduced)Reducing agent (itself oxidised)
Memory aid

OIL RIG: Oxidation Is Loss, Reduction Is Gain (of electrons).

Example: Zn + Cu²⁺ → Zn²⁺ + Cu. Zinc loses electrons (oxidised, reducing agent); Cu²⁺ gains electrons (reduced, oxidising agent).

3. Rules for Oxidation Number

  1. An element in its free state has ON = 0 (Na, O₂, S₈, P₄).
  2. For a monatomic ion, ON = its charge.
  3. Oxygen is −2, except in peroxides (−1, as in H₂O₂ and Na₂O₂), superoxides (−½, as in KO₂) and OF₂ (+2).
  4. Hydrogen is +1, except in metal hydrides (−1, as in NaH and CaH₂).
  5. Fluorine is always −1. Other halogens are −1 except when bonded to oxygen or a more electronegative halogen.
  6. Alkali metals are +1 and alkaline-earth metals are +2 in their compounds.
  7. The sum of ONs equals 0 for a neutral compound and equals the charge for an ion.
SpeciesElementONNEET Note
KMnO₄Mn+7Maximum for Mn
K₂Cr₂O₇Cr+6Orange dichromate
CrO₅Cr+6Has two peroxide bridges; not +10
H₂SO₅ (Caro's acid)S+6One peroxide linkage
Na₂S₂O₃S+2 (average)The two S atoms are in different environments
Na₂S₄O₆S+2.5 (average)Fractional ON is an average
Fe₃O₄Fe+8/3 (average)Mixed Fe(II) and Fe(III)
H₃PO₂, H₃PO₃, H₃PO₄P+1, +3, +5Count only the P–OH hydrogens as acidic

4. Types of Redox Reactions

  • Combination: C + O₂ → CO₂.
  • Decomposition: 2KClO₃ → 2KCl + 3O₂ (not every decomposition is redox; CaCO₃ → CaO + CO₂ is not).
  • Displacement: a more reactive metal displaces a less reactive one (Zn + CuSO₄ → ZnSO₄ + Cu), or a more reactive halogen displaces another (Cl₂ + 2KBr → 2KCl + Br₂).
  • Disproportionation: the same element is both oxidised and reduced. Examples: 2H₂O₂ → 2H₂O + O₂ (O: −1 → −2 and 0); Cl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂O; P₄ + 3OH⁻ + 3H₂O → PH₃ + 3H₂PO₂⁻.
  • Comproportionation is the reverse: two ONs of an element combine to an intermediate ON.
  • Fluorine cannot disproportionate because it has only one negative ON (−1) and no positive ON.
Zn Cu²⁺ 2e⁻ transferred oxidised (ON 0 → +2)reducing agent reduced (+2 → 0)oxidising agent
In Zn + Cu²⁺ → Zn²⁺ + Cu, electrons flow from zinc to copper ions. The substance that loses electrons is the reducing agent.

5. Balancing Redox Equations

Ion–electron (half-reaction) method

  1. Split the reaction into oxidation and reduction half-reactions.
  2. Balance atoms other than O and H.
  3. Balance O by adding H₂O; balance H by adding H⁺ (in acid).
  4. Balance charge by adding electrons.
  5. Multiply the half-reactions so that the electrons lost equal the electrons gained, then add them.
  6. In basic medium, add OH⁻ to both sides to cancel H⁺ (forming H₂O).

Oxidation-number method: find the total increase and total decrease of ON, make them equal with coefficients, then balance O with H₂O and H with H⁺ (or OH⁻).

OxidantMediumChangeElectrons gained
MnO₄⁻Acidic→ Mn²⁺ (colourless)5
MnO₄⁻Neutral / weakly basic→ MnO₂ (brown)3
MnO₄⁻Strongly basic→ MnO₄²⁻ (green)1
Cr₂O₇²⁻Acidic→ 2Cr³⁺ (green)6
H₂O₂Acidic→ H₂O2

6. Redox Titrations and Equivalents

  • In a redox titration, the moles of electrons given by the reducing agent equal the moles of electrons taken by the oxidising agent.
  • Useful relation: n₁ × (moles of oxidant) = n₂ × (moles of reductant), where n is the change in electrons per formula unit (the n-factor).
  • KMnO₄ acts as its own indicator (the first extra drop turns the solution pink). Dichromate titrations use an indicator such as diphenylamine.
  • Redox reactions are also the basis of electrochemical cells: the more negative electrode potential belongs to the stronger reducing agent (see Electrochemistry).

7. Formula Master Table

QuantityRule / formula
Sum of ONs0 (molecule) or ion charge
OxidationON increases; electrons lost
ReductionON decreases; electrons gained
n-factor (redox)change in ON × atoms per formula unit
Titration equalityn₁M₁V₁ = n₂M₂V₂
Equivalent weightmolar mass/n-factor
MnO₄⁻ half-reaction (acid)MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
Cr₂O₇²⁻ half-reaction (acid)Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O

8. Master Summary Table

ConceptKey ideaNEET Note
Oxidising agentGains electronsIts ON falls
Reducing agentLoses electronsIts ON rises
Fractional ONAverage valueFe₃O₄, S₄O₆²⁻
Peroxide trapO = −1CrO₅, H₂SO₅
DisproportionationSame element up and downH₂O₂, Cl₂ in alkali, P₄
KMnO₄ n-factor5 / 3 / 1Acid / neutral / strong base
BalancingElectrons lost = gainedThen atoms and charge

9. Mind Map

Redox Reactions
Definitions
  • Oxygen/hydrogen
  • Electron transfer
  • ON change
Oxidation number
  • Rules
  • Fractional ON
  • Peroxides
Types
  • Combination
  • Displacement
  • Disproportionation
Balancing
  • Half-reaction
  • ON method
  • Titrations

10. Tips and Tricks

Shortcuts
  • A species with an element in its highest ON (KMnO₄, K₂Cr₂O₇, HNO₃) can only act as an oxidising agent.
  • A species with an element in its lowest ON (H₂S, NH₃, I⁻) can only act as a reducing agent.
  • Species in an intermediate ON (H₂O₂, SO₂, HNO₂) can act as both, and can disproportionate.
  • Check a balanced equation by confirming both atoms and total charge are equal.
Common NEET traps
  • Giving Cr in CrO₅ an ON of +10 (it is +6, because of peroxide oxygen).
  • Treating the fractional ON in S₄O₆²⁻ as the ON of each S atom.
  • Assuming every decomposition reaction is a redox reaction.
  • Using n = 5 for KMnO₄ in a neutral solution.

11. Important One-liners

  • Oxidation and reduction always occur together.
  • Fluorine is the strongest oxidising agent and always has ON −1.
  • Oxygen has ON +2 in OF₂.
  • Hydrogen has ON −1 in metal hydrides such as NaH.
  • H₂O₂ can act both as an oxidising and a reducing agent.
  • The reaction Cl₂ + 2NaOH → NaCl + NaOCl + H₂O is a disproportionation.
  • KMnO₄ titrations need no external indicator.

12. Solved Numericals: Average to Advanced

Five average and five advanced numericals. Atomic mass of Fe = 55.85. Click to reveal each answer and working.

Average level

N1Average

Find the oxidation number of Cr in K₂Cr₂O₇, S in H₂SO₄ and Mn in KMnO₄.

Show answer & solution

Concept: the ONs in a neutral compound add up to zero.

  1. K₂Cr₂O₇: 2(+1) + 2x + 7(−2) = 0 ⇒ 2x = 12 ⇒ x = +6.
  2. H₂SO₄: 2(+1) + x + 4(−2) = 0 ⇒ x = +6.
  3. KMnO₄: +1 + x + 4(−2) = 0 ⇒ x = +7.

Answer: Cr +6; S +6; Mn +7

N2Average

Find the average oxidation number of S in Na₂S₄O₆ and of Fe in Fe₃O₄.

Show answer & solution

Concept: the average ON can be a fraction.

  1. Na₂S₄O₆: 2(+1) + 4x + 6(−2) = 0 ⇒ 4x = 10 ⇒ x = +2.5.
  2. Fe₃O₄: 3x + 4(−2) = 0 ⇒ x = +8/3 ≈ +2.67 (really one Fe²⁺ and two Fe³⁺).

Answer: S +2.5; Fe +8/3

N3Average

In the reaction Zn + CuSO₄ → ZnSO₄ + Cu, identify the species oxidised, the species reduced, the oxidising agent and the reducing agent. How many electrons are transferred per zinc atom?

Show answer & solution

Concept: follow the change in ON.

  1. Zn: 0 → +2 (oxidised; it is the reducing agent).
  2. Cu: +2 → 0 (Cu²⁺ is reduced; it is the oxidising agent).
  3. Each Zn atom loses 2 electrons. Sulphate is a spectator ion.

Answer: Zn oxidised (reducing agent); Cu²⁺ reduced (oxidising agent); 2 electrons

N4Average

Balance MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺ in acidic solution.

Show answer & solution

Concept: half-reaction method.

  1. Reduction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O.
  2. Oxidation: Fe²⁺ → Fe³⁺ + e⁻ (multiply by 5).
  3. Add: MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O (charge: +17 = +17 ✓).

Answer: MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O

N5Average

Show that Cl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂O is a disproportionation reaction.

Show answer & solution

Concept: the same element must be both oxidised and reduced.

  1. Cl in Cl₂: 0.
  2. Cl in Cl⁻: −1 (reduced); Cl in ClO⁻: +1 (oxidised).
  3. Chlorine goes both down and up in ON, so this is disproportionation.

Answer: chlorine changes from 0 to −1 and +1 at the same time

Advanced level

N6Advanced

Balance Cr₂O₇²⁻ + SO₂ → Cr³⁺ + SO₄²⁻ in acidic solution.

Show answer & solution

Concept: half-reaction method.

  1. Reduction: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O.
  2. Oxidation: SO₂ + 2H₂O → SO₄²⁻ + 4H⁺ + 2e⁻ (multiply by 3).
  3. Add and cancel: Cr₂O₇²⁻ + 3SO₂ + 2H⁺ → 2Cr³⁺ + 3SO₄²⁻ + H₂O.
  4. Check: O 13 = 13; H 2 = 2; charge 0 = 0 ✓.

Answer: Cr₂O₇²⁻ + 3SO₂ + 2H⁺ → 2Cr³⁺ + 3SO₄²⁻ + H₂O

N7Advanced

Balance MnO₄⁻ + I⁻ → MnO₂ + IO₃⁻ in basic solution.

Show answer & solution

Concept: balance as in acid, then add OH⁻ to both sides.

  1. Reduction: MnO₄⁻ + 4H⁺ + 3e⁻ → MnO₂ + 2H₂O (× 2).
  2. Oxidation: I⁻ + 3H₂O → IO₃⁻ + 6H⁺ + 6e⁻.
  3. Add: 2MnO₄⁻ + I⁻ + 2H⁺ → 2MnO₂ + IO₃⁻ + H₂O.
  4. Add 2OH⁻ to both sides: 2MnO₄⁻ + I⁻ + H₂O → 2MnO₂ + IO₃⁻ + 2OH⁻ (charge −3 = −3 ✓).

Answer: 2MnO₄⁻ + I⁻ + H₂O → 2MnO₂ + IO₃⁻ + 2OH⁻

N8Advanced

What volume of 0.02 M KMnO₄ is needed to oxidise 25 mL of 0.1 M FeSO₄ in acidic solution?

Show answer & solution

Concept: MnO₄⁻ : Fe²⁺ = 1 : 5 (from N4).

  1. Moles of Fe²⁺ = 0.1 × 0.025 = 2.5 × 10⁻³.
  2. Moles of MnO₄⁻ = 2.5 × 10⁻³/5 = 5 × 10⁻⁴.
  3. V = 5 × 10⁻⁴/0.02 = 0.025 L = 25 mL.

Answer: 25 mL

N9Advanced

Find the oxidation number of phosphorus in H₃PO₂, H₃PO₃ and H₄P₂O₇. How many ionisable hydrogens does each acid have?

Show answer & solution

Concept: sum of ONs = 0; only O–H hydrogens are acidic.

  1. H₃PO₂: 3 + x − 4 = 0 ⇒ x = +1; one P–OH, so monobasic.
  2. H₃PO₃: 3 + x − 6 = 0 ⇒ x = +3; two P–OH, so dibasic.
  3. H₄P₂O₇: 4 + 2x − 14 = 0 ⇒ x = +5; four P–OH, so tetrabasic.

Answer: +1, +3, +5; 1, 2 and 4 ionisable H

N10Advanced

20 mL of 0.05 M K₂Cr₂O₇ exactly oxidises an iron(II) solution in acid. Find the mass of iron(II) present.

Show answer & solution

Concept: Cr₂O₇²⁻ gains 6e⁻; each Fe²⁺ loses 1e⁻, so 1 Cr₂O₇²⁻ : 6 Fe²⁺.

  1. Moles of Cr₂O₇²⁻ = 0.05 × 0.020 = 1 × 10⁻³.
  2. Moles of Fe²⁺ = 6 × 10⁻³.
  3. Mass = 6 × 10⁻³ × 55.85 ≈ 0.335 g.

Answer: about 0.335 g

13. NEET-pattern Questions with Detailed Answers

These questions are original and based on concepts that NEET UG asks again and again.

Q1Easy

The oxidation number of oxygen in OF₂ is:

  1. −2
  2. −1
  3. +1
  4. +2
Show solution

F is always −1, so O = +2. This is the classic exception: oxygen is −2 in most compounds, −1 in peroxides and −½ in superoxides.

Answer: (D)

Q2Easy

In the reaction 2Na + Cl₂ → 2NaCl, sodium is:

  1. Reduced
  2. Oxidised
  3. Neither
  4. Both oxidised and reduced
Show solution

Na changes from 0 to +1 by losing an electron, so it is oxidised and acts as the reducing agent. Cl₂ is the species that is reduced.

Answer: (B)

Q3Medium

The oxidation number of Cr in CrO₅ is:

  1. +10
  2. +6
  3. +5
  4. +8
Show solution

CrO₅ has one oxygen at −2 and four peroxide oxygens at −1. So x − 2 − 4 = 0, giving x = +6. The answer +10 assumes every oxygen is −2.

Answer: (B)

Q4Medium

Which reaction is NOT a redox reaction?

  1. CaCO₃ → CaO + CO₂
  2. 2KClO₃ → 2KCl + 3O₂
  3. Zn + 2HCl → ZnCl₂ + H₂
  4. 2H₂O₂ → 2H₂O + O₂
Show solution

In CaCO₃ → CaO + CO₂, no element changes its ON: Ca stays +2, C stays +4 and O stays −2. All the others involve electron transfer.

Answer: (A)

Q5Medium

Which species can act ONLY as a reducing agent?

  1. SO₂
  2. H₂O₂
  3. H₂S
  4. HNO₂
Show solution

In H₂S, sulphur is in its lowest ON (−2), so it can only be oxidised. SO₂, H₂O₂ and HNO₂ have intermediate ONs, so they can act either way.

Answer: (C)

Q6Medium

In a neutral or weakly alkaline medium, the number of electrons gained by one MnO₄⁻ ion is:

  1. 5
  2. 3
  3. 1
  4. 7
Show solution

In neutral or weakly alkaline solution, MnO₄⁻ is reduced to MnO₂ (Mn +7 → +4), a gain of 3 electrons. It gains 5 electrons in acid and 1 electron in strongly basic solution.

Answer: (B)

Q7Hard

In the reaction P₄ + 3NaOH + 3H₂O → PH₃ + 3NaH₂PO₂, phosphorus is:

  1. Only oxidised
  2. Only reduced
  3. Both oxidised and reduced
  4. Neither oxidised nor reduced
Show solution

P starts at 0 in P₄. It goes to −3 in PH₃ (reduced) and to +1 in NaH₂PO₂ (oxidised). This is a disproportionation reaction.

Answer: (C)

Q8Hard

In the balanced equation for Cr₂O₇²⁻ + Fe²⁺ + H⁺ → Cr³⁺ + Fe³⁺ + H₂O, the coefficients of Fe²⁺ and H⁺ are:

  1. 6 and 14
  2. 3 and 7
  3. 6 and 7
  4. 5 and 8
Show solution

Cr₂O₇²⁻ gains 6 electrons and each Fe²⁺ gives 1, so 6 Fe²⁺ are needed. The full equation is Cr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O. (D) gives the coefficients for the permanganate reaction.

Answer: (A)

14. Quick Revision Summary

Remember before the exam
  • OIL RIG; oxidising agent is reduced, reducing agent is oxidised.
  • ON rules: F −1 always; O −2 except peroxide (−1), superoxide (−½), OF₂ (+2); H −1 in hydrides.
  • Fractional ON = average; CrO₅ and H₂SO₅ contain peroxide.
  • Disproportionation: H₂O₂, Cl₂/OH⁻, P₄/OH⁻.
  • Half-reaction balancing; KMnO₄ n = 5/3/1; K₂Cr₂O₇ n = 6.
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