Home › Chemistry Class 11 › Redox Reactions

Redox Reactions

Chemistry · Class 11NCERT pp. 235–2522 sets × 25 questions

Key facts and formulas for Redox Reactions, three worked sample questions, and two full practice sets. Every question has a hint, a reason for every option and a step-by-step solution, with a calculator or animation where one helps. English and Hinglish.

Start the 50-question practice

Formula sheet for this chapter

IdeaRule or fact
Oxidation number rulesFree element = 0; sum in a neutral compound = 0; sum in an ion = its charge. F = −1, H = +1 (−1 in metal hydrides), O = −2 (−1 in peroxides, −½ in superoxides, +2 in OF₂)
Highest and lowest statesHighest = group number − 10 (or valence electrons for s-block and d-block). Lowest = group number − 18. Example N: +5 and −3. Os and Ru reach +8
Oxidation and reductionOxidation = loss of electrons = rise in oxidation number. Reduction = gain of electrons = fall. The oxidant is reduced and the reductant is oxidised
Intermediate state ruleA species at its highest state can only be an oxidant, at its lowest only a reductant, in between it can be both (SO₂, H₂O₂, HNO₂)
DisproportionationOne element in one species is both oxidised and reduced: Cl₂ + 2OH⁻ → ClO⁻ + Cl⁻, 2Cu⁺ → Cu²⁺ + Cu. Comproportionation is the reverse: 2H₂S + SO₂ → 3S
Balancing in acidDichromate: Cr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O. Permanganate: MnO₄⁻ + 5e⁻ → Mn²⁺ in acid, 3 e⁻ to MnO₂ in neutral
Redox titration ratios1 Cr₂O₇²⁻ : 6 Fe²⁺; 2 MnO₄⁻ : 5 C₂O₄²⁻; 1 FeC₂O₄ loses 3 e⁻. Millimoles = volume (mL) × molarity
Halogen displacementOxidising power F₂ > Cl₂ > Br₂ > I₂. Each displaces the heavier halide. F⁻ is oxidised only by electrolysis. I₂ in CCl₄ is violet, Br₂ is orange-brown
Electrode potentialE°cell = E°cathode − E°anode. Higher E° = stronger oxidant. Metals with negative E° release H₂ from acids; Cu (+0.34 V) does not
Daniell cellZn is the anode (oxidation), Cu the cathode (reduction). Electrons flow Zn to Cu in the wire. Salt bridge carries ions. E°cell = 1.10 V

Three sample questions

Q1. The oxidation number of chromium in K₂Cr₂O₇ is
  1. +3
  2. +6
  3. +7
  4. +12
Hint and solution

Hint: The compound is neutral. K is +1 and O is −2. Let Cr be x and make the sum zero.

  1. K = +1, O = −2, Cr = x — Write the known numbers. Potassium is +1 and oxygen is −2 in this compound.
  2. 2(+1) + 2x + 7(−2) = 0 — The sum of all oxidation numbers in a neutral compound is zero. There are 2 K, 2 Cr and 7 O.
  3. 2 + 2x − 14 = 0 → 2x = 12 — Simplify the equation. Move the numbers to one side.
  4. x = +6 — One chromium atom is +6. Option B.
Answer: B. 2(+1) + 2x + 7(−2) = 0, so x = +6.
Q2. Consider: (i) the oxidising agent is itself reduced; (ii) oxidation is an increase in oxidation number; (iii) reduction is loss of electrons; (iv) the reducing agent gains electrons. Which statements are correct?
  1. (i) and (iii)
  2. (ii) and (iii)
  3. (iii) and (iv)
  4. (i) and (ii)
Hint and solution

Hint: Use OIL RIG: oxidation is loss, reduction is gain. An agent does the opposite to itself.

  1. Oxidation = loss of e⁻, number rises — Oxidation means loss of electrons and an increase in oxidation number. Reduction is the opposite.
  2. Agent does the opposite to itself — An oxidising agent causes oxidation of others, so it itself is reduced. A reducing agent is itself oxidised.
  3. Check each statement — (i) true, (ii) true, (iii) false because reduction is gain, (iv) false because a reductant loses electrons.
  4. Correct: (i) and (ii) — So the correct pair is (i) and (ii). Option D.
Answer: D. The oxidant is reduced and oxidation raises the oxidation number. Reduction is gain, and a reductant loses electrons.
Q3. The average oxidation number of sulphur in Na₂S₂O₃ is
  1. +2
  2. +4
  3. +6
  4. −2
Hint and solution

Hint: Na is +1 and O is −2. Solve for the two sulphur atoms together, then divide by two.

  1. Na = +1, O = −2, S = x — Write the known numbers for sodium and oxygen.
  2. 2(+1) + 2x + 3(−2) = 0 — The compound is neutral, so the sum is zero. There are 2 Na, 2 S and 3 O.
  3. 2 + 2x − 6 = 0 → 2x = 4 — Simplify to find the total for both sulphur atoms.
  4. x = +2 (average) — Divide by two atoms: average S is +2. Option A.
Answer: A. 2(+1) + 2x + 3(−2) = 0, so x = +2.

How to use this chapter

Read the NCERT pages first, then attempt Set 1 without the hints. Note every wrong answer in a mistake list, read the reason for each option, and retry Set 2 two days later. A topic is solid when you can say why each wrong option is wrong.

← EquilibriumOrganic Chemistry →
Start practice