Redox Reactions Class 11 Notes

• Classical view: oxidation meant adding oxygen or removing hydrogen; reduction meant removing oxygen or adding hydrogen.
• Modern view (by electron transfer): oxidation loses electrons, reduction gains electrons.
• The species that loses electrons is the reducing agent; the species that gains electrons is the oxidising agent.

• Free elements (O₂, N₂, Fe) have oxidation number 0.
• Monatomic ions carry their charge: Na⁺ is +1, Cl⁻ is −1.
• Hydrogen is +1 except in metal hydrides (NaH, CaH₂) where it is −1.
• Oxygen is −2, except in peroxides (H₂O₂, −1), superoxides, OF₂ (+2) and O₂F₂ (+1).
• Fluorine is always −1; alkali metals +1; alkaline earth metals +2.
• The sum of oxidation numbers in a neutral molecule is 0; in a polyatomic ion it equals the ion's charge.

• Assign oxidation numbers and identify the atoms that change state.
• Balance the total change in oxidation number: the increase on the oxidised atoms must equal the decrease on the reduced atoms.
• Balance atoms other than O and H, then add H₂O, H⁺ (acidic) or OH⁻ (basic) to balance O and H.
• Check that atoms and charge balance on both sides.

• Oxidation half: MnO₄⁻ → Mn²⁺ gains 5 e⁻ per manganese (Mn +7 → +2).
• Reduction half: Fe²⁺ → Fe³⁺ loses 1 e⁻ per iron (+2 → +3).
• Multiply the iron half by 5 so electrons cancel: 5Fe²⁺ + MnO₄⁻ + ... → 5Fe³⁺ + Mn²⁺ + ...
• Balancing in acid: add H₂O to the side short of oxygen, H⁺ to the side short of hydrogen.

• Combination: A + B → product, e.g. Na + ½Cl₂ → NaCl.
• Decomposition: a single compound splits, e.g. 2H₂O₂ → 2H₂O + O₂.
• Displacement: a more reactive element replaces a less reactive one, e.g. Zn + CuSO₄ → ZnSO₄ + Cu.
• Disproportionation: the same species is both oxidised and reduced — e.g. 2H₂O₂ → 2H₂O + O₂, and Cl₂ in alkali gives Cl⁻ and ClO⁻.
• Comproportionation: two compounds of the same element at different oxidation states meet at an intermediate one.

• Standard hydrogen electrode (SHE) is the reference: E° = 0 V for 2H⁺ + 2e⁻ → H₂ under standard conditions.
• Positive E° means the species is more easily reduced than H⁺; negative means it is oxidised more easily.
• A reaction is spontaneous when E°cell > 0.

Example: Find the oxidation number of S in H₂SO₄.

Solution: H is +1 (×2), O is −2 (×4), and the molecule is neutral: 2(+1) + S + 4(−2) = 0, so S = +6.

Example: In MnO₄⁻ → Mn²⁺, is manganese oxidised or reduced, and by how many electrons?

Solution: Mn goes from +7 to +2 — oxidation number decreases by 5, so it is reduced and each MnO₄⁻ gains 5 electrons.

Exercises (NCERT Q7.1–Q7.30)

• 7.1) Assign the oxidation number to the underlined element in each species: (a) NaH₂PO₄, (b) NaHSO₄, (c) H₄P₂O₇, (d) K₂MnO₄, (e) CaO₂, (f) NaBH₄, (g) H₂S₂O₇, (h) KAl(SO₄)₂·12H₂O.
• 7.2) What are the oxidation numbers of the underlined elements in KI₃, H₂S₄O₆, Fe₃O₄, CH₃CH₂OH and CH₃COOH? Rationalise the results, including the difference between an average oxidation number and the actual oxidation states of equivalent atoms.
• 7.3) Justify that the following reactions are redox reactions: (a) CuO + H₂ → Cu + H₂O; (b) Fe₂O₃ + 3CO → 2Fe + 3CO₂; (c) 4BCl₃ + 3LiAlH₄ → 2B₂H₆ + 3LiCl + 3AlCl₃; (d) 2K + F₂ → 2KF; (e) 4NH₃ + 5O₂ → 4NO + 6H₂O.
• 7.4) Fluorine reacts with ice to form HF and HOF: H₂O(s) + F₂(g) → HF(g) + HOF(g). Explain the redox change.
• 7.5) Calculate the oxidation number of sulphur, chromium and nitrogen in H₂SO₅, Cr₂O₇²⁻ and NO₃⁻. Suggest structures and explain the fallacy in assigning oxygen the value −2 to every O atom.
• 7.6) Write formulas for: (a) mercury(II) chloride, (b) nickel(II) sulphate, (c) tin(IV) oxide, (d) thallium(I) sulphate, (e) iron(III) sulphate and (f) chromium(III) oxide.
• 7.7) Suggest a list of substances in which carbon exhibits oxidation states from −4 to +4 and nitrogen exhibits oxidation states from −3 to +5.
• 7.8) SO₂ and H₂O₂ can act as both oxidising and reducing agents, whereas O₃ and HNO₃ act only as oxidants under ordinary conditions. Why?
• 7.9) Consider (a) 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂ and (b) O₃ + H₂O₂ → H₂O + 2O₂. Why is it more appropriate to write (a) as 6CO₂ + 12H₂O → C₆H₁₂O₆ + 6H₂O + 6O₂ and (b) as O₃ + H₂O₂ → H₂O + O₂ + O₂? Suggest a technique to investigate the path of each reaction.
• 7.10) AgF₂ is unstable, but if formed it acts as a very strong oxidising agent. Why?
• 7.11) Justify the statement that a lower oxidation state is favoured when the reducing agent is in excess, while a higher oxidation state is favoured when the oxidising agent is in excess, using three illustrations.
• 7.12) Explain (a) why alcoholic potassium permanganate is used to oxidise toluene to benzoic acid although acidic and alkaline KMnO₄ are also oxidants, and give the balanced redox equation; (b) why concentrated H₂SO₄ gives colourless pungent HCl from a chloride mixture but red bromine vapour from a bromide mixture.
• 7.13) For each reaction, identify the substance oxidised, the substance reduced, the oxidising agent and the reducing agent: (a) 2AgBr + C₆H₆O₂ → 2Ag + 2HBr + C₆H₄O₂; (b) HCHO + 2[Ag(NH₃)₂]⁺ + 3OH⁻ → HCOO⁻ + 2Ag + 4NH₃ + 2H₂O; (c) HCHO + 2Cu²⁺ + 5OH⁻ → Cu₂O + HCOO⁻ + 3H₂O; (d) N₂H₄ + 2H₂O₂ → N₂ + 4H₂O; (e) Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O.
• 7.14) Consider (a) 2S₂O₃²⁻ + I₂ → S₄O₆²⁻ + 2I⁻ and (b) S₂O₃²⁻ + Br₂ in water → SO₄²⁻. Explain the role of thiosulphate and the relative strength of iodine and bromine as oxidising agents.
• 7.15) Justify by reactions that fluorine is the best oxidant among halogens and hydroiodic acid is the best reductant among hydrohalic acids.
• 7.16) Explain: XeO₆⁴⁻ + 2F⁻ + 6H⁺ → XeO₃ + F₂ + 3H₂O. What conclusion about Na₄XeO₆ can be drawn?
• 7.17) Consider (a) H₃PO₂ + 4AgNO₃ + 2H₂O → H₃PO₄ + 4Ag + 4HNO₃; (b) H₃PO₂ + 2CuSO₄ + 2H₂O → H₃PO₄ + 2Cu + H₂SO₄; (c) C₆H₅CHO + 2[Ag(NH₃)₂]⁺ + 3OH⁻ → C₆H₅COO⁻ + 2Ag + 4NH₃ + 2H₂O; and (d) C₆H₅CHO + 2Cu²⁺ + 5OH⁻, for which no change is observed. What do these reactions indicate about Ag⁺ and Cu²⁺?
• 7.18) Balance the following redox reactions by the ion-electron method: (a) MnO₄⁻ + I⁻ → MnO₂ + I₂ in basic medium; (b) MnO₄⁻ + SO₂ → Mn²⁺ + HSO₄⁻ in acidic medium; (c) H₂O₂ + Fe²⁺ → Fe³⁺ + H₂O in acidic medium; (d) Cr₂O₇²⁻ + SO₂ → Cr³⁺ + SO₄²⁻ in acidic medium.
• 7.19) Balance the following equations in basic medium by the ion-electron and oxidation-number methods, and identify the oxidising and reducing agents: (a) P₄ + OH⁻ → PH₃ + HPO₂⁻; (b) N₂H₄ + ClO₃⁻ → NO + Cl⁻; (c) Cl₂O₇ + H₂O₂ → ClO₂⁻ + O₂ + H⁺.
• 7.20) What information can be drawn from (CN)₂ + 2OH⁻ → CN⁻ + CNO⁻ + H₂O?
• 7.21) The Mn³⁺ ion is unstable in solution and disproportionates to Mn²⁺, MnO₂ and H⁺. Write the balanced ionic equation.
• 7.22) For the elements Cs, Ne, I and F, identify (a) the element that exhibits only negative oxidation state, (b) only positive oxidation state, (c) both positive and negative states, and (d) neither positive nor negative state.
• 7.23) Excess chlorine used for water treatment is removed by sulphur dioxide in water. Present the balanced redox equation.
• 7.24) Select (a) possible non-metals and (b) three metals that can show disproportionation reactions.
• 7.25) In the first step of Ostwald’s process, 4NH₃ + 5O₂ → 4NO + 6H₂O. What is the maximum mass of NO obtainable from 10.00 g NH₃ and 20.00 g O₂?
• 7.26) Using E°(Fe³⁺/Fe²⁺) = 0.77 V, E°(I₂/I⁻) = 0.54 V, E°(Ag⁺/Ag) = 0.80 V, E°(Cu²⁺/Cu) = 0.34 V and E°(Br₂/Br⁻) = 1.09 V, predict whether each reaction is feasible: (a) Fe³⁺ and I⁻, (b) Ag⁺ and Cu, (c) Fe³⁺ and Cu, (d) Ag and Fe³⁺, (e) Br₂ and Fe²⁺.
• 7.27) Predict the products of electrolysis of: (i) aqueous AgNO₃ with silver electrodes; (ii) aqueous AgNO₃ with platinum electrodes; (iii) dilute aqueous H₂SO₄ with platinum electrodes; and (iv) aqueous CuCl₂ with platinum electrodes.
• 7.28) Arrange Al, Cu, Fe, Mg and Zn in the order in which they displace each other from solutions of their salts.
• 7.29) Given E°(K⁺/K) = −2.93 V, E°(Ag⁺/Ag) = 0.80 V, E°(Hg²⁺/Hg) = 0.79 V, E°(Mg²⁺/Mg) = −2.37 V and E°(Cr³⁺/Cr) = −0.74 V, arrange the metals in increasing order of reducing power.
• 7.30) For Zn(s) + 2Ag⁺(aq) → Zn²⁺(aq) + 2Ag(s), identify the negative electrode, the current carriers and the reaction at each electrode.

Answers at a glance

1. 1
   Q7.1 → (a) P = +5; (b) S = +6; (c) P = +5; (d) Mn = +6; (e) O = −1; (f) B = +3; (g) S = +6; (h) S = +6.
2. 2
   Q7.2 → (a) I: −1, 0, 0 (average −1/3); (b) S: +5, +5, 0, 0 (average +2.5); (c) Fe: +2, +3, +3 (average +8/3); (d) C: −3 and −1 (average −2); (e) C: −3 and +3 (average 0).
3. 3
   Q7.3 → All five reactions are redox because they contain simultaneous oxidation and reduction.
4. 4
   Q7.4 → Fluorine disproportionates: F(0) → F(−1) in HF and F(0) → F(+1) in HOF.
5. 5
   Q7.5 → S in H₂SO₅ = +6; Cr in Cr₂O₇²⁻ = +6; N in NO₃⁻ = +5. The structures are HO–S(=O)₂–O–O–H, O₃Cr–O–CrO₃, and planar NO₃⁻, respectively.
6. 6
   Q7.6 → (a) HgCl₂; (b) NiSO₄; (c) SnO₂; (d) Tl₂SO₄; (e) Fe₂(SO₄)₃; (f) Cr₂O₃.
7. 7
   Q7.7 → Carbon: −4 CH₄, −3 C₂H₆, −2 CH₃OH, −1 C₂H₂, 0 CH₂Cl₂, +1 ClC≡CCl, +2 CHCl₃ or CO, +3 CCl₃CCl₃, +4 CCl₄ or CO₂. Nitrogen: −3 NH₃, −2 N₂H₄, −1 N₂H₂, 0 N₂, +1 N₂O, +2 NO, +3 N₂O₃, +4 NO₂, +5 N₂O₅.
8. 8
   Q7.8 → SO₂ and H₂O₂ have intermediate oxidation states, so either direction is possible; O₃ and HNO₃ are restricted mainly to reduction under ordinary conditions.
9. 9
   Q7.9 → (a) Water is both used and produced, so the net equation includes 12H₂O on the reactant side and 6H₂O on the product side. (b) O₂ is produced from both O₃ and H₂O₂, so the two O₂ products are written separately. ¹⁸O labelling traces the oxygen atoms.
10. 10
    Q7.10 → AgF₂ contains Ag²⁺, which readily reduces to stable Ag⁺; therefore it is a strong oxidising agent.
11. 11
    Q7.11 → P₄ gives PF₃ or PF₅, K gives K₂O or K₂O₂, and C gives CO or CO₂ depending on whether reductant or oxidant is in excess.
12. 12
    Q7.12 → Alcoholic KMnO₄ oxidises toluene to potassium benzoate with MnO₂ formation. Concentrated H₂SO₄ liberates HCl from chloride, but HBr from bromide is further oxidised to red Br₂.
13. 13
    Q7.13 → (a) C₆H₆O₂ is oxidised and AgBr is reduced; (b) HCHO is oxidised and [Ag(NH₃)₂]⁺ is reduced; (c) HCHO is oxidised and Cu²⁺ is reduced; (d) N₂H₄ is oxidised and H₂O₂ is reduced; (e) Pb is oxidised and PbO₂ is reduced.
14. 14
    Q7.14 → I₂ oxidises S₂O₃²⁻ to S₄O₆²⁻, whereas the stronger oxidant Br₂ oxidises it to SO₄²⁻. Thiosulphate is the reducing agent in both reactions.
15. 15
    Q7.15 → F₂ is the strongest halogen oxidant, and HI is the strongest hydrohalic reducing agent; the orders are F₂ > Cl₂ > Br₂ > I₂ and HF < HCl < HBr < HI.
16. 16
    Q7.16 → Na₄XeO₆ contains Xe(+8) and is a very strong oxidising agent; it oxidises F⁻ to F₂.
17. 17
    Q7.17 → Ag⁺ is a stronger oxidising agent than Cu²⁺ under these conditions: both can be reduced by H₃PO₂, but only Ag⁺ is reduced by benzaldehyde.
18. 18
    Q7.18 → (a) 2MnO₄⁻ + 6I⁻ + 4H₂O → 2MnO₂ + 3I₂ + 8OH⁻; (b) 2MnO₄⁻ + 5SO₂ + 2H₂O + H⁺ → 2Mn²⁺ + 5HSO₄⁻; (c) H₂O₂ + 2Fe²⁺ + 2H⁺ → 2Fe³⁺ + 2H₂O; (d) Cr₂O₇²⁻ + 3SO₂ + 2H⁺ → 2Cr³⁺ + 3SO₄²⁻ + H₂O.
19. 19
    Q7.19 → (a) 5P₄ + 12H₂O + 12OH⁻ → 8PH₃ + 12HPO₂⁻; P₄ disproportionates. (b) 3N₂H₄ + 4ClO₃⁻ → 6NO + 4Cl⁻ + 6H₂O. (c) Cl₂O₇ + 4H₂O₂ + 2OH⁻ → 2ClO₂⁻ + 4O₂ + 5H₂O.
20. 20
    Q7.20 → Cyanogen disproportionates in base: C(+3) is reduced to C(+2) in CN⁻ and oxidised to C(+4) in CNO⁻.
21. 21
    Q7.21 → 2Mn³⁺ + 2H₂O → Mn²⁺ + MnO₂ + 4H⁺.
22. 22
    Q7.22 → (a) F; (b) Cs; (c) I; (d) Ne.
23. 23
    Q7.23 → Cl₂ + SO₂ + 2H₂O → 2Cl⁻ + SO₄²⁻ + 4H⁺; equivalently, Cl₂ + SO₂ + 2H₂O → 2HCl + H₂SO₄.
24. 24
    Q7.24 → Non-metals: P, Cl and S. Metals: Mn, Cu and Ga.
25. 25
    Q7.25 → Maximum mass of NO = 15.00 g; O₂ is the limiting reagent.
26. 26
    Q7.26 → (a) +0.23 V, feasible; (b) +0.46 V, feasible; (c) +0.43 V, feasible; (d) −0.03 V, not feasible; (e) +0.32 V, feasible.
27. 27
    Q7.27 → (i) Ag deposits at the cathode while the Ag anode dissolves; (ii) Ag forms at the cathode and O₂ at the anode, with HNO₃ formed; (iii) H₂ forms at the cathode and O₂ at the anode; (iv) Cu forms at the cathode and Cl₂ at the anode.
28. 28
    Q7.28 → Mg > Al > Zn > Fe > Cu.
29. 29
    Q7.29 → Increasing reducing power: Ag < Hg < Cr < Mg < K.
30. 30
    Q7.30 → Zn electrode: negative anode; Ag electrode: positive cathode. Electrons flow Zn → Ag externally; anions move toward Zn and cations toward Ag in the salt bridge. Half-reactions are Zn → Zn²⁺ + 2e⁻ and 2Ag⁺ + 2e⁻ → 2Ag.

MCQ Questions

1 Mark Questions

1. 1
   Define oxidation in terms of electron transfer. (Ans: loss of electrons / increase in oxidation number)
2. 2
   What is a redox reaction? (Ans: a reaction in which oxidation and reduction occur simultaneously)
3. 3
   Find the oxidation number of Cr in K₂Cr₂O₇. (Ans: +6)
4. 4
   What is a disproportionation reaction? (Ans: a redox reaction in which the same species is simultaneously oxidised and reduced — e.g. 2Cu⁺ → Cu + Cu²⁺)

2 Mark Questions

1. 1
   Differentiate between oxidation and reduction in terms of (i) electron transfer and (ii) oxidation number, with one example each.
2. 2
   Write down the two classical rules for assigning oxidation numbers and apply them to find the oxidation state of Mn in KMnO₄. (Ans: +7)
3. 3
   Give one everyday example each of a redox and a non-redox reaction. (Ans: rusting/combustion vs metathesis like AgNO₃ + NaCl → AgCl)

3 Mark Questions

1. 1
   Balance the equation Cl₂ + OH⁻ → Cl⁻ + ClO₃⁻ (in basic medium) by the oxidation-number method. (Ans: 3Cl₂ + 6OH⁻ → 5Cl⁻ + ClO₃⁻ + 3H₂O; chlorine is both oxidised and reduced — disproportionation)
2. 2
   Explain why the oxidation number of oxygen is −2 in most compounds but −1 in peroxides and +½ in superoxides.
3. 3
   Classify the following as redox or non-redox with reason: (i) NaOH + HCl → NaCl + H₂O (ii) 2Mg + O₂ → 2MgO (iii) CaCO₃ → CaO + CO₂.

5 Mark Questions

1. 1
   Describe the ion-electron (half-reaction) method of balancing redox equations and use it to balance MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺ in acidic medium. (Ans: oxidises Fe²⁺ to Fe³⁺; MnO₄⁻ reduced to Mn²⁺; final equation MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O)
2. 2
   What is oxidation number? State the rules for its determination and calculate the oxidation numbers of the underlined elements: (i) K₂Cr₂O₇, (ii) H₂S₂O₈, (iii) CaO₂, (iv) Na₂S₂O₃. Also explain its use in identifying oxidising/reducing agents. (Ans: Cr +6, S in H₂S₂O₈ average +6 (peroxo — special), O in CaO₂ −1, S in Na₂S₂O₃ +2)