NCERT Solutions for Class 11 Chemistry Chapter 7: Redox Reactions

Exercises

Q 7.1: Assign oxidation numbers to underlined elements

(a) NaH₂PO₄: Na = +1, O = −2, H = +1; let P = x; +1 + 2(+1) + x + 4(−2) = 0 → x = +5. H = +1

(b) NaHSO₄: +1 + 1 + x + 4(−2) = 0 → S = +6

(c) H₄P₂O₇: 4(+1) + 2x + 7(−2) = 0 → 4 + 2x − 14 = 0 → P = +5

(d) K₂MnO₄: 2(+1) + x + 4(−2) = 0 → Mn = +6

(e) CaO₂: Ca = +2, 2x = −2 → O = −1 (peroxide)

(f) NaBH₄: +1 + x + 4(−1) = 0 (H = −1 in hydrides) → B = +3

(g) H₂S₂O₇: 2(+1) + 2x + 7(−2) = 0 → 2 + 2x − 14 = 0 → S = +6

(h) KAl(SO₄)₂·12H₂O: K = +1, SO₄²⁻ each = −2; +1 + x + 2(−2) = 0 → Al = +3

Q 7.2: Oxidation numbers and rationalisation

(a) KI₃: KI₃ is actually K⁺ and I₃⁻ (polyiodide). In I₃⁻: average oxidation state of I = −1/3. However in K⁺I₃⁻ context, one I is −1 and two are 0, or average = −1/3.

(b) H₂S₄O₆: 2(+1) + 4x + 6(−2) = 0 → 4x = 10 → S = +2.5 (average; S exists in two different states: +5 and 0)

(c) Fe₃O₄: Fe₃O₄ = FeO·Fe₂O₃; average = (2+3+3)/3 = +8/3. Contains Fe²⁺ and Fe³⁺.

(d) CH₃CH₂OH: Overall in ethanol: C₂H₅OH, each C has different states; C next to OH: −1; terminal C: −3; H = +1; O = −2.

(e) CH₃COOH: O in carboxyl = −2; the OH oxygen = −2; C of COOH = +3; H = +1.

Q 7.3: Justify the following are redox reactions

(a) CuO(s) + H₂(g) → Cu(s) + H₂O(g): Cu: +2 → 0 (reduced, gain of electrons); H: 0 → +1 (oxidised). Redox reaction. ✓

(b) Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g): Fe: +3 → 0 (reduced); C in CO: +2 → +4 in CO₂ (oxidised). Redox. ✓

(c) 4BCl₃(g) + 3LiAlH₄(s) → 2B₂H₆(g) + 3LiCl(s) + 3AlCl₃(s): B: +3 → −3 (reduced); H: −1 → +1 (wait — in B₂H₆, H = +1, and B = −3? Let's check: B₂H₆, 2x + 6(+1) = 0 → B = −3). H in LiAlH₄: −1 → +1 in B₂H₆ (oxidised). So B is reduced and H is oxidised. Redox. ✓

(d) 2K(s) + F₂(g) → 2K⁺F⁻(s): K: 0 → +1 (oxidised); F: 0 → −1 (reduced). Redox. ✓

(e) 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g): N: −3 → +2 (oxidised); O: 0 → −2 (reduced). Redox. ✓

Q 7.4: H₂O(s) + F₂(g) → HF(g) + HOF(g) is a redox reaction

In H₂O: O = −2, H = +1; In F₂: F = 0

In HF: F = −1 (reduced from 0); In HOF: O = 0 (oxidised from −2 in H₂O), F = −1

F₂ is reduced (0 → −1); O in water is oxidised (−2 → 0 in HOF). Therefore this is a redox reaction. ✓

Q 7.5: Oxidation numbers of S in H₂SO₅, Cr in Cr₂O₇²⁻, N in NO₃⁻

H₂SO₅ (peroxomonosulphuric acid): 2(+1) + x + 5(−2) = 0; fallacy if all O = −2: x = 8 (impossible). Actual: two O are peroxide type (−1 each), three normal O = −2; x + 2(−1) + 3(−2) + 2 = 0; x = +6. S = +6

Cr₂O₇²⁻: 2x + 7(−2) = −2 → 2x = 12 → Cr = +6

NO₃⁻: x + 3(−2) = −1 → x = +5. N = +5

Q 7.6: Formulas for compounds

(a) Mercury(II) chloride: HgCl₂

(b) Nickel(II) sulphate: NiSO₄

(c) Tin(IV) oxide: SnO₂

(d) Thallium(I) sulphate: Tl₂SO₄

(e) Iron(III) sulphate: Fe₂(SO₄)₃

(f) Chromium(III) oxide: Cr₂O₃

Q 7.7: Oxidation states of carbon (−4 to +4) and nitrogen (−3 to +5)

Carbon: CH₄ (−4), CH₃OH (−2), HCHO (0), HCOOH (+2), CO₂ (+4)

Nitrogen: NH₃ (−3), N₂H₄ (−2), NH₂OH (−1), N₂ (0), N₂O (+1), NO (+2), N₂O₃ (+3), NO₂ (+4), N₂O₅ (+5)

Q 7.8: Why SO₂ and H₂O₂ can oxidise/reduce, but O₃ and HNO₃ only oxidise

SO₂ has S at +4 (can go to +6 by oxidation, or to 0 by reduction), so it acts as both oxidant and reductant.

H₂O₂ has O at −1 (can be oxidised to 0 in O₂, or reduced to −2 in H₂O), so both.

O₃ has O at 0 and −2; O₃ → O₂ (O is reduced to −2 in products), always acts as oxidant only.

HNO₃: N is at +5 (maximum for N), it can only be reduced, so only acts as oxidant.

Q 7.9: More appropriate representations of photosynthesis and ozone-H₂O₂ reactions

(a) 6CO₂ + 12H₂O → C₆H₁₂O₆ + 6H₂O + 6O₂ is more appropriate because it tracks the oxygen atoms specifically — the 6O₂ released comes from H₂O (not CO₂), which can be proven by using H₂¹⁸O isotopic labelling. Writing 6H₂O as product makes the water molecule tracking explicit.

(b) O₃ + H₂O₂ → H₂O + O₂ + O₂ is more appropriate to show that the two O₂ molecules come from different sources (one from O₃, one from H₂O₂), provable by isotopic labelling with ¹⁸O.

Q 7.10: AgF₂ as very strong oxidant

In AgF₂, Ag is at +2 (unusual state). Ag²⁺ is unstable and tends to revert to Ag⁺ (more stable). In doing so, it oxidises other species (gains electrons), making it a strong oxidant.

Q 7.11: Lower state if reducing agent in excess; higher state if oxidising agent in excess

Example 1: P reacts with Cl₂

Reducing agent excess (P excess): 2P + 3Cl₂ → 2PCl₃ (P in +3 state)

Oxidising agent excess (Cl₂ excess): 2P + 5Cl₂ → 2PCl₅ (P in +5 state)

Example 2: S reacts with O₂

S excess: 2S + O₂ → 2SO (lower state); O₂ excess: S + O₂ → SO₂ (higher state)

Example 3: Na reacts with O₂

Na excess: 4Na + O₂ → 2Na₂O (O in −2); O₂ excess: 2Na + O₂ → Na₂O₂ (O in −1, higher)

Q 7.12: Observations explained

(a) Alcoholic KMnO₄ for toluene oxidation: Alkaline and acidic KMnO₄ both oxidise toluene. Alcoholic KMnO₄ is used to avoid over-oxidation in acidic conditions, since alcohol as solvent provides mild conditions. Reaction: C₆H₅CH₃ + 2KMnO₄ → C₆H₅COOK + 2MnO₂ + KOH + H₂O (in alkaline medium, benzoate forms).

(b) Cl⁻ gives HCl, Br⁻ gives Br₂ with conc. H₂SO₄: Conc. H₂SO₄ oxidises Br⁻ to Br₂ (red vapour) because Br⁻ is a stronger reducing agent than Cl⁻. Cl⁻ is not oxidised by H₂SO₄ and only gives HCl gas.

Q 7.13: Identify oxidised/reduced substances, oxidising/reducing agents

(a) 2AgBr(s) + C₆H₆O₂(aq) → 2Ag(s) + 2HBr(aq) + C₆H₄O₂(aq):

Oxidised: C₆H₆O₂ (hydroquinone) → C₆H₄O₂ (quinone), reducing agent

Reduced: Ag⁺ in AgBr → Ag⁰, oxidising agent = AgBr

(b) HCHO + 2[Ag(NH₃)₂]⁺ + 3OH⁻ → 2Ag + HCOO⁻ + 4NH₃ + 2H₂O:

Oxidised: HCHO (C: 0 in HCHO → +2 in HCOO⁻), reducing agent = HCHO

Reduced: Ag⁺ → Ag⁰, oxidising agent = [Ag(NH₃)₂]⁺

(c) HCHO + 2Cu²⁺ + 5OH⁻ → Cu₂O + HCOO⁻ + 3H₂O:

Oxidised: HCHO, reducing agent; Reduced: Cu²⁺ → Cu⁺, oxidising agent = Cu²⁺

(d) N₂H₄(l) + 2H₂O₂(l) → N₂(g) + 4H₂O(l):

Oxidised: N in N₂H₄ (−2 → 0), reducing agent = N₂H₄

Reduced: O in H₂O₂ (−1 → −2), oxidising agent = H₂O₂

(e) Pb(s) + PbO₂(s) + 2H₂SO₄ → 2PbSO₄(s) + 2H₂O

Oxidised: Pb⁰ → Pb²⁺ in PbSO₄, reducing agent = Pb

Reduced: Pb⁴⁺ in PbO₂ → Pb²⁺ in PbSO₄, oxidising agent = PbO₂

This is a disproportionation where both Pb⁰ and Pb⁴⁺ go to Pb²⁺.

Q 7.14: Why S₂O₃²⁻ reacts differently with I₂ and Br₂?

S₂O₃²⁻ + I₂: 2S₂O₃²⁻ + I₂ → S₄O₆²⁻ + 2I⁻ (S₂O₃²⁻ is oxidised from +2 to +2.5 in S₄O₆²⁻). I₂ is a mild oxidant; thiosulphate is only partially oxidised.

S₂O₃²⁻ + Br₂: S₂O₃²⁻ + 4Br₂ + 5H₂O → 2SO₄²⁻ + 8HBr (S: +2 → +6). Br₂ is a stronger oxidant than I₂, so it completely oxidises S₂O₃²⁻ to SO₄²⁻.

Q 7.15: F₂ best oxidant; HI best reductant among halides

F₂ has highest standard electrode potential (+2.87 V), making it the strongest oxidant. It can oxidise all other halide ions: 2F₂ + 2H₂O → 4HF + O₂.

HI has the lowest bond dissociation energy and I⁻ has the largest radius (most polarisable), making it the easiest to lose electrons. Standard oxidation potential of I⁻/I₂ = −0.54 V (vs. more negative for F⁻/F₂). HI reduces Fe³⁺, Cu²⁺ etc.

Q 7.16: XeO₄²⁻(aq) + 2F⁻(aq) + 6H⁺ → XeO₃(g) + F₂(g) + 3H₂O

In this reaction, F⁻ is being oxidised to F₂ (−1 → 0), which is very unusual since F₂ is the strongest oxidant. This means XeO₄⁴⁻ (in Na₄XeO₆) is an even stronger oxidant than F₂. Conclusion: Na₄XeO₆ is an extraordinarily powerful oxidising agent.

Q 7.17: Ag⁺ oxidises H₃PO₂ but Cu²⁺ does not in similar conditions

H₃PO₂ is oxidised by Ag⁺ (reactions a, b) and silver reagent for benzaldehyde (c).

Cu²⁺ does not oxidise benzaldehyde (d) under the conditions used.

Conclusion: Ag⁺ is a stronger oxidant than Cu²⁺ in these reactions; Ag⁺ can oxidise compounds that Cu²⁺ cannot. Tollen's reagent (Ag⁺) detects aldehydes; Fehling's solution (Cu²⁺) detects only aliphatic aldehydes, not aromatic ones.

Q 7.18: Balance by ion-electron method (acidic/basic)

(a) MnO₄⁻ + I⁻ → MnO₂ + I₂ (basic medium):

2MnO₄⁻(aq) + 6I⁻(aq) + 4H₂O → 2MnO₂(s) + 3I₂(s) + 8OH⁻(aq)

(b) MnO₄⁻ + SO₂ → Mn²⁺ + HSO₄⁻ (acidic):

Reduction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O (×2)

Oxidation: SO₂ + 2H₂O → HSO₄⁻ + 3H⁺ + 2e⁻ (×5)

2MnO₄⁻ + 5SO₂ + 2H₂O → 2Mn²⁺ + 5HSO₄⁻ + H⁺

(c) H₂O₂ + Fe²⁺ → Fe³⁺ + H₂O (acidic):

Reduction: H₂O₂ + 2H⁺ + 2e⁻ → 2H₂O; Oxidation: Fe²⁺ → Fe³⁺ + e⁻ (×2)

H₂O₂ + 2Fe²⁺ + 2H⁺ → 2Fe³⁺ + 2H₂O

(d) Cr₂O₇²⁻ + SO₂ → Cr³⁺ + SO₄²⁻ (acidic):

Reduction: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O

Oxidation: SO₂ + 2H₂O → SO₄²⁻ + 4H⁺ + 2e⁻ (×3)

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

Q 7.19: Balance in basic medium by ion-electron and oxidation number methods

(a) P₄ + OH⁻ → PH₃ + HPO₂⁻:

P: 0 → −3 in PH₃ (reduced, ×1); 0 → +1 in HPO₂⁻ (oxidised, ×3)

Balance: P₄ + 3OH⁻ + 3H₂O → PH₃ + 3HPO₂⁻

OA = P₄ (where P → PH₃); RA = P₄ (where P → HPO₂⁻) — disproportionation

(b) N₂H₄ + ClO₃⁻ → NO + Cl⁻:

N: −2 → +2 in NO (oxidised); Cl: +5 in ClO₃⁻ → −1 in Cl⁻ (reduced)

Each N loses 4e⁻; N₂H₄ loses 8e⁻; Each Cl gains 6e⁻; balance e⁻: 3N₂H₄ with 4ClO₃⁻

3N₂H₄ + 4ClO₃⁻ → 6NO + 4Cl⁻ + 6H₂O; Oxidising agent = ClO₃⁻; Reducing agent = N₂H₄

(c) Cl₂O₇ + H₂O₂ → ClO₂⁻ + O₂ + H⁺:

Cl: +7 in Cl₂O₇ → +3 in ClO₂⁻ (reduced); O in H₂O₂: −1 → 0 in O₂ (oxidised)

Each Cl gains 4e⁻; Cl₂O₇ gains 8e⁻; H₂O₂ → O₂ + 2H⁺ + 2e⁻ (×4)

Cl₂O₇ + 4H₂O₂ → 2ClO₂⁻ + 4O₂ + 2H⁺ + 3H₂O

Q 7.20: (CN)₂(g) + 2OH⁻ → CN⁻ + CNO⁻ + H₂O

This is a disproportionation reaction. In (CN)₂: C is at +2 (average). In CN⁻: C = +2; in CNO⁻: C = +4 (if N = −3, O = −2, charge = −1: C + (−3) + (−2) = −1, C = +4). One CN₂ molecule undergoes simultaneous oxidation and reduction. Information: cyanogen behaves like a halogen (disproportionates in base) — called a pseudohalogen.

Q 7.21: Disproportionation of Mn³⁺

Mn³⁺ is unstable: Mn³⁺ → Mn²⁺ (reduced) and Mn³⁺ → MnO₂ (oxidised)

Q 7.22: Cs, Ne, I and F

(a) Only negative oxidation state: F (most electronegative, always −1 or 0)

(b) Only positive oxidation state: Cs (alkali metal, always +1)

(c) Both positive and negative oxidation states: I (can be −1, 0, +1, +3, +5, +7)

(d) Neither positive nor negative: Ne (noble gas, zero oxidation state always)

Q 7.23: Cl₂ + SO₂ reaction in water

Cl₂ + SO₂ + 2H₂O → H₂SO₄ + 2HCl

Or in ionic form: Cl₂(g) + SO₂(g) + 2H₂O(l) → SO₄²⁻(aq) + 2Cl⁻(aq) + 4H⁺(aq)

Cl is reduced: 0 → −1; S is oxidised: +4 → +6

Q 7.24: Non-metals and metals for disproportionation

(a) Non-metals showing disproportionation: Cl (Cl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂O), S, P

(b) Metals showing disproportionation: Cu (2Cu⁺ → Cu + Cu²⁺), Mn (Mn³⁺ → Mn²⁺ + MnO₂), Pb (Pb₃O₄ = PbO + 2PbO₂ in a sense)

Q 7.25: Ostwald's process: max NO from 10.00 g NH₃ and 20.00 g O₂

Moles O₂ required for 0.5882 mol NH₃ = 0.5882 × 5/4 = 0.7353 mol > 0.625 mol available

O₂ is the limiting reagent. Moles NH₃ that react = 0.625 × 4/5 = 0.5 mol

Moles NO = 0.5 mol; Mass NO = 0.5 × 30 = 15.0 g

Q 7.26: Feasibility of reactions using standard electrode potentials

(a) Fe³⁺ + I⁻: E°(Fe³⁺/Fe²⁺) = +0.77 V; E°(I₂/I⁻) = +0.54 V; ΔE° = 0.77 − 0.54 = +0.23 V → Feasible

(b) Ag⁺ + Cu(s): E°(Ag⁺/Ag) = +0.80 V; E°(Cu²⁺/Cu) = +0.34 V; ΔE° = 0.80 − 0.34 = +0.46 V → Feasible

(c) Fe³⁺ + Cu(s): E°(Fe³⁺/Fe²⁺) = +0.77 V; E°(Cu²⁺/Cu) = +0.34 V; ΔE° = 0.77 − 0.34 = +0.43 V → Feasible

(d) Ag(s) + Fe³⁺: Ag is reducing agent; E°cell = E°(Fe³⁺/Fe²⁺) − E°(Ag⁺/Ag) = 0.77 − 0.80 = −0.03 V → Not feasible

(e) Br₂ + Fe²⁺: E°(Br₂/Br⁻) = +1.07 V; E°(Fe³⁺/Fe²⁺) = +0.77 V; ΔE° = 1.07 − 0.77 = +0.30 V → Feasible

Q 7.27: Products of electrolysis

(i) AgNO₃(aq) with Ag electrodes: Cathode: Ag⁺ + e⁻ → Ag(s); Anode: Ag(s) → Ag⁺ + e⁻ (electrode dissolves). Net: Ag transfers from anode to cathode.

(ii) AgNO₃(aq) with Pt electrodes: Cathode: Ag⁺ + e⁻ → Ag(s); Anode: H₂O → ½O₂ + 2H⁺ + 2e⁻. Products: Ag metal at cathode, O₂ at anode.

(iii) Dilute H₂SO₄ with Pt: Cathode: 2H⁺ + 2e⁻ → H₂(g); Anode: H₂O → ½O₂ + 2H⁺ + 2e⁻. Products: H₂ and O₂.

(iv) CuCl₂(aq) with Pt: Cathode: Cu²⁺ + 2e⁻ → Cu(s); Anode: 2Cl⁻ → Cl₂(g) + 2e⁻. Products: Cu at cathode, Cl₂ at anode.

Q 7.28: Order of displacement: Al, Cu, Fe, Mg, Zn

Higher in reactivity series displaces lower: Mg > Al > Zn > Fe > Cu

Q 7.29: Increasing order of reducing power

Higher reduction potential = weaker reducing agent; Lower reduction potential = stronger reducing agent.

E° values: K⁺/K = −2.93 V; Mg²⁺/Mg = −2.37 V; Cr³⁺/Cr = −0.74 V; Hg²⁺/Hg = +0.79 V; Ag⁺/Ag = +0.80 V

Increasing reducing power (decreasing E°): Ag < Hg < Cr < Mg < K

Q 7.30: Galvanic cell for Zn(s) + 2Ag⁺(aq) → Zn²⁺(aq) + 2Ag(s)

Cell notation: Zn(s) | Zn²⁺(aq) || Ag⁺(aq) | Ag(s)

(i) Negatively charged electrode: Zn electrode (anode, oxidation occurs, electrons produced)

(ii) Current carriers: In solution: Zn²⁺ and Ag⁺ ions (and NO₃⁻ from salt bridge); In external circuit: electrons

(iii) Individual reactions:

Anode: Zn(s) → Zn²⁺(aq) + 2e⁻

Cathode: 2Ag⁺(aq) + 2e⁻ → 2Ag(s)

Important Extra Questions

Q A: What is disproportionation reaction? Give two examples.

Disproportionation is a redox reaction in which the same element undergoes both oxidation and reduction simultaneously.

Q B: Explain the concept of oxidising agent and reducing agent with electrode potentials.

An oxidising agent accepts electrons (is itself reduced). High standard reduction potential = strong oxidising agent (e.g., F₂: E° = +2.87 V).

A reducing agent donates electrons (is itself oxidised). Low standard reduction potential = strong reducing agent (e.g., Li: E° = −3.04 V).

E°cell = E°(cathode) − E°(anode); positive E°cell indicates spontaneous reaction.

Q C: Distinguish between oxidation number method and ion-electron method for balancing redox equations.

Oxidation number method: 1) Assign oxidation numbers. 2) Find change in oxidation number. 3) Equalise electrons lost and gained. 4) Balance remaining atoms and charges.

Ion-electron method (half-reaction): 1) Write separate half-reactions. 2) Balance atoms and charges in each half-reaction using H₂O, H⁺ (acidic) or OH⁻ (basic). 3) Equalise electrons. 4) Add half-reactions.

Ion-electron method is preferred for ionic equations; both give same final result.

Q D: Explain why fluorine cannot act as a reducing agent while iodine cannot act as an oxidising agent as strongly.

Fluorine (F₂): highest electronegativity (4.0), highest standard reduction potential (+2.87 V). F always exists as F⁻ (−1) after reduction. F has no d-orbitals and cannot expand its valency. There is no higher oxidation state for F; it cannot donate electrons, so it cannot be oxidised. Hence it cannot be a reducing agent.

Iodine: lower electronegativity; I₂/I⁻ E° = +0.54 V (weaker oxidant than Cl₂ or Br₂). I⁻ is a better reducing agent than Cl⁻ or Br⁻ due to larger size and lower charge density.

Q E: Calculate the EMF of the cell Zn | Zn²⁺(0.1 M) || Cu²⁺(0.1 M) | Cu at 298 K. E°(Zn²⁺/Zn) = −0.76 V, E°(Cu²⁺/Cu) = +0.34 V.