NCERT Solutions for Class 11 Chemistry Chapter 7: Redox Reactions (NCERT 2026–27)

These Class 11 Chemistry Chapter 7 solutions cover Redox Reactions completely — every numbered NCERT exercise question (7.1–7.30) is reproduced verbatim and solved step by step, with oxidation numbers assigned, equations balanced by both the oxidation-number and ion–electron (half-reaction) methods, and the numerical answer cross-checked against the official NCERT key. Updated for session 2026–27.

Class: 11 Subject: Chemistry Chapter: 7 Title: Redox Reactions Exercises: 7.1–7.30 Session: 2026–27
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Class 11 Chemistry Chapter 7 Solutions – Overview

Chapter 7, Redox Reactions, builds the idea that oxidation and reduction always happen together. The chapter develops three connected pictures of the same process: the classical view (addition/removal of oxygen or hydrogen), the electronic view (loss/gain of electrons), and the most powerful tool, the oxidation number. Using oxidation numbers you can spot the oxidant and reductant, classify a redox change as combination, decomposition, displacement or disproportionation, and balance any redox equation by the oxidation-number method or the ion–electron (half-reaction) method. The unit closes with redox couples, the standard electrode potential table and the Daniell cell, linking chemistry to electricity and previewing electrochemistry in Class 12.

Key Concepts & Definitions

Oxidation: loss of electron(s), OR an increase in the oxidation number of an element. (Classical: addition of O/electronegative element or removal of H/electropositive element.)

Reduction: gain of electron(s), OR a decrease in oxidation number. (Classical: removal of O or addition of H.)

Oxidising agent (oxidant): a species that accepts electrons / increases another element’s oxidation number and is itself reduced.

Reducing agent (reductant): a species that donates electrons / decreases another element’s oxidation number and is itself oxidised.

Oxidation number (state): the charge an atom would have if every bond were fully ionic, assigned by a fixed set of rules.

Disproportionation: a single element in an intermediate oxidation state is simultaneously oxidised and reduced (e.g. 2H2O2 → 2H2O + O2).

Redox couple: the oxidised and reduced forms of a species in one half-reaction, written oxidised/reduced (e.g. Zn2+/Zn).

Important Rules for Oxidation Numbers

1. Free element (uncombined) → oxidation number = 0 (e.g. H2, O2, P4, Na).

2. Monatomic ion → equals its charge (Na+ = +1, O2− = −2).

3. Oxygen = −2 usually; −1 in peroxides (H2O2), −½ in superoxides (KO2), positive with F (+2 in OF2, +1 in O2F2).

4. Hydrogen = +1, except −1 in metal hydrides (NaH, CaH2).

5. Fluorine = −1 always; other halogens = −1 except with O or a more electronegative halogen.

6. Σ(oxidation numbers) = 0 for a neutral molecule, = the ion charge for a polyatomic ion.

n-factor (oxidant/reductant): total change in oxidation number per formula unit = number of electrons gained/lost.

NCERT Exercises (7.1–7.30) — Full Solutions

Questions reproduced verbatim from the NCERT textbook (Reprint 2026–27). Answers are original and expert-checked; the numerical answer to Q7.25 matches the official NCERT key (15 g).

7.1 Assign oxidation number to the underlined elements in each of the following species: (a) NaP…   NaH2PO4    (b) NaHSO4    (c) H4P2O7    (d) K2MnO4 (e) CaO2    (f) NaBH4    (g) H2S2O7    (h) KAl(SO4)2·12H2O

ANSWER (a) NaH2PO4 → P: +1 + 2(+1) + x + 4(−2) = 0 ⇒ x = +5. (b) NaHSO4 → S: +1 + 1 + x − 8 = 0 ⇒ x = +6. (c) H4P2O7 → P: 4(+1) + 2x + 7(−2) = 0 ⇒ 2x = 10 ⇒ x = +5. (d) K2MnO4 → Mn: 2(+1) + x + 4(−2) = 0 ⇒ x = +6. (e) CaO2 → O: CaO2 is a peroxide, Ca = +2, so 2x = −2 ⇒ O = −1. (f) NaBH4 → B: here H is hydridic (−1): +1 + x + 4(−1) = 0 ⇒ x = +3. (g) H2S2O7 → S: 2(+1) + 2x + 7(−2) = 0 ⇒ 2x = 12 ⇒ x = +6. (h) KAl(SO4)2·12H2O → Al: K = +1, each SO4 = −2, so +1 + x + 2(−2) = 0 ⇒ x = +3.

7.2 What are the oxidation number of the underlined elements in each of the following and how do you rationalise your results? (a) KI3    (b) H2S4O6    (c) Fe3O4    (d) CH3CH2OH    (e) CH3COOH

ANSWER (a) KI3 → I = −⅓. K = +1, so 3 I share −1, average = −⅓. Structurally KI3 is K+(I3)−; one I− ion (−1) is bonded to an I2 molecule (0), so the fractional value is just the average. (b) H2S4O6 → S = +2½. 2(+1) + 4x + 6(−2) = 0 ⇒ 4x = 10 ⇒ average = +2.5. Structurally two terminal S are +5 and two middle S are 0 (a fractional average, not a real single state). (c) Fe3O4 → Fe = +8⁄3. 3x + 4(−2) = 0 ⇒ x = +8/3. It is a mixed oxide FeO·Fe2O3: one Fe is +2 and two Fe are +3, so the average is +8/3. (d) CH3CH2OH → C = −2 (average). 2x + 6(+1) + (−2) = 0 ⇒ 2x = −4 ⇒ average = −2. (CH3 carbon = −3, CH2OH carbon = −1.) (e) CH3COOH → C = 0 (average). 2x + 4(+1) + 2(−2) = 0 ⇒ x = 0. (CH3 carbon = −3, COOH carbon = +3.)

7.3 Justify that the following reactions are redox reactions: (a) CuO(s) + H2(g) → Cu(s) + H2O(g) (b) Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g) (c) 4BCl3(g) + 3LiAlH4(s) → 2B2H6(g) + 3LiCl(s) + 3AlCl3(s) (d) 2K(s) + F2(g) → 2K+F−(s) (e) 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)

ANSWER (a) Cu: +2 → 0 (reduced); H: 0 → +1 (oxidised). Both oxidation numbers change ⇒ redox. CuO is the oxidant, H2 the reductant. (b) Fe: +3 → 0 (reduced); C: +2 → +4 (oxidised). ⇒ redox; Fe2O3 oxidant, CO reductant. (c) B: +3 → −3 (reduced); H of LiAlH4: −1 → +1 (oxidised) within B2H6. Oxidation numbers change ⇒ redox. (d) K: 0 → +1 (oxidised); F: 0 → −1 (reduced). ⇒ redox; F2 oxidant, K reductant. (e) N: −3 → +2 (oxidised); O: 0 → −2 (reduced). ⇒ redox; O2 oxidant, NH3 reductant.

7.4 Fluorine reacts with ice and results in the change: H2O(s) + F2(g) → HF(g) + HOF(g). Justify that this reaction is a redox reaction.

ANSWER In F2 the oxidation number of fluorine is 0. In the products, fluorine in HF is −1 (reduced) while fluorine in HOF is +1 (oxidised). Thus the same element fluorine is both oxidised (0 → +1) and reduced (0 → −1) — a disproportionation, which is a special redox reaction. (Oxygen stays −2 and H stays +1 throughout.)

7.5 Calculate the oxidation number of sulphur, chromium and nitrogen in H2SO5, Cr2O72− and NO3−. Suggest structure of these compounds. Count for the fallacy.

ANSWER H2SO5 (Caro’s acid): by the simple rule, 2(+1) + x + 5(−2) = 0 ⇒ x = +8, which is impossible (S has only 6 valence electrons). Fallacy: H2SO5 contains a peroxide (−O−O−) linkage. Taking 2 oxygens as −1 (peroxide) and 3 as −2: 2(+1) + x + 2(−1) + 3(−2) = 0 ⇒ x = +6. Cr2O72−: all 7 O are normal (−2). 2x + 7(−2) = −2 ⇒ 2x = 12 ⇒ Cr = +6. (No fallacy; structure has two CrO4 tetrahedra sharing a corner O.) NO3−: x + 3(−2) = −1 ⇒ N = +5. (Planar, N at centre with three O; no peroxide, so no fallacy.)

7.6 Write formulas for the following compounds: (a) Mercury(II) chloride    (b) Nickel(II) sulphate    (c) Tin(IV) oxide    (d) Thallium(I) sulphate    (e) Iron(III) sulphate    (f) Chromium(III) oxide

ANSWER (a) HgCl2   (b) NiSO4   (c) SnO2   (d) Tl2SO4   (e) Fe2(SO4)3   (f) Cr2O3. The Roman numeral gives the metal’s oxidation state, which is then balanced against the anion charge (Cl−, SO42−, O2−).

7.7 Suggest a list of the substances where carbon can exhibit oxidation states from −4 to +4 and nitrogen from −3 to +5.

ANSWER Carbon: CH4 (−4), C2H6/CH3Cl (−3 to −2), CH3OH/CH2Cl2 (−2 to 0), CH2O (0), HCOOH/CHCl3 (+2), CO (+2), CO2/CCl4 (+4). Nitrogen: NH3 (−3), N2H4 (−2), NH2OH (−1), N2 (0), N2O (+1), NO (+2), N2O3/HNO2 (+3), NO2 (+4), N2O5/HNO3 (+5).

7.8 While sulphur dioxide and hydrogen peroxide can act as oxidising as well as reducing agents in their reactions, ozone and nitric acid act only as oxidants. Why?

ANSWER In SO2, sulphur is in the intermediate state +4 (range −2 to +6), so it can be oxidised to +6 (acts as reductant) or reduced to lower states (acts as oxidant). In H2O2, oxygen is −1 (intermediate between −2 and 0), so it too can go either way. In O3 oxygen is at 0/−2 acting as a strong electron acceptor, and in HNO3 nitrogen is at its highest state +5, which cannot increase further. Hence both can only be reduced — they act only as oxidants.

7.9 Consider the reactions: (a) 6CO2(g) + 6H2O(l) → C6H12O6(aq) + 6O2(g) (b) O3(g) + H2O2(l) → H2O(l) + 2O2(g) Why it is more appropriate to write these reactions as: (a) 6CO2 + 12H2O → C6H12O6 + 6H2O + 6O2; (b) O3 + H2O2 → H2O + O2 + O2? Also suggest a technique to investigate the path.

ANSWER The rewritten equations make clear which species supplies the O2. In (a) the O2 released in photosynthesis comes from water, not CO2; the extra 6H2O shows water is both consumed and produced. In (b) one O2 comes from O3 and the other from H2O2, so writing them separately shows both oxidants are reduced. Technique: use the radioactive (isotopic) tracer method — label oxygen as the heavy isotope 18O in one reactant (e.g. H218O) and track where it appears in the products, confirming the origin of the evolved O2.

7.10 The compound AgF2 is unstable compound. However, if formed, the compound acts as a very strong oxidising agent. Why?

ANSWER In AgF2, silver is in the unusually high +2 state, but its stable state is +1. Therefore Ag2+ has a strong tendency to gain one electron and revert to the stable Ag+ (+1) state. By gaining an electron it oxidises other species, so AgF2 behaves as a very strong oxidising agent.

7.11 Whenever a reaction between an oxidising agent and a reducing agent is carried out, a compound of lower oxidation state is formed if the reducing agent is in excess and a compound of higher oxidation state is formed if the oxidising agent is in excess. Justify this statement giving three illustrations.

ANSWER (i) P4 + Cl2: excess reductant P4 → PCl3 (P = +3, lower); excess oxidant Cl2 → PCl5 (P = +5, higher). (ii) C + O2: limited O2 (C in excess) → CO (C = +2, lower); excess O2 → CO2 (C = +4, higher). (iii) Na + O2: excess Na → Na2O (O = −2, oxide); excess O2 → Na2O2 (O = −1, peroxide, higher state of O). The available amount of oxidant/reductant decides the final oxidation state.

7.12 How do you count for the following observations? (a) Though alkaline KMnO4 and acidic KMnO4 both are used as oxidants, yet in the manufacture of benzoic acid from toluene we use alcoholic potassium permanganate as an oxidant. Why? Write a balanced redox equation for the reaction. (b) When concentrated sulphuric acid is added to an inorganic mixture containing chloride, we get colourless pungent smelling gas HCl, but if the mixture contains bromide then we get red vapour of bromine. Why?

ANSWER (a) In aqueous (acidic/alkaline) medium, water would react with and over-oxidise the product; alcoholic KMnO4 provides a controlled, mild medium that oxidises the –CH3 of toluene cleanly to –COOH without further attack. Balanced equation:
C6H5CH3 + 2KMnO4 → C6H5COOK + 2MnO2 + KOH + H2O, then C6H5COOK + HCl → C6H5COOH + KCl. (b) Conc. H2SO4 is a strong acid but a weak/moderate oxidant toward Cl−, so it only displaces colourless HCl (no oxidation of Cl−). But it can oxidise the more easily oxidised Br− to Br2 (red vapour): 2Br− + 2H2SO4 → Br2 + SO2 + 2H2O + SO42−. Br− is a stronger reductant than Cl−.

7.13 Identify the substance oxidised, reduced, oxidising agent and reducing agent for each of the following reactions: (a) 2AgBr(s) + C6H6O2(aq) → 2Ag(s) + 2HBr(aq) + C6H4O2(aq) (b) HCHO(l) + 2[Ag(NH3)2]+(aq) + 3OH−(aq) → 2Ag(s) + HCOO−(aq) + 4NH3(aq) + 2H2O(l) (c) HCHO(l) + 2Cu2+(aq) + 5OH−(aq) → Cu2O(s) + HCOO−(aq) + 3H2O(l) (d) N2H4(l) + 2H2O2(l) → N2(g) + 4H2O(l) (e) Pb(s) + PbO2(s) + 2H2SO4(aq) → 2PbSO4(s) + 2H2O(l)

ANSWER (a) Oxidised: C6H6O2 (quinol → quinone), the reducing agent. Reduced: AgBr (Ag+ → Ag), the oxidising agent. (b) Oxidised: HCHO (C: 0 → +2), reducing agent. Reduced: [Ag(NH3)2]+ (Ag+ → Ag), oxidising agent. (c) Oxidised: HCHO, reducing agent. Reduced: Cu2+ → Cu+ (Cu2O), the oxidising agent. (d) Oxidised: N2H4 (N: −2 → 0), reducing agent. Reduced: H2O2 (O: −1 → −2), oxidising agent. (e) Oxidised: Pb (0 → +2), reducing agent. Reduced: PbO2 (Pb: +4 → +2), oxidising agent.

7.14 Consider the reactions: 2S2O32−(aq) + I2(s) → S4O62−(aq) + 2I−(aq) S2O32−(aq) + 2Br2(l) + 5H2O(l) → 2SO42−(aq) + 4Br−(aq) + 10H+(aq) Why does the same reductant, thiosulphate react differently with iodine and bromine?

ANSWER Bromine is a stronger oxidising agent than iodine. With the weaker oxidant I2, thiosulphate (S = +2) is only mildly oxidised to tetrathionate S4O62− (S = +2.5). With the stronger oxidant Br2, thiosulphate is oxidised much further, all the way to sulphate SO42− (S = +6). The extent of oxidation depends on the strength of the oxidant.

7.15 Justify giving reactions that among halogens, fluorine is the best oxidant and among hydrohalic compounds, hydroiodic acid is the best reductant.

ANSWER Fluorine is the best oxidant: it has the most positive E° (F2 + 2e− → 2F−, E° = +2.87 V) and displaces all other halide ions, e.g. F2 + 2Cl− → 2F− + Cl2. The reverse is never spontaneous. HI is the best reductant: the H–I bond is the weakest among HX, so I− loses its electron most easily. HI readily reduces, e.g. 2HI + H2SO4 → I2 + SO2 + 2H2O, while HF/HCl do not. (I2/I− has the least positive E° among the halogens, so I− is the most readily oxidised.)

7.16 Why does the following reaction occur? XeO64−(aq) + 2F−(aq) + 6H+(aq) → XeO3(g) + F2(g) + 3H2O(l). What conclusion about the compound Na4XeO6 (of which XeO64− is a part) can be drawn from the reaction.

ANSWER Here Xe goes from +8 in XeO64− to +6 in XeO3 (reduced), while F goes from −1 to 0 in F2 (oxidised). The reaction proceeds because XeO64− is able to oxidise even F− to F2. Conclusion: since F− is normally extremely hard to oxidise, Na4XeO6 (containing Xe in its highest state +8) must be an even stronger oxidising agent than F2 — one of the strongest known oxidants.

7.17 Consider the reactions: (a) H3PO2(aq) + 4AgNO3(aq) + 2H2O(l) → H3PO4(aq) + 4Ag(s) + 4HNO3(aq) (b) H3PO2(aq) + 2CuSO4(aq) + 2H2O(l) → H3PO4(aq) + 2Cu(s) + 2H2SO4(aq) (c) C6H5CHO(l) + 2[Ag(NH3)2]+(aq) + 3OH−(aq) → C6H5COO−(aq) + 2Ag(s) + 4NH3(aq) + 2H2O(l) (d) C6H5CHO(l) + 2Cu2+(aq) + 5OH−(aq) → No change observed. What inference do you draw about the behaviour of Ag+ and Cu2+ from these reactions?

ANSWER In (a) and (c), Ag+ is reduced to Ag by H3PO2 and by benzaldehyde. In (b), Cu2+ is reduced by the strong reductant H3PO2, but in (d) the weaker reductant benzaldehyde fails to reduce Cu2+ (no change). Inference: Ag+ is a stronger (milder-requirement) oxidising agent than Cu2+. Ag+ is reduced even by a mild reductant, whereas Cu2+ needs a strong reductant. (Consistent with E°Ag+/Ag = +0.80 V > E°Cu2+/Cu = +0.34 V.)

7.18 Balance the following redox reactions by ion-electron method: (a) MnO4−(aq) + I−(aq) → MnO2(s) + I2(s) (in basic medium) (b) MnO4−(aq) + SO2(g) → Mn2+(aq) + HSO4−(aq) (in acidic solution) (c) H2O2(aq) + Fe2+(aq) → Fe3+(aq) + H2O(l) (in acidic solution) (d) Cr2O72− + SO2(g) → Cr3+(aq) + SO42−(aq) (in acidic solution)

ANSWER (a) Basic medium. Oxidation: 2I− → I2 + 2e− (×3). Reduction: MnO4− + 2H2O + 3e− → MnO2 + 4OH− (×2). Adding:
2MnO4− + 6I− + 4H2O → 2MnO2 + 3I2 + 8OH−. (b) Acidic medium. Oxidation: SO2 + 2H2O → HSO4− + 3H+ + 2e− (×5). Reduction: MnO4− + 8H+ + 5e− → Mn2+ + 4H2O (×2). Adding and cancelling:
2MnO4− + 5SO2 + 2H2O → 2Mn2+ + 5HSO4− + H+. (c) Acidic medium. Oxidation: Fe2+ → Fe3+ + e− (×2). Reduction: H2O2 + 2H+ + 2e− → 2H2O. Adding:
H2O2 + 2Fe2+ + 2H+ → 2Fe3+ + 2H2O. (d) Acidic medium. Oxidation: SO2 + 2H2O → SO42− + 4H+ + 2e− (×3). Reduction: Cr2O72− + 14H+ + 6e− → 2Cr3+ + 7H2O. Adding and cancelling:
Cr2O72− + 3SO2 + 2H+ → 2Cr3+ + 3SO42− + H2O.

7.19 Balance the following equations in basic medium by ion-electron method and oxidation number methods and identify the oxidising agent and the reducing agent. (a) P4(s) + OH−(aq) → PH3(g) + HPO2−(aq) (b) N2H4(l) + ClO3−(aq) → NO(g) + Cl−(aq) (c) Cl2O7(g) + H2O2(aq) → ClO2−(aq) + O2(g) + H+

ANSWER (a) Disproportionation of P (0 → −3 in PH3 and 0 → +1 in HPO2−). Balanced:
P4 + 3OH− + 3H2O → PH3 + 3HPO2−. Here P4 is both oxidant and reductant (disproportionation). (b) N: −2 → +2 (loses 4e−, oxidised); Cl: +5 → −1 (gains 6e−, reduced). LCM 12 ⇒ 3 N2H4 and 2 ClO3−:
3N2H4 + 2ClO3− → 6NO + 2Cl− + 6H2O. Oxidant: ClO3−; reductant: N2H4. (c) Cl: +7 → +3 in ClO2− (gains 4e− per Cl, reduced); O of H2O2: −1 → 0 (oxidised). Balanced:
Cl2O7 + 4H2O2 + 2OH− → 2ClO2− + 4O2 + 5H2O. Oxidant: Cl2O7; reductant: H2O2.

7.20 What sorts of informations can you draw from the following reaction? (CN)2(g) + 2OH−(aq) → CN−(aq) + CNO−(aq) + H2O(l)

ANSWER In cyanogen (CN)2 each CN group has an average oxidation state of +3 (for carbon). In the products, carbon is +2 in CN− (reduced) and +4 in CNO− (oxidised). Information: cyanogen behaves like a halogen (a “pseudohalogen”) and undergoes disproportionation in alkali — the same element is simultaneously oxidised and reduced, exactly as Cl2 does with OH−.

7.21 The Mn3+ ion is unstable in solution and undergoes disproportionation to give Mn2+, MnO2, and H+ ion. Write a balanced ionic equation for the reaction.

ANSWER Mn: +3 → +2 (gain 1e−, reduced) and +3 → +4 in MnO2 (lose 1e−, oxidised). Electrons already balance (1 each); ratio 2 Mn3+ → 1 Mn2+ + 1 MnO2. Balancing O with 2H2O and charge with 4H+: 2Mn3+ + 2H2O → Mn2+ + MnO2 + 4H+. (Check: left charge +6 = right charge +2 + 0 + 4 = +6.)

7.22 Consider the elements: Cs, Ne, I and F. (a) Identify the element that exhibits only negative oxidation state. (b) Identify the element that exhibits only positive oxidation state. (c) Identify the element that exhibits both positive and negative oxidation states. (d) Identify the element which exhibits neither the negative nor does the positive oxidation state.

ANSWER (a) F — most electronegative element, shows only −1. (b) Cs — an alkali metal, shows only +1. (c) I — shows −1 (in iodides) and positive states (+1, +3, +5, +7 with oxygen/fluorine). (d) Ne — an inert noble gas, shows oxidation state 0 only.

7.23 Chlorine is used to purify drinking water. Excess of chlorine is harmful. The excess of chlorine is removed by treating with sulphur dioxide. Present a balanced equation for this redox change taking place in water.

ANSWER Cl2 (0 → −1, reduced) oxidises SO2 (S: +4 → +6, oxidised) in water: Cl2 + SO2 + 2H2O → 2HCl + H2SO4 (ionic: Cl2 + SO2 + 2H2O → 2Cl− + SO42− + 4H+). The harmful excess chlorine is converted to harmless chloride.

7.24 Refer to the periodic table given in your book and now answer the following questions: (a) Select the possible non-metals that can show disproportionation reaction. (b) Select three metals that can show disproportionation reaction.

ANSWER (a) Non-metals: P, Cl, S (and Br, I) — they have at least three oxidation states and an intermediate state available, e.g. P4, Cl2, S8 disproportionate in alkali. (b) Metals: Mn, Cu and Ga (also Mn3+, Cu+, etc.) — these exist in intermediate states that disproportionate, e.g. 2Cu+ → Cu2+ + Cu.

7.25 In Ostwald’s process for the manufacture of nitric acid, the first step involves the oxidation of ammonia gas by oxygen gas to give nitric oxide gas and steam. What is the maximum weight of nitric oxide that can be obtained starting only with 10.00 g of ammonia and 20.00 g of oxygen?

ANSWER Reaction: 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g). Moles: NH3 = 10.00 ÷ 17 = 0.588 mol; O2 = 20.00 ÷ 32 = 0.625 mol. Limiting reagent: 4 mol NH3 needs 5 mol O2, so 0.588 mol NH3 needs 0.588 × 5/4 = 0.735 mol O2. Only 0.625 mol O2 is available ⇒ O2 is the limiting reagent. NO formed: 5 mol O2 → 4 mol NO, so 0.625 mol O2 → 0.625 × 4/5 = 0.500 mol NO. Mass of NO = 0.500 × 30 = 15 g. (Matches the NCERT answer key.)

7.26 Using the standard electrode potentials given in the Table 7.1, predict if the reaction between the following is feasible: (a) Fe3+(aq) and I−(aq)    (b) Ag+(aq) and Cu(s)    (c) Fe3+(aq) and Cu(s)    (d) Ag(s) and Fe3+(aq)    (e) Br2(aq) and Fe2+(aq)

ANSWER A reaction is feasible if the overall cell potential E°cell = E°cathode − E°anode is positive (oxidant’s E° > species being oxidised). Values used: Fe3+/Fe2+ = +0.77, I2/I− = +0.54, Ag+/Ag = +0.80, Cu2+/Cu = +0.34, Br2/Br− = +1.09 V. (a) Fe3+ (+0.77) oxidises I− (+0.54): E° = +0.23 V ⇒ feasible. (b) Ag+ (+0.80) oxidises Cu (+0.34): E° = +0.46 V ⇒ feasible. (c) Fe3+ (+0.77) oxidises Cu (+0.34): E° = +0.43 V ⇒ feasible. (d) Fe3+ (+0.77) oxidising Ag (+0.80): E° = −0.03 V ⇒ not feasible. (e) Br2 (+1.09) oxidises Fe2+ (+0.77): E° = +0.32 V ⇒ feasible.

7.27 Predict the products of electrolysis in each of the following: (i) An aqueous solution of AgNO3 with silver electrodes (ii) An aqueous solution of AgNO3 with platinum electrodes (iii) A dilute solution of H2SO4 with platinum electrodes (iv) An aqueous solution of CuCl2 with platinum electrodes

ANSWER (i) Cathode: Ag+ + e− → Ag (silver deposits). Anode: silver dissolves, Ag → Ag+ + e− (active electrode). Net: silver transfers from anode to cathode. (ii) Cathode: Ag deposited. Anode (inert Pt): water oxidised → O2 + H+ (2H2O → O2 + 4H+ + 4e−). (iii) Cathode: 2H+ + 2e− → H2. Anode: 2H2O → O2 + 4H+ + 4e−. Net: electrolysis of water. (iv) Cathode: Cu2+ + 2e− → Cu. Anode: 2Cl− → Cl2 + 2e−.

7.28 Arrange the following metals in the order in which they displace each other from the solution of their salts. Al, Cu, Fe, Mg and Zn.

ANSWER A metal displaces any metal below it in the activity (electrochemical) series. Using their standard reduction potentials (more negative = more reactive = better displacer): Mg > Al > Zn > Fe > Cu. So Mg displaces all the others, and Cu is displaced by all of them.

7.29 Given the standard electrode potentials, K+/K = −2.93 V, Ag+/Ag = 0.80 V, Hg2+/Hg = 0.79 V, Mg2+/Mg = −2.37 V, Cr3+/Cr = −0.74 V, arrange these metals in their increasing order of reducing power.

ANSWER The more negative the standard reduction potential, the greater the reducing power. Increasing reducing power means going from the most positive E° to the most negative E°: Ag (0.80) < Hg (0.79) < Cr (−0.74) < Mg (−2.37) < K (−2.93). Thus K is the strongest reducing agent and Ag the weakest.

7.30 Depict the galvanic cell in which the reaction Zn(s) + 2Ag+(aq) → Zn2+(aq) + 2Ag(s) takes place. Further show: (i) which of the electrode is negatively charged, (ii) the carriers of the current in the cell, and (iii) individual reaction at each electrode.

ANSWER Cell representation: Zn(s) | Zn2+(aq) || Ag+(aq) | Ag(s). (i) The zinc electrode (anode) is the negative electrode, because oxidation releases electrons there. (ii) Current carriers: electrons in the external metallic wire, and ions (through the solutions and the salt bridge) inside the cell. (iii) Anode (oxidation): Zn(s) → Zn2+(aq) + 2e−. Cathode (reduction): 2Ag+(aq) + 2e− → 2Ag(s).

Extra Practice Questions

Short Answer Type Questions

Q1. What is the oxidation number of Cr in K2Cr2O7?

ANSWER2(+1) + 2x + 7(−2) = 0 ⇒ 2x = 12 ⇒ Cr = +6.

Q2. Define a disproportionation reaction with one example.

ANSWERA redox reaction in which the same element in one oxidation state is simultaneously oxidised and reduced. Example: 2H2O2 → 2H2O + O2 (O goes from −1 to −2 and to 0).

Q3. Why does the oxidation number of oxygen become −1 in peroxides?

ANSWERIn peroxides (e.g. H2O2, Na2O2) two oxygen atoms are directly bonded (−O−O−). The shared O–O bond pair is split equally, so each oxygen carries only −1 instead of the usual −2.

Q4. Identify the oxidant and reductant in: Zn + CuSO4 → ZnSO4 + Cu.

ANSWERZn (0 → +2) is oxidised, so Zn is the reducing agent; Cu2+ (+2 → 0) is reduced, so CuSO4 is the oxidising agent.

Q5. Why is fluorine unable to show a positive oxidation state?

ANSWERFluorine is the most electronegative element and has no d-orbitals, so it always attracts electrons and can only show 0 (in F2) or −1 (in compounds) — never a positive state.

Long Answer Type Questions

Q1. Explain the four types of redox reactions with one balanced example each.

ANSWERCombination: two species combine and at least one is elemental, e.g. C + O2 → CO2 (C: 0→+4). Decomposition: a compound breaks down giving at least one element, e.g. 2KClO3 → 2KCl + 3O2 (O: −2→0). Displacement: one element replaces another, e.g. CuSO4 + Zn → ZnSO4 + Cu (metal displacement), or Cl2 + 2KBr → 2KCl + Br2 (non-metal displacement). Disproportionation: one element is both oxidised and reduced, e.g. 2H2O2 → 2H2O + O2. In each case oxidation numbers change, confirming the redox nature.

Q2. Describe the ion-electron (half-reaction) method to balance MnO4− + Fe2+ in acidic medium.

ANSWERWrite the half-reactions: oxidation Fe2+ → Fe3+ + e−; reduction MnO4− + 8H+ + 5e− → Mn2+ + 4H2O. Balance O with H2O, H with H+, and charge with electrons. Multiply the oxidation half by 5 to equalise electrons, then add: MnO4− + 5Fe2+ + 8H+ → Mn2+ + 5Fe3+ + 4H2O. A final check of atoms and charge (left +17 = right +17) confirms the equation is balanced.

Q3. Explain how a Daniell cell converts a redox reaction into electrical energy.

ANSWERA Daniell cell separates the spontaneous reaction Zn + Cu2+ → Zn2+ + Cu into two half-cells: a Zn rod in ZnSO4 (anode) and a Cu rod in CuSO4 (cathode), joined by a salt bridge. At the anode Zn is oxidised (Zn → Zn2+ + 2e−); the electrons travel through the external wire to the cathode where Cu2+ is reduced (Cu2+ + 2e− → Cu). The salt bridge maintains electrical neutrality by ion migration. Because the electron transfer now happens through the wire rather than directly, the energy of the redox reaction is delivered as an electric current. This is the basis of all galvanic cells studied further in Class 12.

MCQs & Assertion–Reason

1. The oxidation number of S in H2SO4 is:

(a) +2    (b) +4    (c) +6    (d) −2

2. In which compound does oxygen have an oxidation number of −1?

(a) H2O    (b) H2O2    (c) OF2    (d) CO2

3. The oxidation number of H in NaH is:

(a) +1    (b) 0    (c) −1    (d) +2

4. 2H2O2 → 2H2O + O2 is an example of:

(a) combination    (b) displacement    (c) disproportionation    (d) combustion

5. An oxidising agent is a species that:

(a) loses electrons    (b) gains electrons    (c) is itself oxidised    (d) decreases its own oxidation number is impossible

6. The average oxidation number of Fe in Fe3O4 is:

(a) +2    (b) +3    (c) +8/3    (d) +4

7. Which halogen is the strongest oxidising agent?

(a) F2    (b) Cl2    (c) Br2    (d) I2

8. In the reaction Cl2 + 2KI → 2KCl + I2, chlorine acts as:

(a) reducing agent    (b) oxidising agent    (c) both    (d) neither

9. The standard electrode potential of the hydrogen electrode is taken as:

(a) +1.00 V    (b) −1.00 V    (c) 0.00 V    (d) +0.34 V

10. In a galvanic cell, oxidation occurs at the:

(a) cathode    (b) anode    (c) salt bridge    (d) wire

Answer key: 1-(c), 2-(b), 3-(c), 4-(c), 5-(b), 6-(c), 7-(a), 8-(b), 9-(c), 10-(b).

For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.

A-R 1. Assertion: Oxidation and reduction always occur simultaneously.

Reason: The electrons lost by the species being oxidised are gained by the species being reduced.

A-R 2. Assertion: Fluorine cannot show a positive oxidation state.

Reason: Fluorine is the most electronegative element and has no available d-orbitals.

A-R 3. Assertion: HNO3 can act both as an oxidising and a reducing agent.

Reason: Nitrogen in HNO3 is in its highest oxidation state, +5.

A-R 4. Assertion: In a disproportionation reaction one element is both oxidised and reduced.

Reason: The disproportionating element must exist in at least three oxidation states and start in an intermediate state.

A-R 5. Assertion: Zinc displaces copper from copper sulphate solution.

Reason: Zinc has a more negative standard reduction potential than copper, making it a stronger reducing agent.

Answer key: 1-(A), 2-(A), 3-(D), 4-(A), 5-(A).

Common Mistakes to Avoid

Watch out for these

  • Taking oxygen as −2 in peroxides/superoxides — it is −1 in H2O2, CaO2 and −½ in KO2.
  • Taking hydrogen as +1 in metal hydrides (NaH, CaH2) — there it is −1.
  • Reporting an impossible state (e.g. S = +8 in H2SO5) instead of spotting the peroxide linkage.
  • Forgetting to add OH− in basic medium (and to convert H+ + OH− to H2O) when balancing.
  • Not equalising electrons before adding the two half-reactions.
  • Confusing reducing power with E° — remember more negative E° = stronger reducing agent.
  • In stoichiometry problems, skipping the limiting-reagent check (as in Q7.25).

How to score full marks in this chapter

Memorise the six oxidation-number rules and apply them every time. For balancing, state the medium first (acidic vs basic), write clean half-reactions, balance O with water and H with H+, then electrons, and always do a final atom-and-charge check. For E° questions, write E°cell = E°cathode − E°anode and conclude “feasible if positive.” For numericals like Q7.25, convert grams to moles, identify the limiting reagent using the mole ratio, then compute the product mass and check against units.

Frequently Asked Questions

What is Class 11 Chemistry Chapter 7 about?

Chapter 7, Redox Reactions, explains reactions in which oxidation and reduction occur together. It covers the classical, electronic and oxidation-number views, the rules for assigning oxidation numbers, the four types of redox reactions, balancing by the oxidation-number and ion–electron methods, redox couples, standard electrode potentials and the Daniell cell.

How many exercises are there in Chapter 7 Redox Reactions?

There are 30 numbered exercise questions (7.1 to 7.30), covering oxidation numbers, identifying oxidants and reductants, balancing redox equations, electrode-potential feasibility, electrolysis products and one stoichiometry numerical (Q7.25). All are solved on this page.

What is the answer to NCERT Q7.25 (Ostwald’s process)?

Using 4NH3 + 5O2 → 4NO + 6H2O, oxygen (0.625 mol) is the limiting reagent, giving 0.500 mol NO = 15 g of nitric oxide, matching the official NCERT answer key.

Are these Class 11 Chemistry Chapter 7 solutions free?

Yes. All solutions are free and follow the official NCERT Chemistry Part II textbook for session 2026–27.

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