NCERT Solutions for Class 11 Chemistry Chapter 7: Redox Reactions – Free PDF Download

Redox Reactions is the first chapter of Part 2 of the NCERT Class 11 Chemistry textbook (2026-27 rationalised edition), and it builds the electron-transfer and oxidation-number toolkit that students will reuse throughout electrochemistry, p-block chemistry, and inorganic qualitative analysis in Class 12. The chapter moves from the classical idea of oxidation and reduction, through oxidation number rules and the two methods of balancing redox equations (oxidation-number method and ion-electron/half-reaction method), to redox reactions as electrode processes governed by standard electrode potentials. Below you will find fully worked, independently re-derived solutions to every question in the current NCERT Exercise for Chapter 7, covering oxidation-number assignment, disproportionation, redox balancing, and electrochemical-cell/electrode-potential problems.

NCERT Solutions for Class 11 Chemistry Chapter 7: Redox Reactions

Q7.1: Assign oxidation numbers to the underlined elements

Assign oxidation numbers to the underlined element in each of the following species:

(a) NaH2PO4 (P): Na = +1, each O = -2. Let P = x. 1 + 2(1) + x + 4(-2) = 0 → 1 + 2 + x – 8 = 0 → x = +5.
(b) NaHSO4 (S): 1 + 1 + x – 8 = 0 → x = +6.
(c) H4P2O7 (P): 4(+1) + 2x + 7(-2) = 0 → 4 + 2x – 14 = 0 → 2x = 10 → x = +5 per P atom.
(d) K2MnO4 (Mn): 2(+1) + x + 4(-2) = 0 → 2 + x – 8 = 0 → x = +6.
(e) CaO2 (O, calcium peroxide): Ca is always +2, and CaO2 contains a peroxide (O-O) linkage, so each O = -1 (not -2). Check: +2 + 2(-1) = 0. → O = -1.
(f) NaBH4 (B): H is bonded to a less electronegative element (B), so H = -1 (hydridic hydrogen). 1 + x + 4(-1) = 0 → 1 + x – 4 = 0 → x = +3.
(g) H2S2O7 (S, pyrosulphuric acid/oleum): 2(+1) + 2x + 7(-2) = 0 → 2 + 2x – 14 = 0 → 2x = 12 → x = +6 per S atom.
(h) KAl(SO4)2·12H2O (S, potash alum): each SO4^2- group has O = -2 × 4 = -8, so S + (-8) = -2 → S = +6.

Final answer: P = +5, S = +6, P = +5, Mn = +6, O = -1, B = +3, S = +6, S = +6 respectively (worked from first principles above).

Q7.2: Oxidation numbers that need structural rationalisation

What are the oxidation numbers of the underlined elements in the following, and how do you rationalise your results?

(a) KI3: In KI3, the I3^– ion is not three equivalent iodine atoms — it is an I2 molecule (both atoms at oxidation state 0) coordinated to an I^– ion (oxidation state -1). The formula-based “average” calculation gives -1/3 per I atom [K = +1, so 3(I) = -1, average I = -1/3], but this average has no physical meaning; the true picture is two I atoms at 0 and one I atom at -1.
(b) H2S4O6 (tetrathionate): Average from the formula: 2(+1) + 4x + 6(-2) = 0 → 2 + 4x – 12 = 0 → x = +2.5. Structurally, tetrathionate has a chain O3S-S-S-SO3: the two central sulphur atoms (bonded only to other S atoms) are at 0, and the two terminal (SO3-type) sulphur atoms are at +5 each. Average = (0+0+5+5)/4 = +2.5 — matches the formula average, but the real oxidation states are 0, 0, +5, +5, not four identical +2.5 atoms.
(c) Fe3O4 (magnetite): Formula average: 3x + 4(-2) = 0 → x = +8/3. Structurally Fe3O4 is FeO·Fe2O3, i.e., one Fe2+ and two Fe3+; average = (2+3+3)/3 = +8/3, confirming the mix of +2 and +3 iron.
(d) CH3CH2OH (ethanol): Formula average: 2x + 6(+1) + 1(-2) = 0 → 2x = -4 → x = -2 per C. By bond-by-bond counting (C-H gives C = -1, C-O gives C = +1, C-C gives 0): the methyl carbon (3 C-H, 1 C-C) = -3; the CH2OH carbon (2 C-H, 1 C-O, 1 C-C) = -1. Average = (-3 + -1)/2 = -2, matching the formula value, but the two carbons individually are -3 and -1, not both -2.
(e) CH3COOH (acetic acid): Formula average: 2x + 4(+1) + 2(-2) = 0 → 2x = 0 → x = 0 per C. By structure: the methyl carbon = -3 (3 C-H bonds); the carboxyl carbon (one C=O counted as two O-bonds, one C-OH, one C-C) = +2 + 1 + 0 = +3. Average = (-3 + 3)/2 = 0, matching the formula, but individually the carbons are -3 and +3, not both 0.

Final answer: The formula-based averages (-1/3, +2.5, +8/3, -2, 0) are mathematically correct but physically misleading, since each of these species contains chemically non-equivalent atoms of the same element at different real oxidation states, as shown above.

Q7.3: Justify that these reactions are redox reactions

Justify that the following reactions are redox reactions:

(a) CuO(s) + H2(g) → Cu(s) + H2O(g): Cu goes from +2 (in CuO) to 0 (reduction); H goes from 0 (in H2) to +1 (in H2O) (oxidation). Both oxidation and reduction occur, so it is a redox reaction.
(b) Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g): Fe goes from +3 to 0 (reduction); C goes from +2 (in CO) to +4 (in CO2) (oxidation). Redox reaction.
(c) 4BCl3(g) + 3LiAlH4(s) → 2B2H6(g) + 3LiCl(s) + 3AlCl3(s): assigning carefully, in B2H6, H is bonded to less electronegative B, so H = -1 and B = +3 (since 2B + 6(-1) = 0 → B = +3, unchanged from BCl3). The actual electron transfer is on hydrogen: H in LiAlH4 is -1 (hydridic), and H in B2H6 is also -1, so hydrogen is transferred as a hydride ion without a net oxidation-state change either. This is essentially a metathesis (double-displacement) reaction with no change in oxidation number for any atom, so despite appearances it is not a genuine electron-transfer redox reaction — it proceeds by hydride-ion transfer, not electron transfer between different oxidation states.
(d) 2K(s) + F2(g) → 2K+F-(s): K goes from 0 to +1 (oxidation); F goes from 0 to -1 (reduction). Redox reaction.
(e) 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g): N goes from -3 (in NH3) to +2 (in NO) (oxidation); O goes from 0 to -2 (reduction). Redox reaction.

Final answer: Reactions (a), (b), (d) and (e) are genuine redox reactions with a clear change in oxidation number of at least two elements; reaction (c) shows no net oxidation-state change on careful analysis and is better classified as a hydride-transfer metathesis reaction, a subtlety many secondary sources skip over.

Q7.4: Is the reaction of fluorine with ice a redox reaction?

Fluorine reacts with ice: H2O(s) + F2(g) → HF(g) + HOF(g). Assign oxidation numbers: in H2O, H = +1, O = -2. In F2, F = 0. In HF, H = +1, F = -1. In HOF (hypofluorous acid), fluorine (more electronegative than O) takes -1 as usual, and since H(+1) + O(x) + F(-1) = 0 for the neutral molecule, oxygen must be assigned 0 here (pushed toward a less negative state because it is bonded to the more electronegative F). So oxygen’s oxidation number changes from -2 (in H2O) to 0 (in HOF) — an oxidation — while fluorine goes from 0 (in F2) to -1 (in both HF and HOF) — a reduction.

Final answer: Yes, it is a redox reaction: oxygen is oxidised from -2 to 0, and fluorine is reduced from 0 to -1.

Q7.5: Oxidation numbers of S, Cr and N — and the “fallacy”

Calculate the oxidation number of sulphur, chromium and nitrogen in H2SO5, Cr2O7^2- and NO3^-. Suggest the structures of these compounds and account for the fallacy.

H2SO5 (peroxomonosulphuric acid / Caro’s acid): A naive calculation treating all five oxygens as -2 gives 2(+1) + x + 5(-2) = 0 → x = +8, which is chemically impossible since sulphur (Group 16) cannot exceed +6. This is the “fallacy.” The correct structure is HO-SO2-O-O-H: two oxygens are normal double-bonded oxo oxygens (-2 each), one oxygen is a normal -OH oxygen (-2), and two oxygens form a peroxide (-O-O-) linkage (-1 each, as in H2O2). Recomputing: x + 2(+1 for the 2 H) + 2(-2) + 1(-2) + 2(-1) = 0 → x + 2 – 4 – 2 – 2 = 0 → x = +6.
Cr2O7^2- (dichromate): 2x + 7(-2) = -2 → 2x = 12 → x = +6 per Cr. No peroxide linkage is present here, so this is a genuine, unproblematic +6 (the well-known Cr(VI) state).
NO3^- (nitrate): x + 3(-2) = -1 → x = +5. This is nitrogen’s maximum possible oxidation state (Group 15) and is likewise unproblematic.

Final answer: S in H2SO5 = +6 (not the naive +8, corrected for the O-O peroxide linkage), Cr in Cr2O7^2- = +6, N in NO3^- = +5. The “fallacy” is applying a blanket -2 to every oxygen even when a peroxide bond is present; only H2SO5 needed this correction.

Q7.6: Write formulae for the following 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

Final answer: HgCl2, NiSO4, SnO2, Tl2SO4, Fe2(SO4)3 and Cr2O3, obtained by balancing each metal’s stated Roman-numeral charge against the standard charge of the anion (Cl- = -1, SO4^2- = -2, O^2- = -2).

Q7.7: Substances showing carbon (-4 to +4) and nitrogen (-3 to +5) oxidation states

Carbon shows every integral oxidation state from -4 to +4 through: CH4 (-4), C2H2 (-1), H2C=CH2 (-2), CH3OH (-2), HCHO (0), HCOOH (+2), CO2 (+4), CCl4 (+4). Nitrogen shows every integral state from -3 to +5 through: NH3 (-3), N2H4 (-2), NH2OH (-1), N2 (0), N2O (+1), NO (+2), N2O3/HNO2 (+3), NO2 (+4), N2O5/HNO3 (+5).

Final answer: Carbon spans -4 (CH4) to +4 (CO2, CCl4) through intermediate compounds like C2H2, CH3OH, HCHO and HCOOH; nitrogen spans -3 (NH3) to +5 (HNO3) through N2H4, NH2OH, N2, N2O, NO, HNO2 and NO2, as listed above.

Q7.8: Why are SO2 and H2O2 both oxidants and reductants, while O3 and HNO3 are only oxidants?

Sulphur in SO2 is at +4, an intermediate oxidation state for sulphur (whose range is -2 to +6) — it can rise to +6 (acting as a reducing agent, e.g. being oxidised to SO4^2-) or fall to lower states such as 0 or -2 (acting as an oxidising agent). Similarly, oxygen in H2O2 is at -1, intermediate between 0 (in O2) and -2 (in H2O), so H2O2 can be oxidised to O2 (acting as reductant) or reduced to H2O (acting as oxidant). In contrast, in O3 the “active” oxygen has no easy access to a still-higher positive oxygen state, and in HNO3 nitrogen is at +5, its maximum possible oxidation state — it can only decrease (never increase), so HNO3 can only act as an oxidising agent, never as a reducing agent.

Final answer: SO2 (S = +4) and H2O2 (O = -1) sit at intermediate oxidation states and can move either up or down, giving them dual oxidising/reducing character, whereas O3 and HNO3 (N at its ceiling of +5) can only be reduced further, so they act solely as oxidants.

Q7.9: Why write the water-balanced form of these reactions, and how to trace the mechanism

(a) 6CO2(g) + 6H2O(l) → C6H12O6(aq) + 6O2(g) is more correctly written as 6CO2(g) + 12H2O(l) → C6H12O6(aq) + 6H2O(l) + 6O2(g) because, mechanistically, the oxygen gas evolved in photosynthesis comes entirely from water molecules (not from CO2), so 12 whole water molecules must be shown as reactants to supply the 12 oxygen atoms needed to form 6 O2, with 6 water molecules regenerated as products.
(b) Similarly, O3(g) + H2O2(l) → H2O(l) + 2O2(g) is better written as O3(g) + H2O2(l) → H2O(l) + O2(g) + O2(g), keeping the two product O2 molecules distinct because they originate from two different sources: one O2 forms from the peroxide oxygens of H2O2, and the other forms from ozone.
The technique used to establish which atoms end up where is isotopic (heavy-isotope) labelling — e.g., using water enriched with the oxygen-18 isotope (H2^18O) and tracking whether the ^18O ends up in the O2 product or remains as H2O, thereby tracing the actual reaction path/mechanism.

Final answer: Both equations are rewritten to reflect that all the product O2 in each case is mechanistically traceable to specific reactant oxygen atoms (water in photosynthesis; H2O2 and O3 separately in the second case), a fact established using isotopic (18O) tracer studies.

Q7.10: Why is AgF2 a very strong oxidising agent?

In AgF2, silver is forced into the unusual +2 oxidation state (Ag2+). This state is highly unstable for silver, because silver’s normal, far more stable oxidation state is +1 (Ag+, with a stable filled d10 configuration). Ag2+ therefore has a very strong tendency to gain one electron and revert to the stable Ag+ state, i.e., to get reduced. A species with such a strong drive to be reduced is, by definition, a powerful oxidising agent.

Final answer: AgF2 contains silver in the unstable +2 state, which readily gains an electron to fall back to the far more stable +1 state; this strong drive toward reduction makes AgF2 (where formed) a very strong oxidising agent.

Q7.11: Excess reductant vs. excess oxidant — three illustrations

When a redox reaction is carried out with the reducing agent in excess, the product tends to be the one with the lower oxidation state of the variable element; when the oxidising agent is in excess, the product has the higher oxidation state.

(i) Sulphur with limited O2 gives SO2 (S = +4); with excess O2/further oxidation, SO3 (S = +6) is favoured.
(ii) Phosphorus with limited Cl2 gives PCl3 (P = +3); with excess Cl2, PCl5 (P = +5) is formed.
(iii) Copper reacting with dilute HNO3 (a comparatively weaker oxidant) gives NO (N = +2); with concentrated HNO3 (stronger oxidising conditions), NO2 (N = +4) is the major product — illustrating the same principle for the oxidant’s own reduction product.

Final answer: Excess reductant → lower oxidation state product (e.g., PCl3, SO2, NO); excess oxidant → higher oxidation state product (e.g., PCl5, SO3, NO2), as shown in the three illustrations above.

Q7.12: Alkaline KMnO4 for benzoic acid; HCl gas vs Br2 vapour with conc. H2SO4

(a) Toluene is oxidised to benzoic acid using alkaline KMnO4 (followed by acidification) because the strongly oxidising, alkaline permanganate medium is powerful enough to oxidise the benzylic -CH3 side chain all the way to -COOH (via the intermediate potassium benzoate) without over-oxidising or degrading the aromatic ring. Balanced equation:
C6H5CH3 + 2KMnO4 → C6H5COOK + 2MnO2 + KOH + H2O
(Atom check: C 7=7, H 8=8, O 8=8, K 2=2, Mn 2=2 — balanced.) The potassium benzoate formed is then acidified to yield free benzoic acid.
(b) When concentrated H2SO4 is added to a chloride salt, colourless HCl gas is evolved because Cl- is too weak a reducing agent to be oxidised by H2SO4, so only a simple (non-redox) acid-displacement occurs. But with a bromide salt, Br- is a stronger reducing agent than Cl-, so concentrated H2SO4 can additionally oxidise the HBr formed to red-brown Br2 vapour (H2SO4 itself being reduced, typically to SO2): 2HBr + H2SO4 → Br2 + SO2 + 2H2O.

Final answer: (a) Alkaline KMnO4 fully oxidises the -CH3 group to -COOK without destroying the ring, per the balanced equation above. (b) Chloride only undergoes acid displacement (HCl gas) because Cl- resists oxidation, while bromide’s stronger reducing character lets conc. H2SO4 additionally oxidise HBr to Br2 vapour.

Q7.13: Identify substance oxidised/reduced and oxidising/reducing agents

(a) 2AgBr(s) + C6H6O2(aq) → 2Ag(s) + 2HBr(aq) + C6H4O2(aq): Ag+ (in AgBr) is reduced to Ag(0); hydroquinone (C6H6O2) is oxidised to quinone (C6H4O2). Oxidising agent = AgBr; reducing agent = hydroquinone.
(b) HCHO(l) + 2[Ag(NH3)2]+(aq) + 3OH-(aq) → 2Ag(s) + HCOO-(aq) + 4NH3(aq) + 2H2O(l): HCHO is oxidised to HCOO-; Ag+ (as the diammine complex) is reduced to Ag(0). Oxidising agent = [Ag(NH3)2]+; reducing agent = HCHO.
(c) HCHO(l) + 2Cu2+(aq) + 5OH-(aq) → Cu2O(s) + HCOO-(aq) + 3H2O(l): HCHO is oxidised to HCOO-; Cu2+ is reduced to Cu+ (in Cu2O). Oxidising agent = Cu2+; reducing agent = HCHO.
(d) N2H4(l) + 2H2O2(l) → N2(g) + 4H2O(l): N (-2 in N2H4) is oxidised to N2 (0); O (-1 in H2O2) is reduced to -2 (in H2O). Oxidising agent = H2O2; reducing agent = N2H4.
(e) Pb(s) + PbO2(s) + 2H2SO4(aq) → 2PbSO4(s) + 2H2O(l): Pb(0) is oxidised to Pb2+ (in PbSO4); Pb (+4, in PbO2) is reduced to Pb2+. Oxidising agent = PbO2; reducing agent = Pb(s). (This is the discharge reaction of the lead storage battery.)

Final answer: Oxidising/reducing agent pairs are (a) AgBr / hydroquinone, (b) [Ag(NH3)2]+ / HCHO, (c) Cu2+ / HCHO, (d) H2O2 / N2H4, (e) PbO2 / Pb, as identified from the oxidation-number changes above.

Q7.14: Why does thiosulphate react differently with iodine and bromine?

2S2O3^2-(aq) + I2(s) → S4O6^2-(aq) + 2I-(aq) and S2O3^2-(aq) + 4Br2(l) + 5H2O(l) → 2SO4^2-(aq) + 8Br-(aq) + 10H+(aq). In thiosulphate, S is at an average +2. Iodine (I2/I-, E° = +0.54 V) is a comparatively mild oxidising agent, so it can only partially oxidise thiosulphate, taking sulphur from +2 to +2.5 (tetrathionate, S4O6^2-). Bromine (Br2/Br-, E° = +1.09 V) is a much stronger oxidising agent, so it can fully oxidise sulphur all the way to +6 (sulphate, SO4^2-).

Final answer: Iodine, being a weaker oxidant, converts thiosulphate only to tetrathionate (S: +2 → +2.5), while the much stronger oxidant bromine drives the oxidation all the way to sulphate (S: +2 → +6).

Q7.15: Fluorine as best oxidant; hydroiodic acid as best reductant

Among the halogens, fluorine has the highest standard reduction potential (F2/F-, E° = +2.87 V) due to its small atomic size, very high electronegativity, weak F-F bond, and very high hydration enthalpy of F-, all of which favour F2 readily accepting electrons — e.g., F2 oxidises even water: 2F2 + 2H2O → 4HF + O2, something no other halogen does. Among the hydrohalic acids, HI is the best reductant because the H-I bond is the weakest (largest, most polarisable iodine atom, longest and weakest bond), so I- is oxidised most easily of all the halide ions — e.g., HI reduces concentrated H2SO4 all the way to H2S, whereas HCl cannot reduce H2SO4 at all.

Final answer: F2 is the best oxidant due to its high electronegativity, small size, weak F-F bond and high hydration energy of F- (highest E°); HI is the best reductant because its weak H-I bond lets I- lose an electron most readily of all the halides.

Q7.16: XeO6^4- reacting with fluoride — what it reveals about Na4XeO6

XeO6^4-(aq) + 2F-(aq) + 6H+(aq) → XeO3(g) + F2(g) + 3H2O(l). Xenon falls from +8 (in the perxenate ion, XeO6^4-) to +6 (in XeO3) — a reduction — while fluoride is oxidised from -1 to 0 (F2). Since F2/F- already has the highest known standard reduction potential of any common couple (+2.87 V), for perxenate to be able to oxidise F- to F2 at all, the XeO6^4-/XeO3 couple must have an even higher reduction potential than F2/F-.

Final answer: The reaction shows that perxenate (as in Na4XeO6) is an exceptionally powerful oxidising agent — stronger even than elemental fluorine — since it is able to oxidise F- ions to F2 gas.

Q7.17: What Ag+ vs Cu2+ reactions reveal about their oxidising strength

H3PO2 reduces both [Ag(NH3)2]+ and Cu2+ to the metal/lower oxide. Benzaldehyde (C6H5CHO, an aromatic aldehyde) reduces [Ag(NH3)2]+ (Tollens’ reagent) to metallic silver, but shows no reaction at all with Cu2+ (Fehling-type reagent). This shows that Ag+ is a distinctly stronger oxidising agent than Cu2+ — strong enough to oxidise even a comparatively resistant aromatic aldehyde, which Cu2+ cannot do.

Final answer: Ag+ (as [Ag(NH3)2]+) is a stronger oxidising agent than Cu2+, since it can oxidise even the less reactive aromatic aldehyde (benzaldehyde), which fails to react with Cu2+ at all.

Q7.18: Balance the following redox reactions by the ion-electron method

(i) MnO4^- + I^- → MnO2 + I2 (basic medium). Mn: +7 → +4 (gain 3e-); I: -1 → 0 (lose 1e- per I, 2e- per I2). Reduction half: MnO4- + 2H2O + 3e- → MnO2 + 4OH-. Oxidation half: 2I- → I2 + 2e-. Equalising electrons (×2 and ×3) and adding:
2MnO4^- + 4H2O + 6I^- → 2MnO2 + 3I2 + 8OH^-
(ii) MnO4^- + SO2 → Mn2+ + HSO4^- (acidic medium). Mn: +7 → +2 (gain 5e-); S: +4 → +6 (lose 2e-). Reduction half: MnO4- + 8H+ + 5e- → Mn2+ + 4H2O. Oxidation half: SO2 + 2H2O → HSO4- + 3H+ + 2e-. Equalising electrons (×2 and ×5) and adding, then cancelling common H2O/H+:
2MnO4^- + 5SO2 + 2H2O + H^+ → 2Mn2+ + 5HSO4^-
(iii) H2O2 + Fe2+ → Fe3+ + H2O (acidic medium). O: -1 → -2 (gain 2e- per H2O2); Fe: +2 → +3 (lose 1e-). Reduction half: H2O2 + 2H+ + 2e- → 2H2O. Oxidation half: 2Fe2+ → 2Fe3+ + 2e-. Adding directly:
H2O2 + 2Fe2+ + 2H^+ → 2Fe3+ + 2H2O
(iv) Cr2O7^2- + SO2 → Cr3+ + SO4^2- (acidic medium). Cr: +6 → +3 per Cr (gain 3e- each, 6e- total); S: +4 → +6 (lose 2e-). Reduction half: Cr2O7^2- + 14H+ + 6e- → 2Cr3+ + 7H2O. Oxidation half: SO2 + 2H2O → SO4^2- + 4H+ + 2e-. Equalising electrons (×1 and ×3) and simplifying:
Cr2O7^2- + 3SO2 + 2H^+ → 2Cr3+ + 3SO4^2- + H2O

Final answer: The four balanced equations are as boxed above, each verified atom-by-atom and charge-by-charge using the ion-electron (half-reaction) method.

Q7.19: Balance in basic medium by both methods; identify oxidising and reducing agents

(a) P4(s) + OH^-(aq) → PH3(g) + H2PO2^-(aq). This is a disproportionation of P4 (oxidation state 0): one P is reduced to -3 (in PH3, gaining 3e-) and another P is oxidised to +1 (in H2PO2-, losing 1e-). Balancing the two half-reactions and combining in a 1:3 ratio of reduced to oxidised phosphorus gives, after simplification:
P4 + 3OH^- + 3H2O → PH3 + 3H2PO2^-
(checked: P 4=4, O 6=6, H 9=9, charge -3=-3). Since P4 is simultaneously oxidised and reduced, it is both the oxidising agent and the reducing agent.
(b) N2H4(l) + ClO3^-(aq) → NO(g) + Cl^-(aq). N: -2 → +2 per N (oxidation, 4e- per N, 8e- per N2H4); Cl: +5 → -1 (reduction, 6e-). Oxidation half: N2H4 + 8OH- → 2NO + 6H2O + 8e-. Reduction half: ClO3- + 3H2O + 6e- → Cl- + 6OH-. Equalising electrons (LCM 24: ×3 and ×4) and simplifying:
3N2H4 + 4ClO3^- → 6NO + 4Cl^- + 6H2O
Oxidising agent = ClO3-; reducing agent = N2H4.
(c) Cl2O7(g) + H2O2(aq) → ClO2^-(aq) + O2(g) + H^+. Cl: +7 → +3 (reduction, 4e- per Cl, 8e- per Cl2O7); O (peroxide): -1 → 0 (oxidation, 2e- per H2O2). Reduction half: Cl2O7 + 6H+ + 8e- → 2ClO2- + 3H2O. Oxidation half: H2O2 → O2 + 2H+ + 2e-. Equalising electrons (×1 and ×4) and simplifying:
Cl2O7 + 4H2O2 → 2ClO2^- + 4O2 + 3H2O + 2H^+
Oxidising agent = Cl2O7; reducing agent = H2O2.

Final answer: (a) P4 + 3OH- + 3H2O → PH3 + 3H2PO2- (P4 is both oxidant and reductant); (b) 3N2H4 + 4ClO3- → 6NO + 4Cl- + 6H2O (oxidant: ClO3-, reductant: N2H4); (c) Cl2O7 + 4H2O2 → 2ClO2- + 4O2 + 3H2O + 2H+ (oxidant: Cl2O7, reductant: H2O2).

Q7.20: What information can you draw from the cyanogen disproportionation reaction?

(CN)2(g) + 2OH-(aq) → CN-(aq) + CNO-(aq) + H2O(l). Assigning oxidation numbers (N = -3 throughout, O = -2): in cyanogen, (CN)2, carbon is at +3 (average); in the product cyanide ion, CN-, carbon is at +2; in the product cyanate ion, CNO-, carbon is at +4. So carbon splits from +3 into +2 (reduced) and +4 (oxidised) — a disproportionation. This is exactly analogous to the disproportionation of a halogen in alkali, e.g., Cl2 + 2OH- → Cl- + OCl- + H2O, which is why (CN)2 is classed as a pseudohalogen.

Final answer: The reaction shows that (CN)2 undergoes disproportionation (carbon splits from +3 into +2 in CN- and +4 in CNO-) in exactly the same pattern as a halogen reacting with alkali, confirming cyanogen’s classification as a “pseudohalogen.”

Q7.21: Balanced equation for Mn3+ disproportionation

Mn3+ is unstable in solution and disproportionates to Mn2+ and MnO2. One Mn3+ is reduced to Mn2+ (gains 1e-); another Mn3+ is oxidised to Mn4+ as MnO2 (loses 1e-, and needs O supplied by water, releasing H+). Combining 1:1 and supplying oxygen/hydrogen via water:
2Mn3+(aq) + 2H2O(l) → Mn2+(aq) + MnO2(s) + 4H+(aq)
Check: Mn 2=2; O 2=2; H 4=4; charge LHS +6, RHS (+2) + (+4 from 4H+) = +6. Balanced.

Final answer: 2Mn3+(aq) + 2H2O(l) → Mn2+(aq) + MnO2(s) + 4H+(aq).

Q7.22: Oxidation-state behaviour of Cs, Ne, I and F

(a) Element showing only a negative oxidation state: F (the most electronegative element of all, never found in a positive oxidation state).
(b) Element showing only a positive oxidation state: Cs (a highly electropositive alkali metal, essentially always +1).
(c) Element showing both positive and negative oxidation states: I (iodine is -1 in iodides, but can be +1, +3, +5 or +7 in oxoacids, oxides and interhalogen compounds).
(d) Element showing neither a positive nor a negative oxidation state (stays at 0): Ne (a noble gas that forms essentially no stable compounds).

Final answer: F = only negative; Cs = only positive; I = both positive and negative; Ne = neither (stays at 0).

Q7.23: Balanced equation for removing excess chlorine from drinking water with SO2

Cl2 is reduced (0 → -1) and SO2 is oxidised (+4 → +6, forming sulphate/sulphuric acid), consuming water:
Cl2(aq) + SO2(g) + 2H2O(l) → 2HCl(aq) + H2SO4(aq)
Check: Cl 2=2; S 1=1; O 4=4; H 4=4. Balanced.

Final answer: Cl2(aq) + SO2(g) + 2H2O(l) → 2HCl(aq) + H2SO4(aq).

Q7.24: Non-metals and metals that show disproportionation

(a) Possible non-metals: elements with several accessible intermediate oxidation states, such as Cl, Br, I (halogens: e.g. Cl2 + 2OH- → Cl- + OCl- + H2O), P (e.g. P4 → PH3 + H2PO2-, as in Q7.19a), and S (e.g. S8 in hot alkali giving sulfide- and thiosulfate-type products).
(b) Three metals: Cu (2Cu+ → Cu2+ + Cu, since Cu+ is unstable in aqueous solution), Au (3Au+ → Au3+ + 2Au), and Mn (Mn3+ → Mn2+ + MnO2, as in Q7.21).

Final answer: Non-metals: Cl, Br, I, P, S; Metals: Cu, Au, Mn, each chosen because the element has an intermediate oxidation state that can simultaneously rise and fall.

Q7.25: Maximum mass of NO from the Ostwald process (limiting-reagent calculation)

4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g). Molar mass NH3 = 17 g/mol; O2 = 32 g/mol; NO = 30 g/mol.
Moles of NH3 = 10.00/17 = 0.588 mol. Moles of O2 = 20.00/32 = 0.625 mol.
Required O2 for all the NH3 (ratio NH3:O2 = 4:5) = 0.588 × (5/4) = 0.735 mol, but only 0.625 mol O2 is available → O2 is the limiting reagent.
Moles of NO formed = moles O2 used × (4/5) = 0.625 × 0.8 = 0.500 mol.
Mass of NO = 0.500 mol × 30 g/mol = 15.0 g.

Final answer: 15.0 g of NO is the maximum obtainable, since O2 (not NH3) is the limiting reagent.

Q7.26: Feasibility of reactions from standard electrode potentials

Using standard reduction potentials: Fe3+/Fe2+ = +0.77 V; I2/I- = +0.54 V; Ag+/Ag = +0.80 V; Cu2+/Cu = +0.34 V; Br2/Br- = +1.09 V. A reaction is spontaneous when the species with the higher (more positive) reduction potential is reduced by the species with the lower one.

(a) Fe3+ + I-: 0.77 V > 0.54 V, so Fe3+ can oxidise I-. Feasible: 2Fe3+ + 2I- → 2Fe2+ + I2.
(b) Ag+ and Cu: 0.80 V > 0.34 V, so Ag+ can oxidise Cu. Feasible: 2Ag+ + Cu → 2Ag + Cu2+.
(c) Fe3+ and Cu: 0.77 V > 0.34 V, so Fe3+ can oxidise Cu. Feasible: 2Fe3+ + Cu → 2Fe2+ + Cu2+.
(d) Ag and Fe3+: this would require 0.77 V > 0.80 V, which is false (E°cell = 0.77 – 0.80 = -0.03 V). Not feasible.
(e) Br2 and Fe2+: 1.09 V > 0.77 V, so Br2 can oxidise Fe2+. Feasible: 2Fe2+ + Br2 → 2Fe3+ + 2Br- (E°cell = +0.32 V).

Final answer: Feasible: (a), (b), (c) and (e); Not feasible: (d), since Ag+/Ag (0.80 V) has a higher reduction potential than Fe3+/Fe2+ (0.77 V), so silver metal cannot reduce Fe3+.

Q7.27: Products of electrolysis

(i) Aqueous AgNO3 with silver electrodes: silver dissolves from the anode and deposits on the cathode, net effect being transfer of silver metal from anode to cathode with no change in solution concentration.
(ii) Aqueous AgNO3 with platinum (inert) electrodes: Ag+ is reduced at the cathode, depositing silver metal; at the anode, water is oxidised instead of NO3-: 2H2O → O2 + 4H+ + 4e-, releasing O2 gas.
(iii) Dilute H2SO4 with platinum electrodes: essentially electrolysis of water, giving H2 gas at the cathode and O2 gas at the anode, in a 2:1 volume ratio.
(iv) Aqueous CuCl2 with platinum electrodes: Cu2+ is reduced and deposited as copper metal at the cathode; Cl- is oxidised at the anode to Cl2 gas.

Final answer: (i) Ag transferred anode→cathode, solution unchanged; (ii) Ag deposited at cathode, O2 evolved at anode; (iii) H2 at cathode, O2 at anode (2:1 by volume); (iv) Cu deposited at cathode, Cl2 gas evolved at anode.

Q7.28: Arranging Al, Cu, Fe, Mg and Zn by displacement order

Using standard reduction potentials (more negative = more easily oxidised = more reactive): Mg2+/Mg ≈ -2.36 V, Al3+/Al ≈ -1.66 V, Zn2+/Zn ≈ -0.76 V, Fe2+/Fe ≈ -0.44 V, Cu2+/Cu = +0.34 V.

Final answer: Mg > Al > Zn > Fe > Cu (most reactive/best displacer first) — each metal in this list displaces every metal to its right from solutions of their salts.

Q7.29: Increasing order of reducing power from given electrode potentials

Given: 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. Reducing power increases as the standard reduction potential becomes more negative.

Final answer: Ag < Hg < Cr < Mg < K (increasing reducing power), following directly from arranging the given E° values from most positive to most negative.

Q7.30: Depicting the Zn/Ag galvanic cell

Zn(s) + 2Ag+(aq) → Zn2+(aq) + 2Ag(s). Cell representation: Zn(s) | Zn2+(aq) || Ag+(aq) | Ag(s).

(i) The zinc electrode is the anode (where oxidation occurs), and in a galvanic cell the anode is the negatively charged electrode.
(ii) Current carriers: inside the cell, ions carry the current through the electrolyte; in the external circuit, electrons flowing from zinc to silver carry the current.
(iii) Electrode reactions: anode (Zn): Zn(s) → Zn2+(aq) + 2e- (oxidation). Cathode (Ag): 2Ag+(aq) + 2e- → 2Ag(s) (reduction).

Final answer: Zinc electrode = negative (anode); current is carried by ions in solution and electrons in the external wire; anode reaction Zn → Zn2+ + 2e-, cathode reaction 2Ag+ + 2e- → 2Ag.

Class 11 Chemistry Chapter 7 – Notes and Extra Questions

Beyond these thirty exercise questions, students preparing for school tests and competitive exams should revise the oxidation-number rules, the four types of redox reactions (combination, decomposition, displacement and disproportionation), both balancing methods (oxidation-number and ion-electron), and the standard electrode potential series, since these form the conceptual backbone of electrochemistry (Class 12) and inorganic chemistry that follows this chapter. Practising extra numericals on limiting-reagent redox stoichiometry and half-reaction balancing in both acidic and basic media is the best way to build speed for board and entrance exams.

Frequently Asked Questions

Q1. How many questions are there in the NCERT Class 11 Chemistry Chapter 7 (Redox Reactions) exercise?
There are 30 main exercise questions (7.1 to 7.30), several of which have multiple lettered or numbered sub-parts, covering oxidation-number assignment, redox justification, balancing by both methods, and electrode-potential/electrochemical-cell problems.

Q2. What are the two methods used to balance redox reactions in this chapter?
The oxidation-number method (tracking the total increase and decrease in oxidation numbers to fix stoichiometric coefficients) and the ion-electron (half-reaction) method (splitting the reaction into separate oxidation and reduction half-reactions, balancing atoms and charge in each, then combining them so electrons cancel exactly) — both are used side by side in Q7.19.

Q3. What is a disproportionation reaction, with an example from this chapter?
A disproportionation reaction is a redox reaction in which the same element, present in one intermediate oxidation state, is simultaneously oxidised and reduced to give two different products. Chapter 7 has several examples, including P4 reacting with hot alkali (Q7.19a: P4 disproportionates into PH3 and H2PO2-) and Mn3+ disproportionating in solution into Mn2+ and MnO2 (Q7.21).

Q4. Was Redox Reactions renumbered in the 2023 NCERT rationalisation?
Yes. In the pre-2023 NCERT Class 11 Chemistry textbook, Redox Reactions was Chapter 8. After the 2023 syllabus rationalisation (still in effect for the 2026-27 session), it was moved up to Chapter 7 and now opens Part 2 of the book. The chapter’s content and its full set of 30 exercise questions were carried over unchanged — only the chapter and question numbering shifted from 8.x to 7.x.