NCERT Solutions for Class 11 Chemistry Chapter 7: Redox Reactions

Concept. The oxidation number of an atom in a neutral compound is found by assigning the known values of the other atoms (Na, K, Ca +1/+1/+2; H +1 except in metal hydrides; O −2 except in peroxides) and using the rule that the sum of all oxidation numbers equals zero in a neutral molecule (or the charge on an ion). In the printed question the underline is lost in the text; the elements normally asked, which we solve for, are P, S, P, Mn, O, B, S, S — the other elements are also given below for completeness.

1. (a) NaH₂PO₄: (+1) + 2(+1) + P + 4(−2) = 0 ⇒ P = 8 − 3 = +5.
2. (b) NaHSO₄: (+1) + (+1) + S + 4(−2) = 0 ⇒ S = +6.
3. (c) H₄P₂O₇: 4(+1) + 2P + 7(−2) = 0 ⇒ 2P = 10 ⇒ P = +5.
4. (d) K₂MnO₄: 2(+1) + Mn + 4(−2) = 0 ⇒ Mn = +6.
5. (e) CaO₂: this is calcium peroxide (Ca²⁺ and O₂²⁻), so Ca = +2 and O = −1 (2 + 2x = 0 ⇒ x = −1).
6. (f) NaBH₄: Na = +1 and H is a hydride (−1): (+1) + B + 4(−1) = 0 ⇒ B = +3.
7. (g) H₂S₂O₇: 2(+1) + 2S + 7(−2) = 0 ⇒ 2S = 12 ⇒ S = +6.
8. (h) KAl(SO₄)₂·12H₂O: (+1) + (+3) + 2S + 8(−2) = 0 (water is neutral) ⇒ 2S = 12 ⇒ S = +6.

Extra depth. Note the pattern: sulphur and phosphorus in these oxo-compounds are always at their group-maximum states (+6 for S, +5 for P), because oxygen is present in excess. The two special cases are CaO₂ and NaBH₄, where the usual rule (O = −2, H = +1) is overruled by the exception (peroxide oxygen −1, hydride hydrogen −1). Remember the rule order: first fix the ‘known’ atoms, then solve for the unknown. For H₄P₂O₇ (pyrophosphoric acid) and H₂S₂O₇ (oleum component) the oxidation number is the average over two atoms of the same element; both atoms are in the same state here, so the average equals the real value.

Conclusion: (a) P +5; (b) S +6; (c) P +5; (d) Mn +6; (e) O −1; (f) B +3; (g) S +6; (h) S +6.

Concept. In these cases the algebraic average gives a fractional oxidation number for the underlined element. Since electrons are not transferred in fractions, the fraction is only the average; the structure shows that the atoms of that element are present in different whole-number oxidation states (section ‘The paradox of fractional oxidation number’). The underlined elements here are I in KI₃, S in H₂S₄O₆, Fe in Fe₃O₄, and C in the two organic compounds.

1. (a) KI₃: K = +1, so the three iodine atoms (I₃⁻) total −1: average I = −1/3. Rationalisation: KI₃ is K⁺ and the triiodide ion I₃⁻ formed by I₂ + I⁻ → I₃⁻: two I are in the 0 state (the I₂ unit) and one I⁻ is in the −1 state; average = (0 + 0 − 1)/3 = −1/3.
2. (b) H₂S₄O₆ (tetrathionic acid): 2(+1) + 4S + 6(−2) = 0 ⇒ S = +2.5. The structure HO₃S–S–S–SO₃H has two terminal sulphurs at +5 and two middle sulphurs at 0 (they are linked only to other S atoms): (+5 + 0 + 0 + 5)/4 = +2.5.
3. (c) Fe₃O₄: 3Fe + 4(−2) = 0 ⇒ Fe = +8/3. It is a mixed oxide FeO·Fe₂O₃: one Fe is +2 and two Fe are +3; average (2 + 3 + 3)/3 = 8/3.
4. (d) CH₃CH₂OH: overall 2C + 6(+1) + (−2) = 0 ⇒ C = −2 (average). The two carbons are different: in CH₃ carbon is −3 (three C–H bonds), in CH₂OH carbon is −1 (two C–H bonds, one C–O bond): (−3 − 1)/2 = −2.
5. (e) CH₃COOH: 2C + 4(+1) + 2(−2) = 0 ⇒ C = 0 (average). The CH₃ carbon is −3 and the carboxylic carbon is +3 (C=O counts +2 and C–OH +1): (−3 + 3)/2 = 0.

Extra depth. Notice how the same reasoning explains very different molecules: organic compounds always contain carbon atoms in different environments, and carbon’s ‘average’ oxidation number is just a bookkeeping value. The same applies to iron oxide (a combination of Fe(II) and Fe(III)) and to polythionates (sulphur atoms in different bonding states). The idea becomes important in organic redox: oxidation of ethanol (C avg −2) to ethanoic acid (C avg 0) means the carbinol carbon goes from −1 to +3, an increase of 4 — four electrons lost per carbon.

Conclusion: (a) I −1/3; (b) S +2.5; (c) Fe +8/3; (d) C −2; (e) C 0 — all averages of different integer oxidation states.

Concept. A reaction is a redox reaction if some atoms change their oxidation number (equivalently, if electrons are transferred, or oxygen/hydrogen are added or removed). The method: assign oxidation numbers to every atom on both sides, identify the atoms that increase (oxidised) and those that decrease (reduced), and name the oxidant and reductant.

1. (a) CuO + H₂ → Cu + H₂O: Cu: +2 → 0 (reduced; CuO is the oxidant); H: 0 → +1 (oxidised; H₂ is the reductant). Classically: oxygen is removed from CuO and added to hydrogen.
2. (b) Fe₂O₃ + 3CO → 2Fe + 3CO₂: Fe: +3 → 0 (reduced; Fe₂O₃ oxidant); C: +2 → +4 (oxidised; CO reductant).
3. (c) 4BCl₃ + 3LiAlH₄ → 2B₂H₆ + 3LiCl + 3AlCl₃: B: +3 (in BCl₃) → −3 (in B₂H₆, with H counted as +1 by the rules) — reduced; H: −1 (hydride in LiAlH₄) → +1 (in B₂H₆) — oxidised. BCl₃ is the oxidant, LiAlH₄ the reductant.
4. (d) 2K + F₂ → 2K⁺F⁻: half reactions K → K⁺ + e⁻ (oxidation, 0 → +1) and F₂ + 2e⁻ → 2F⁻ (reduction, 0 → −1). K is the reductant, F₂ the oxidant.
5. (e) 4NH₃ + 5O₂ → 4NO + 6H₂O: N: −3 → +2 (oxidised; NH₃ reductant); O: 0 → −2 (reduced; O₂ oxidant). Total increase 5 × 4 = 20 = total decrease 2 × 10 = 20.

Extra depth. Reaction (c) deserves a remark: it is the preparation of diborane. If one used electronegativity values strictly (B 2.0 < H 2.1) one might argue that hydrogen stays −1 in B₂H₆ and boron +3, making the change ambiguous — but the rules of section 7.3 (H = +1 except in metal hydrides) are the ones followed in the NCERT answer, giving B: +3 → −3. In the same way (b) is the central reaction of the blast furnace and (e) is the first step of Ostwald’s process for nitric acid. Reaction (a) is both a reduction of the oxide and an oxidation of hydrogen — the cleanest example of ‘oxidation and reduction go together’.

Conclusion: In every case there is a change in oxidation number of at least two elements, hence each reaction is a redox reaction.

Concept. Fluorine is the most electronegative element, so in every compound its oxidation number is −1. Hydrogen is +1 in water and in HOF. Oxygen’s oxidation number in HOF has to be calculated — this is the key to the question: O is bonded to H and to F (more electronegative than oxygen).

1. H₂O: H = +1, O = −2. F₂: F = 0. HF: H = +1, F = −1.
2. HOF: H = +1 and F = −1, so (+1) + O + (−1) = 0 ⇒ O = 0.
3. Changes: oxygen goes from −2 (in H₂O) to 0 (in HOF) — oxidation; fluorine goes from 0 (in F₂) to −1 (in both HF and HOF) — reduction.
4. Conclusion: since oxidation (O) and reduction (F) occur at the same time, the reaction is a redox reaction. Fluorine is the oxidising agent (and it is itself reduced); water (its oxygen) is the reducing agent.

Extra depth. HOF (hypofluorous acid) is unusual: it is the only known oxoacid of fluorine, and oxygen in it is formally 0 — that is why it is a very unstable compound that decomposes into HF and O₂. The reaction also supports the statement in the chapter that fluorine is so reactive it can ‘displace oxygen’ of water. A check on the electron count: 2 electrons are lost by one O atom (−2 → 0) while two F atoms gain 1 each — the total 2 electrons lost = 2 electrons gained.

Conclusion: O: −2 → 0 (oxidation) and F: 0 → −1 (reduction): redox reaction.

Concept. The usual calculation assigns −2 to every oxygen. This works for Cr₂O₇²⁻ and NO₃⁻, but for H₂SO₅ it gives an impossible answer, because H₂SO₅ (peroxomonosulphuric acid, Caro’s acid) contains a peroxide (O–O) linkage. The structures reveal the fallacy.

1. H₂SO₅ (naive): 2(+1) + S + 5(−2) = 0 ⇒ S = +8 — impossible, since sulphur has only 6 valence electrons and cannot exceed +6.
2. H₂SO₅ (correct): the structure is HO–O–S(=O)₂–OH, so two oxygens (the peroxide pair) are −1 and three oxygens are −2: 2(+1) + S + 2(−1) + 3(−2) = 0 ⇒ S = +6.
3. Cr₂O₇²⁻: 2Cr + 7(−2) = −2 ⇒ 2Cr = +12 ⇒ Cr = +6. Structure: two CrO₄ tetrahedra sharing one oxygen corner, O₃Cr–O–CrO₃ (six terminal Cr–O, one bridging O); no O–O bond, so the calculation is valid.
4. NO₃⁻: N + 3(−2) = −1 ⇒ N = +5. Structure: planar triangular with three equivalent N–O bonds (resonance hybrid of three forms with one N=O and two N–O⁻), N sp²; the highest state of N (+5, group 15) — valid.

Extra depth. The lesson: the oxidation number rules are a quick shortcut, but an impossible result (+8 for sulphur) signals a hidden peroxide (or similar) linkage — the maximum possible oxidation number of an element is its group number (S: 6, Cr: 6, N: 5). The same fallacy would arise in CrO₅ (Cr +6, with a peroxo chromium structure) or K₂S₂O₈ (peroxodisulphate: S +6). In H₂SO₅ the peroxide oxygen atoms have oxidation number −1; so as a Caro’s acid it is a strong oxidising agent.

Conclusion: S = +6 in H₂SO₅ (peroxide O–O present), Cr = +6 in Cr₂O₇²⁻, N = +5 in NO₃⁻.

Concept. The Roman numeral in brackets (Stock notation) gives the oxidation number of the metal. To write the formula, balance the total positive charge of the metal ions against the total negative charge of the anion (Cl⁻ −1, SO₄²⁻ −2, O²⁻ −2) — the ‘criss-cross’ method.

1. (a) Mercury(II) chloride: Hg²⁺ with two Cl⁻ ⇒ HgCl₂.
2. (b) Nickel(II) sulphate: Ni²⁺ with SO₄²⁻ (charges balance 1 : 1) ⇒ NiSO₄.
3. (c) Tin(IV) oxide: Sn⁴⁺ with two O²⁻ ⇒ SnO₂.
4. (d) Thallium(I) sulphate: Tl⁺ needs two ions to balance one SO₄²⁻ ⇒ Tl₂SO₄.
5. (e) Iron(III) sulphate: 2Fe³⁺ (+6) with 3SO₄²⁻ (−6) ⇒ Fe₂(SO₄)₃.
6. (f) Chromium(III) oxide: 2Cr³⁺ (+6) with 3O²⁻ (−6) ⇒ Cr₂O₃.

Extra depth. Stock notation removes ambiguity for metals with more than one oxidation state: mercury(II) chloride (corrosive sublimate, HgCl₂) differs from mercury(I) chloride (calomel, Hg₂Cl₂, which contains the Hg₂²⁺ ion), just as tin(IV) oxide SnO₂ differs from tin(II) oxide SnO. Thallium is a classic case where the +1 state is more stable than +3 (inert-pair effect), so Tl(I) compounds are common. Thallium compounds are highly poisonous.

Conclusion: HgCl₂; NiSO₄; SnO₂; Tl₂SO₄; Fe₂(SO₄)₃; Cr₂O₃.

Concept. Carbon has four valence electrons, so its oxidation number can range from −4 (all four bonds to less electronegative atoms, e.g. H) to +4 (all four bonds to more electronegative atoms such as O, Cl). Nitrogen has five valence electrons: its states run from −3 (three bonds to H) to +5 (as in HNO₃). A good list gives one example for each integer oxidation state.

1. Carbon: −4 in CH₄; −3 in C₂H₆; −2 in C₂H₄ (or CH₃Cl); −1 in C₂H₂; 0 in C (graphite/diamond) and HCHO, CH₂Cl₂; +1 in CH₃CHO (the CHO carbon); +2 in CO and HCOOH; +3 in (COOH)₂ (oxalic acid); +4 in CO₂, CCl₄.
2. Nitrogen: −3 in NH₃; −2 in N₂H₄; −1 in NH₂OH; 0 in N₂; +1 in N₂O; +2 in NO; +3 in N₂O₃ and HNO₂; +4 in NO₂; +5 in N₂O₅ and HNO₃.

Extra depth. Textbook note: the printed question has the minus and plus signs and the number missing from the text (‘from − to +’); it asks for carbon from −4 to +4 as given above. Check the entries: in CH₃OH carbon is −2 (3 H = −3, one O = +1); in C₂H₂ each carbon is −1; in HCHO carbon is 0 (2 H = −2, O= = +2); in HCOOH carbon is +2; in oxalic acid each carbon is +3. Carbon and nitrogen are the elements with the widest range of oxidation states among the first-period p-block, which is why their chemistry is so rich (carbon cycle, nitrogen cycle).

Conclusion: C from −4 (CH₄) to +4 (CO₂); N from −3 (NH₃) to +5 (HNO₃), with intermediate examples as in the table.

Concept. A species can act as a reducing agent if it contains an element that can be oxidised (raise its oxidation number), and as an oxidising agent if it contains an element that can be reduced (lower its oxidation number). An element in an intermediate oxidation state can do either; an element in its highest state can only be reduced; in its lowest state only oxidised.

1. SO₂: S is +4, intermediate between its maximum (+6) and minimum (−2). It can be oxidised to +6 (SO₄²⁻: reducing agent, e.g. with MnO₄⁻) or reduced to 0 or −2 (oxidising agent, e.g. SO₂ + 2H₂S → 3S + 2H₂O).
2. H₂O₂: O is −1, between 0 (in O₂) and −2 (in H₂O). Reduced to −2 (oxidant: H₂O₂ + 2Fe²⁺ + 2H⁺ → 2Fe³⁺ + 2H₂O) or oxidised to 0 (reductant: 2MnO₄⁻ + 5H₂O₂ + 6H⁺ → 2Mn²⁺ + 5O₂ + 8H₂O).
3. O₃: oxygen is 0. It cannot go to a positive state (positive oxidation states of oxygen occur only with fluorine, the only more electronegative element), so it can only be reduced (0 → −2) — only an oxidant.
4. HNO₃: N is +5, the highest state of group 15 (5 valence electrons). It can only be reduced (to NO₂, NO, N₂O, N₂, NH₄⁺) — only an oxidant.

Extra depth. The statement for HNO₃ applies to the nitrate nitrogen. Note a subtlety: in the same oxidation-state logic, concentrated H₂SO₄ (S +6) and KMnO₄ (Mn +7) are also pure oxidants, and Cl⁻/I⁻ (lowest state) are pure reductants. A useful shortcut: highest state ⇒ only oxidant; lowest state ⇒ only reductant; intermediate ⇒ both (and possibly disproportionation, e.g. H₂O₂ → H₂O + O₂).

Conclusion: SO₂ (S +4) and H₂O₂ (O −1) are in intermediate states so they can be oxidised or reduced; O₃ (O 0, cannot rise) and HNO₃ (N +5, highest) can only be reduced.

Concept. The rewritten equations are chemically identical in net result but they show the source of each atom. The first form of each hides the mechanism; the second form records which reactant supplies which product. A technique that follows atoms through a reaction is isotopic (radioactive-tracer) labelling.

1. (a) Photosynthesis. In the second form 12 H₂O are used and 6 H₂O are produced: net 6 H₂O consumed, the same as in the first form. The point is that all six O₂ molecules come from the oxidation of water (12H₂O → 6O₂ + 24H⁺ + 24e⁻ in the light reaction), while the oxygen atoms of CO₂ end up in the water (6H₂O) and in glucose. This was proven by Ruben and Kamen’s ¹⁸O experiments.
2. (b) O₃ + H₂O₂. In the second form two separate O₂ molecules are written: one is formed from ozone (which loses one O atom, reduced to water), the other from hydrogen peroxide (oxygen −1 → 0, oxidised). It makes the redox picture clear: O₃ is the oxidant (O: 0 → −2 for the atom ending up in H₂O) and H₂O₂ the reductant (O: −1 → 0).
3. Technique to investigate the path: isotopic labelling. For (a) use water enriched with ¹⁸O (H₂¹⁸O) and ordinary CO₂; if the evolved oxygen is ¹⁸O₂, it comes from water. Using C¹⁸O₂ with ordinary water, the evolved oxygen is ordinary ¹⁶O₂ and the label appears in H₂¹⁸O and in glucose. For (b) label H₂O₂ (H₂¹⁸O₂) and analyse the O₂ and H₂O produced by mass spectrometry to show that the label appears in O₂ only.

Extra depth. Isotopic tracer experiments are a standard tool to decide mechanisms (also used to show that in esterification the water formed takes its oxygen from the acid, using an alcohol labelled with ¹⁸O, and to map the Calvin cycle with ¹⁴CO₂). In the context of this question: in (b), the first form conceals that the 2 O₂ molecules have different origins, even though the total atoms are conserved: 3 + 2 = 5 O atoms on the left; 1 + 4 = 5 on the right.

Conclusion: The second forms show the origin of atoms; use ¹⁸O isotopic labelling (e.g. H₂¹⁸O) and mass-spectrometric analysis of the O₂ to trace the paths.

Concept. Silver almost always exists as Ag(I) (4d¹⁰, a stable filled d-shell); the +2 state is unusual. In AgF₂ silver is in the +2 oxidation state, which is why the compound is unstable and strongly oxidising.

1. Oxidation state of Ag: fluorine is always −1, so in AgF₂: Ag + 2(−1) = 0 ⇒ Ag = +2.
2. Stability of states: the common, stable oxidation state of silver is +1 (Ag⁺, d¹⁰). The +2 state (d⁹) is very uncommon and unstable because the removal of a second electron from Ag⁺ requires a lot of energy.
3. Tendency to revert: Ag(II) therefore tends to gain one electron to become stable Ag(I): Ag²⁺ + e⁻ → Ag⁺.
4. Conclusion: since it readily takes electrons from other species (oxidising them) and is reduced to the stable Ag(I), AgF₂ acts as a very strong oxidising agent, e.g. it can oxidise many substances that ordinary oxidants cannot (it is used as a powerful fluorinating agent).

Extra depth. Fluoride stabilises high oxidation states better than any other ligand because F⁻ is the most difficult anion to oxidise — that is why AgF₂, CoF₃ and similar compounds exist at all; with other halides (Cl⁻, I⁻) Ag(II) would immediately oxidise the halide. Even so, AgF₂ reacts violently with water and oxidises it. The 2.87 V potential for F₂/F⁻ (Table 7.1) shows how hard it is to oxidise F⁻; this ensures that the only product that can be formed from AgF₂ by electron capture is Ag(I).

Conclusion: In AgF₂ silver is in the unstable +2 state; it readily gains an electron to reach the stable +1 state, so it is a very strong oxidising agent.

Concept. The final oxidation state reached by the element being oxidised depends on how much oxidant is available. If the reductant is in excess, there is not enough oxidant to take each atom all the way up, so each atom of the reductant element is oxidised only a little (a lower oxidation state). If the oxidant is in excess, every atom of the reductant element can be oxidised fully to the highest state it can reach with that oxidant.

1. Illustration 1 — phosphorus and oxygen. With a limited supply of oxygen (phosphorus in excess): P₄ + 3O₂ → P₄O₆ (P = +3, lower state). With excess oxygen: P₄ + 5O₂ → P₄O₁₀ (P = +5, higher state).
2. Illustration 2 — carbon and oxygen. Limited oxygen (carbon in excess): 2C + O₂ → 2CO (C = +2). Excess oxygen: C + O₂ → CO₂ (C = +4).
3. Illustration 3 — iron and chlorine (iron in solution). With excess chlorine, 2Fe + 3Cl₂ → 2FeCl₃ (Fe = +3). With excess iron (reductant), the Fe(III) formed is reduced back by the metal: 2FeCl₃ + Fe → 3FeCl₂ (Fe = +2).
4. Extra example — phosphorus and chlorine. P₄ + 6Cl₂ → 4PCl₃ (Cl₂ limited, P = +3); P₄ + 10Cl₂ → 4PCl₅ (Cl₂ excess, P = +5).

Extra depth. The principle is of great practical importance: incomplete combustion of fuels in limited air produces poisonous CO (lower oxidation state of carbon) instead of CO₂, and phosphorus trichloride versus pentachloride are prepared by controlling the supply of chlorine. The same logic applies to titrations: adding the oxidant dropwise to excess reductant gives the lower-state product until the equivalence point, after which the higher-state product (or excess oxidant colour) appears. It also explains why iron(III) salts are kept with iron filings to prevent them from being oxidised, and vice-versa.

Conclusion: Reductant in excess ⇒ lower state (P₄O₆, CO, FeCl₂); oxidant in excess ⇒ higher state (P₄O₁₀, CO₂, FeCl₃).

Concept. (a) Toluene (C₆H₅CH₃) is an organic liquid immiscible with water, while KMnO₄ is an ionic salt dissolved in water. Alkaline or acidic aqueous permanganate would give a two-layer system and slow, uncontrolled reaction. (b) Concentrated H₂SO₄ is an oxidising agent as well as an acid; whether it merely displaces the hydrohalic acid or also oxidises it depends on how easily the halide ion is oxidised (Cl⁻ < Br⁻ < I⁻ in reducing power).

1. (a) Why alcoholic KMnO₄: alcohol is miscible with both water and toluene, so alcoholic permanganate gives a homogeneous medium in which toluene and MnO₄⁻ come in close contact; the reaction is smooth and in an essentially neutral (mild) medium, so the side-chain CH₃ is oxidised to COO⁻ without breaking the benzene ring. In a strongly acidic or alkaline aqueous solution the oxidation would be heterogeneous and uncontrolled (and permanganate is a stronger oxidant in acid, risking degradation).
2. Balanced equation (oxidation number check): C of CH₃: −3 → +3 in COO⁻ (increase 6); Mn +7 → +4 (decrease 3 per Mn, so 2 MnO₄⁻): C₆H₅CH₃ + 2MnO₄⁻ → C₆H₅COO⁻ + 2MnO₂ + OH⁻ + H₂O. Verify: C 7 = 7; H 8 = 5 + 1 + 2; O 8 = 2 + 4 + 1 + 1; charge −2 = −1 − 1. Acidifying the benzoate gives benzoic acid.
3. (b) Chloride: NaCl + H₂SO₄ → NaHSO₄ + HCl. Cl⁻ is a very poor reducing agent (E° Cl₂/Cl⁻ = +1.36 V), so conc. H₂SO₄ cannot oxidise it; the colourless pungent HCl gas escapes.
4. Bromide: Br⁻ is a better reducing agent (E° Br₂/Br⁻ = +1.09 V), so conc. H₂SO₄ first liberates HBr (NaBr + H₂SO₄ → NaHSO₄ + HBr) and then oxidises it: 2HBr + H₂SO₄ → Br₂ + SO₂ + 2H₂O. The red-brown vapour is bromine (and SO₂ is also formed).

Extra depth. This is the basis of two lab tests: the formation of red bromine vapour when a bromide is warmed with concentrated sulphuric acid, and the violet iodine vapour (together with H₂S and sulphur, the further reduction products of the acid) given by iodides, whose reducing power is greatest. It also explains why HI cannot be prepared by heating NaI with conc. H₂SO₄; non-oxidising H₃PO₄ is used instead. Toluene → benzoic acid is an industrial and laboratory route often asked as a balanced equation.

Conclusion: (a) Alcohol dissolves both toluene and KMnO₄; C₆H₅CH₃ + 2MnO₄⁻ → C₆H₅COO⁻ + 2MnO₂ + OH⁻ + H₂O. (b) Br⁻ (not Cl⁻) is oxidised by conc. H₂SO₄ to Br₂.

Concept. Assign oxidation numbers to every atom that might change. The substance containing the atom whose oxidation number increases is oxidised (and is the reducing agent); the substance containing the atom whose oxidation number decreases is reduced (and is the oxidising agent).

1. (a) Ag: +1 (AgBr) → 0 (Ag): AgBr is reduced; it is the oxidising agent. Hydroquinone C₆H₆O₂ → quinone C₆H₄O₂ (loses two H, i.e. 2 H atoms removed = oxidation): oxidised, and it is the reducing agent.
2. (b) HCHO → HCOO⁻: C goes from 0 to +2: HCHO is oxidised (reducing agent). [Ag(NH₃)₂]⁺: Ag +1 → 0: reduced (oxidising agent, Tollens’ reagent).
3. (c) HCHO → HCOO⁻ (C 0 → +2): oxidised, reducing agent. Cu²⁺ → Cu₂O (Cu +2 → +1): reduced, oxidising agent (Fehling-type reagent).
4. (d) N₂H₄ → N₂: N −2 → 0: oxidised, reducing agent (hydrazine). H₂O₂ → H₂O: O −1 → −2: reduced, oxidising agent.
5. (e) Pb(s) (0) → PbSO₄ (+2): Pb is oxidised, the reducing agent. PbO₂ (+4) → PbSO₄ (+2): PbO₂ is reduced, the oxidising agent. (This is the discharge reaction of the lead storage battery — lead is both reductant and a component of the product.)

Extra depth. Reactions (b) and (c) are the classic tests for aldehydes: Tollens’ reagent gives a silver mirror and Fehling’s solution gives red Cu₂O. Reaction (a) is the chemistry of photographic developing (hydroquinone reduces silver halide grains to silver). Reaction (d) is used in rocket propellants (hydrazine + H₂O₂), and (e) is a comproportionation — Pb(0) and Pb(IV) produce Pb(II), the reverse of disproportionation.

Conclusion: (a) AgBr is the oxidant, hydroquinone the reductant; (b) and (c) HCHO is oxidised/reductant, Ag⁺ complex or Cu²⁺ reduced/oxidant; (d) N₂H₄ reductant, H₂O₂ oxidant; (e) Pb reductant, PbO₂ oxidant.

Concept. The extent to which a reductant is oxidised depends on the strength of the oxidising agent. A stronger oxidant takes the reductant to a higher oxidation state; a weaker oxidant only takes it a little way. Compare the two oxidants through their standard electrode potentials (Table 7.1): Br₂ + 2e⁻ → 2Br⁻ (E° = +1.09 V) and I₂ + 2e⁻ → 2I⁻ (E° = +0.54 V).

1. Oxidation number of S in thiosulphate: S₂O₃²⁻: 2S + 3(−2) = −2 ⇒ average S = +2.
2. With iodine: S₂O₃²⁻ → S₄O₆²⁻, average S = +2.5 — only a small increase (0.5 per S).
3. With bromine: S₂O₃²⁻ → SO₄²⁻, S = +6 — a large increase (4 per S).
4. Reason: Br₂ is a much stronger oxidising agent than I₂ (1.09 V against 0.54 V), so it oxidises sulphur completely to the highest state (+6, sulphate), while the weak oxidant I₂ oxidises it only to the average +2.5 of tetrathionate.

Extra depth. The mild behaviour of iodine is exploited in iodometric titrations: because I₂ oxidises thiosulphate cleanly and quantitatively to tetrathionate (2 mol S₂O₃²⁻ : 1 mol I₂), thiosulphate is used as a standard solution with a starch end point (section 7.3.3). If the oxidant were bromine or chlorine the stoichiometry would be different and not reproducible, which is why such halogens (that over-oxidise) are not used in this titration. Chlorine, being a stronger oxidant still (E° 1.36 V), also oxidises thiosulphate to sulphate — the basis of its use as an ‘antichlor’ in the textile industry.

Conclusion: Br₂ is a stronger oxidising agent than I₂, so it oxidises thiosulphate to sulphate (+6) whereas I₂ gives only tetrathionate (+2.5).

Concept. The strength of an oxidising agent corresponds to the tendency to gain electrons (a high positive E°); the strength of a reducing agent to the tendency of its reduced form to lose electrons (a low E°). Both trends in the halogens can be proved by displacement reactions and by the E° values in Table 7.1: F₂ 2.87 V, Cl₂ 1.36 V, Br₂ 1.09 V, I₂ 0.54 V.

1. Fluorine is the best oxidant. It oxidises the other halide ions: F₂ + 2Cl⁻ → 2F⁻ + Cl₂, F₂ + 2Br⁻ → 2F⁻ + Br₂, F₂ + 2I⁻ → 2F⁻ + I₂. It even oxidises water: 2H₂O + 2F₂ → 4HF + O₂. The reverse never happens: Cl₂, Br₂, I₂ cannot oxidise F⁻.
2. Order among the other halogens: Cl₂ displaces Br⁻ and I⁻ (Cl₂ + 2KBr → 2KCl + Br₂; Cl₂ + 2KI → 2KCl + I₂), Br₂ displaces I⁻ (Br₂ + 2I⁻ → 2Br⁻ + I₂); so oxidising power F₂ > Cl₂ > Br₂ > I₂.
3. HI is the best reductant. I⁻ loses an electron most easily (E° I₂/I⁻ = 0.54 V is the lowest). Concentrated H₂SO₄ oxidises HI all the way to I₂ and itself goes to H₂S: 8HI + H₂SO₄ → 4I₂ + H₂S + 4H₂O; it oxidises HBr only to Br₂ (with SO₂): 2HBr + H₂SO₄ → Br₂ + SO₂ + 2H₂O; and it does not oxidise HCl at all.
4. Another test: Fe³⁺ (E° 0.77 V) oxidises I⁻ (2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂) but not Br⁻ or Cl⁻, confirming that I⁻ is the strongest reductant: HI > HBr > HCl > HF.

Extra depth. The two statements are two sides of the same trend: the stronger the halogen is as an oxidant, the weaker its halide ion is as a reductant. F⁻ cannot be oxidised by any chemical oxidant — F₂ can only be prepared by electrolytic oxidation (Moissan). The unusually high oxidising power of fluorine in water is mainly due to the weak F–F bond (easy to break) and the very high hydration enthalpy of the small F⁻ ion, which together favour F₂ + 2e⁻ → 2F⁻(aq).

Conclusion: F₂ (E° 2.87 V) oxidises Cl⁻, Br⁻, I⁻ and water; HI (I⁻, E° 0.54 V) reduces conc. H₂SO₄ to H₂S while HBr gives SO₂ and HCl none.

Concept. Textbook note: the charge on the perxenate ion is 4− (the superscript is lost in the printed question); check: left side charge = −4 − 2 + 6 = 0; right side = 0, so the equation is balanced only with XeO₆⁴⁻. In a redox reaction the reaction occurs if the oxidant is a stronger oxidising agent than the product-side oxidant. Here F⁻ is oxidised to F₂ — something no ordinary oxidant can do, because F₂ is the strongest oxidising agent known (E° = +2.87 V).

1. Oxidation numbers: Xe in XeO₆⁴⁻: Xe + 6(−2) = −4 ⇒ Xe = +8; in XeO₃: Xe + 3(−2) = 0 ⇒ Xe = +6. F: −1 (F⁻) → 0 (F₂).
2. Change: Xe decreases by 2 (reduction); F increases by 1 per atom, 2 for two F⁻ (oxidation). The two changes match, so the reaction is a balanced redox reaction.
3. Why it occurs: Xe(VIII) in perxenate is a very powerful electron acceptor; in the presence of acid it is able to take electrons even from F⁻, the most difficult ion to oxidise, producing F₂.
4. Conclusion: since XeO₆⁴⁻ oxidises F⁻ to F₂, the perxenate ion (and so Na₄XeO₆) is a stronger oxidising agent than fluorine, i.e. an exceptionally strong oxidant.

Extra depth. Without this reaction we would conclude that no chemical reagent can oxidise fluoride to fluorine (so F₂ is made only by electrolysis of KHF₂/HF). The perxenate reaction shows that Xe(VIII) can do this in acidic solution. This also shows that the chemistry of noble gases (Xe), long considered inert, includes extremely high oxidation states bonded to the most electronegative elements (F and O).

Conclusion: XeO₆⁴⁻ (Xe +8) oxidises F⁻ to F₂; hence Na₄XeO₆ is a stronger oxidising agent than fluorine.

Concept. Compare which reductants are able to reduce each metal ion. Hypophosphorous acid (H₃PO₂, P = +1) is a strong reducing agent that reduces both Ag⁺ and Cu²⁺ to the metals (P: +1 → +5). Benzaldehyde (an aromatic aldehyde, a weaker reductant than aliphatic aldehydes) reduces Tollens’ reagent (Ag⁺) but gives no reaction with the Cu²⁺ reagent.

1. (a) and (b): H₃PO₂ is oxidised (P +1 → +5) and reduces both Ag⁺ (to Ag) and Cu²⁺ (to Cu). So both Ag⁺ and Cu²⁺ can be reduced by a strong reductant.
2. (c): benzaldehyde is oxidised to benzoate (C: +1 → +3) by the Ag⁺ complex, which is reduced to metallic silver (silver mirror) — Tollens’ test is positive for the aromatic aldehyde.
3. (d): with Cu²⁺ in alkaline solution (the Fehling-type reagent) benzaldehyde gives no change: Cu²⁺ is not a strong enough oxidant to oxidise an aromatic aldehyde.
4. Inference: Ag⁺ is a stronger oxidising agent than Cu²⁺: it oxidises a weaker reducing agent (benzaldehyde) which Cu²⁺ cannot. Table 7.1 supports this: E°(Ag⁺/Ag) = +0.80 V > E°(Cu²⁺/Cu) = +0.34 V.

Extra depth. This is why aliphatic aldehydes give both Tollens’ and Fehling’s tests, while aromatic aldehydes (and ketones) give only Tollens’ (aromatic aldehydes) or none. The reason in terms of E°: the more positive the E°, the better the oxidant; with a weak reductant only the stronger oxidant (Ag⁺) reacts. The behaviour of the oxidising agent is therefore ranked Ag⁺ > Cu²⁺ even though both can be used to detect aldehydes. In (a) and (b) the formula H₃PO₂ is a monobasic acid with two P–H bonds, strongly reducing because it has P–H bonds that can be oxidised.

Conclusion: Ag⁺ is a stronger oxidising agent than Cu²⁺; Cu²⁺ fails to oxidise the weaker reductant benzaldehyde.

Concept. In the ion–electron (half-reaction) method, balance each half reaction separately: atoms other than O and H, then O (with H₂O), then H (with H⁺), then charge (with e⁻); equalise electrons; add; cancel; and (for basic medium) convert H⁺ to H₂O by adding OH⁻ on both sides.

1. (a) basic. Oxidation: 2I⁻ → I₂ + 2e⁻. Reduction: MnO₄⁻ + 4H⁺ + 3e⁻ → MnO₂ + 2H₂O; add 4OH⁻ to both sides: MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻. Multiply oxidation by 3, reduction by 2 and add: 2MnO₄⁻ + 6I⁻ + 4H₂O → 2MnO₂ + 3I₂ + 8OH⁻.
2. (b) acidic. Reduction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O. Oxidation: SO₂ + 2H₂O → HSO₄⁻ + 3H⁺ + 2e⁻. Multiply reduction by 2 and oxidation by 5: 2MnO₄⁻ + 16H⁺ + 5SO₂ + 10H₂O → 2Mn²⁺ + 8H₂O + 5HSO₄⁻ + 15H⁺. Cancel: 2MnO₄⁻ + 5SO₂ + 2H₂O + H⁺ → 2Mn²⁺ + 5HSO₄⁻.
3. (c) acidic. Reduction: H₂O₂ + 2H⁺ + 2e⁻ → 2H₂O. Oxidation: Fe²⁺ → Fe³⁺ + e⁻ (×2). Add: H₂O₂ + 2Fe²⁺ + 2H⁺ → 2Fe³⁺ + 2H₂O.
4. (d) acidic. Reduction: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O. Oxidation: SO₂ + 2H₂O → SO₄²⁻ + 4H⁺ + 2e⁻ (×3). Add: Cr₂O₇²⁻ + 14H⁺ + 3SO₂ + 6H₂O → 2Cr³⁺ + 7H₂O + 3SO₄²⁻ + 12H⁺. Cancel: Cr₂O₇²⁻ + 3SO₂ + 2H⁺ → 2Cr³⁺ + 3SO₄²⁻ + H₂O.

Extra depth. Always finish with the three-way check: each atom, then total charge, then (as a bonus) that electrons cancelled. In (b) notice that the question wants HSO₄⁻ (not SO₄²⁻) as the product, which changes the H⁺ count — a frequent source of mistakes. Equations (b) and (d) show SO₂ acting as a reductant, the basis of the reaction used to test for SO₂ (decolourisation of purple KMnO₄ or orange K₂Cr₂O₇ turning green). Reaction (a) is also Problem 7.10 of the text.

Conclusion: (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.

Concept. Textbook note: in (a) the product is hypophosphite, H₂PO₂⁻ (as in eq. 7.46 of the chapter), because the printed HPO₂⁻ has lost a subscript; in (b) Cl⁻ is aqueous. In basic medium use OH⁻ and H₂O: balance as in acid first and then add OH⁻ to both sides to convert every H⁺ into H₂O. Two methods are required, so each part is solved twice.

1. (a) Oxidation-number method. P₄ (0) → PH₃ (−3): decrease of 3 per P; P₄ (0) → H₂PO₂⁻ (+1): increase of 1 per P. One P goes to −3 (decrease 3), three P go to +1 (increase 3): P₄ → PH₃ + 3H₂PO₂⁻. Right charge −3, so 3OH⁻ on the left; then 3H₂O on the left to balance H and O: P₄ + 3OH⁻ + 3H₂O → PH₃ + 3H₂PO₂⁻.
2. (a) Ion–electron method. Reduction: P₄ + 12H₂O + 12e⁻ → 4PH₃ + 12OH⁻. Oxidation: P₄ + 8OH⁻ → 4H₂PO₂⁻ + 4e⁻; multiply by 3: 3P₄ + 24OH⁻ → 12H₂PO₂⁻ + 12e⁻. Add: 4P₄ + 12H₂O + 12OH⁻ → 4PH₃ + 12H₂PO₂⁻ (12OH⁻ cancelled); divide by 4: the same equation. P₄ is both the oxidising and the reducing agent (disproportionation).
3. (b) Oxidation-number method. N₂H₄: N −2 → +2 (increase 4 per N, 8 per N₂H₄); ClO₃⁻: Cl +5 → −1 (decrease 6). LCM = 24: take 3N₂H₄ (24) and 4ClO₃⁻ (24): 3N₂H₄ + 4ClO₃⁻ → 6NO + 4Cl⁻. Charge is already balanced (−4 = −4); O: 12 on the left, 6 on the right, so add 6H₂O on the right (H 12 = 12): 3N₂H₄ + 4ClO₃⁻ → 6NO + 4Cl⁻ + 6H₂O.
4. (b) Ion–electron method. Oxidation: N₂H₄ + 8OH⁻ → 2NO + 6H₂O + 8e⁻. Reduction: ClO₃⁻ + 3H₂O + 6e⁻ → Cl⁻ + 6OH⁻. Multiply oxidation by 3 and reduction by 4 (24 electrons each): 3N₂H₄ + 24OH⁻ → 6NO + 18H₂O + 24e⁻; 4ClO₃⁻ + 12H₂O + 24e⁻ → 4Cl⁻ + 24OH⁻. Add and cancel 24OH⁻ and 12H₂O: the same equation as above. Reductant N₂H₄; oxidant ClO₃⁻.
5. (c) Oxidation-number method. Cl₂O₇: Cl +7 → +3 (decrease 4 per Cl, 8 per Cl₂O₇); H₂O₂: O −1 → 0 (increase 1 per O, 2 per H₂O₂): 4H₂O₂ per Cl₂O₇. Skeleton: Cl₂O₇ + 4H₂O₂ → 2ClO₂⁻ + 4O₂. Charge: right −2, so 2OH⁻ on the left; then 5H₂O on the right balances H and O: Cl₂O₇ + 4H₂O₂ + 2OH⁻ → 2ClO₂⁻ + 4O₂ + 5H₂O (O: 7 + 8 + 2 = 17 = 4 + 8 + 5; H: 8 + 2 = 10 = 10).
6. (c) Ion–electron method. Oxidation: H₂O₂ + 2OH⁻ → O₂ + 2H₂O + 2e⁻. Reduction: Cl₂O₇ + 3H₂O + 8e⁻ → 2ClO₂⁻ + 6OH⁻. Multiply oxidation by 4: 4H₂O₂ + 8OH⁻ → 4O₂ + 8H₂O + 8e⁻. Add and cancel 6OH⁻ and 3H₂O: Cl₂O₇ + 4H₂O₂ + 2OH⁻ → 2ClO₂⁻ + 4O₂ + 5H₂O — the same result. Oxidant Cl₂O₇; reductant H₂O₂.

Extra depth. Reaction (a) is the disproportionation of white phosphorus in alkali — one P₄ molecule serves as oxidising agent and reducing agent at once (phosphine PH₃ is very toxic and the impure gas ignites spontaneously). In (c) the formation of O₂ shows hydrogen peroxide acting as a reducing agent towards the very strong oxidant Cl₂O₇ (the anhydride of perchloric acid). A two-method answer is a built-in check: both methods must lead to the identical balanced equation.

Conclusion: (a) P₄ + 3OH⁻ + 3H₂O → PH₃ + 3H₂PO₂⁻; (b) 3N₂H₄ + 4ClO₃⁻ → 6NO + 4Cl⁻ + 6H₂O; (c) Cl₂O₇ + 4H₂O₂ + 2OH⁻ → 2ClO₂⁻ + 4O₂ + 5H₂O.

Concept. Look at the oxidation numbers of carbon and nitrogen on both sides (nitrogen being the more electronegative element in the C–N group, it takes the −3 in cyanide, cyanate and cyanogen alike). The reaction is a disproportionation of cyanogen, (CN)₂, in alkali, exactly like Cl₂ + 2OH⁻ → Cl⁻ + OCl⁻ + H₂O.

1. Oxidation numbers: (CN)₂: N = −3 and C = +3 (each C–C bond contributes 0). CN⁻: N = −3 so C = +2. CNO⁻: N = −3, O = −2, so C = +4.
2. Redox change: carbon goes from +3 → +2 (reduction, product CN⁻) and +3 → +4 (oxidation, product CNO⁻). The same species is oxidised and reduced simultaneously, so it is a disproportionation reaction.
3. Intermediate oxidation state: carbon in (CN)₂ is in the intermediate state +3, with a lower state (+2 in CN⁻) and a higher state (+4 in CNO⁻) available, which is the condition for disproportionation.
4. Analogy with halogens: (CN)₂ behaves like Cl₂ (Cl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂O), CN⁻ like a halide ion and CNO⁻ like hypohalite: (CN)₂ is a pseudohalogen and CN⁻ a pseudohalide.

Extra depth. Cyanogen is a poisonous gas that resembles chlorine in many ways: it forms the cyanide (CN⁻) as the analogue of Cl⁻ (AgCN is insoluble like AgCl), it forms an acid HCN (a weak acid), and it dissolves in alkali by disproportionation. In reaction terms: 1 mol of (CN)₂ gains 2 e⁻ (to make 2CN⁻) in the half reaction (CN)₂ + 2e⁻ → 2CN⁻ and loses 2 e⁻ in (CN)₂ + 4OH⁻ → 2CNO⁻ + 2H₂O + 2e⁻; half goes each way. The reaction is balanced: C 2 = 2; N 2 = 2; O 2 = 1 + 1; H 2 = 2; charge −2 = −2.

Conclusion: The reaction is a disproportionation of cyanogen: C (+3) is both reduced to CN⁻ (+2) and oxidised to CNO⁻ (+4); (CN)₂ is a pseudohalogen.

Concept. In disproportionation the same species is oxidised and reduced: Mn³⁺ (+3) goes to Mn²⁺ (+2, reduction) and to MnO₂ (+4, oxidation). The reaction is in aqueous (acidic) solution, so H₂O and H⁺ balance O and H.

1. Skeleton: Mn³⁺(aq) → Mn²⁺(aq) + MnO₂(s) + H⁺(aq).
2. Half reactions: reduction: Mn³⁺ + e⁻ → Mn²⁺. Oxidation: Mn³⁺ + 2H₂O → MnO₂ + 4H⁺ + e⁻ (O: balanced by 2H₂O; H by 4H⁺; charge +3 = +4 − 1).
3. Add (one electron each, so they cancel): 2Mn³⁺ + 2H₂O → Mn²⁺ + MnO₂ + 4H⁺.
4. Check: Mn 2 = 2; O 2 = 2; H 4 = 4; charge +6 = +2 + 4 = +6. Balanced.

Extra depth. Manganese shows oxidation states from +2 to +7 in common compounds, so Mn(III), lying between Mn(II) and Mn(IV), can disproportionate. The reaction produces H⁺ (acid), so disproportionation is favoured in less acidic solution; in strongly acidic solution (large [H⁺]) Mn³⁺ is more stable. Similarly Cu⁺ disproportionates in water (2Cu⁺ → Cu²⁺ + Cu), as does MnO₄²⁻ (3MnO₄²⁻ + 4H⁺ → 2MnO₄⁻ + MnO₂ + 2H₂O).

Conclusion: 2Mn³⁺ + 2H₂O → Mn²⁺ + MnO₂ + 4H⁺.

Concept. The possible oxidation states depend on the electronegativity and the electronic configuration of the element. Fluorine, the most electronegative element, only gains electrons in compounds. An alkali metal has a single outer electron which it loses. A noble gas has a complete octet and is generally inert. Iodine, a less electronegative halogen, can show −1 and positive states with more electronegative atoms (O, F, Cl).

1. (a) Only negative: F — F is the most electronegative element, shows only −1 in all compounds (and 0 in F₂).
2. (b) Only positive: Cs — an alkali metal that always forms Cs⁺ (+1) in compounds.
3. (c) Both positive and negative: I — −1 in iodides (KI), and +1, +3, +5, +7 with more electronegative elements (ICl, ICl₃, IO₃⁻, IO₄⁻, IF₇).
4. (d) Neither: Ne — a noble gas with a closed-shell 2s²2p⁶ configuration; it forms no compounds under ordinary conditions, so its oxidation number is only 0 in the free state.

Extra depth. The case of Ne shows that oxidation number is meaningful only for elements that form compounds. Heavier noble gases such as Xe form compounds with F and O (XeF₂, XeO₃, with Xe +2, +4, +6, +8 — see Exercise 7.16), so Xe would belong in group (b) — only positive states — while neon forms no stable compounds. Fluorine shows 0 in F₂, which is a free-element state rather than an oxidation state in a compound.

Conclusion: (a) F; (b) Cs; (c) I; (d) Ne.

Concept. Chlorine (an oxidising agent) is removed by sulphur dioxide (a reducing agent) in water. Cl₂ is reduced to Cl⁻ (0 → −1), SO₂ is oxidised to sulphate (S: +4 → +6). This is the process of dechlorination.

1. Skeleton: Cl₂ + SO₂ → Cl⁻ + SO₄²⁻ (in water).
2. Half reactions: reduction: Cl₂ + 2e⁻ → 2Cl⁻; oxidation: SO₂ + 2H₂O → SO₄²⁻ + 4H⁺ + 2e⁻.
3. Add (2e⁻ each): Cl₂(g) + SO₂(g) + 2H₂O(l) → 2Cl⁻(aq) + SO₄²⁻(aq) + 4H⁺(aq), i.e. Cl₂ + SO₂ + 2H₂O → 2HCl + H₂SO₄.
4. Check: Cl 2 = 2; S 1 = 1; O 2 + 2 = 4 = 4; H 4 = 4; charge 0 = −2 − 2 + 4 = 0.

Extra depth. The products (dilute HCl and H₂SO₄, as Cl⁻ and SO₄²⁻ ions) are harmless at the very low concentrations involved. Sodium thiosulphate and sodium sulphite are also used as dechlorinating agents (antichlor) in water treatment and in textile bleaching. Remember that chlorine disinfects water through the HOCl/OCl⁻ formed in Cl₂ + H₂O → HCl + HOCl; the oxidising power of HOCl kills the germs, but any excess must be removed to avoid the taste and the harmful chlorinated by-products.

Conclusion: Cl₂ + SO₂ + 2H₂O → 2Cl⁻ + SO₄²⁻ + 4H⁺.

Concept. A species can disproportionate only if the element is in an intermediate oxidation state, with a lower and a higher state both available (the element must exist in at least three oxidation states, including the one in the reactant). So we look for elements with several oxidation states.

1. (a) Non-metals: P, Cl, S (also Br and I, but not F, which has no positive state). Examples: P₄ + 3OH⁻ + 3H₂O → PH₃ + 3H₂PO₂⁻; Cl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂O; S₈ + 12OH⁻ → 4S²⁻ + 2S₂O₃²⁻ + 6H₂O.
2. (b) Metals: Cu, Mn and Hg (also Au, Tl, Ga, Pb can). Examples: 2Cu⁺ → Cu²⁺ + Cu (Cu(I) in the middle: 0, +1, +2); 2Mn³⁺ + 2H₂O → Mn²⁺ + MnO₂ + 4H⁺ (Exercise 7.21); Hg₂²⁺ → Hg²⁺ + Hg (Hg(I) in the middle: 0, +1, +2).

Extra depth. Fluorine cannot disproportionate (it shows only −1 and 0), whereas nitrogen can (2NO₂ + 2OH⁻ → NO₂⁻ + NO₃⁻ + H₂O, Problem 7.6(d)) and so can oxygen in hydrogen peroxide (2H₂O₂ → 2H₂O + O₂, eq. 7.45). Metals with only one common oxidation state (Na, Mg, Al, Zn) cannot disproportionate. The rule of thumb: an element in its highest state (e.g. ClO₄⁻) or lowest state cannot disproportionate.

Conclusion: (a) P, Cl, S (and Br, I); (b) Cu, Mn, Hg (also Au, Tl, Ga, Pb).

Concept. This is a limiting-reagent problem. The balanced equation (found by the oxidation-number method) is 4NH₃ + 5O₂ → 4NO + 6H₂O. The maximum amount of NO is fixed by whichever reactant runs out first — the limiting reagent.

1. Moles available: NH₃: 10.00 g/17 g mol⁻¹ = 0.588 mol; O₂: 20.00 g/32 g mol⁻¹ = 0.625 mol.
2. Which is limiting? 4 mol NH₃ need 5 mol O₂, so 0.588 mol NH₃ would need 0.588 × 5/4 = 0.735 mol O₂. Only 0.625 mol O₂ is available (less than 0.735), so O₂ is the limiting reagent (NH₃ is in excess).
3. Moles of NO from O₂: 5 mol O₂ → 4 mol NO, so 0.625 mol O₂ gives 0.625 × 4/5 = 0.500 mol NO.
4. Mass of NO = 0.500 mol × 30 g mol⁻¹ = 15.0 g.

Extra depth. NH₃ left over = 0.588 − 0.500 = 0.088 mol (1.5 g). Check by the NH₃-based yield: 0.588 mol NO (17.6 g) would be the answer if oxygen were plentiful — the smaller of the two (15.0 g) is the maximum possible. This reaction (4NH₃ + 5O₂ → 4NO + 6H₂O, Pt–Rh gauze catalyst, about 800 °C) is followed by 2NO + O₂ → 2NO₂ and 3NO₂ + H₂O → 2HNO₃ + NO. The redox balance: N from −3 to +2 (5 electrons per N, 20 per 4 N) and O from 0 to −2 (2 per O, 20 per 10 O atoms).

Conclusion: Maximum NO = 0.625 mol O₂ × 4/5 × 30 g mol⁻¹ = 15.0 g.

Concept. A redox reaction is feasible (spontaneous) if the overall cell potential is positive: E°_cell = E°(reduction couple) − E°(oxidation couple) > 0. In words: the oxidised form of the couple with the higher E° oxidises the reduced form of the couple with the lower E°. (The printed question says ‘Table 8.1’; the data are in Table 7.1 of this chapter.)

1. (a) Fe³⁺ + I⁻: Fe³⁺/Fe²⁺ = +0.77 V, I₂/I⁻ = +0.54 V. Fe³⁺ (higher E°) oxidises I⁻: 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂, E°_cell = 0.77 − 0.54 = +0.23 V → feasible.
2. (b) Ag⁺ + Cu(s): Ag⁺/Ag = +0.80 V, Cu²⁺/Cu = +0.34 V: 2Ag⁺ + Cu → 2Ag + Cu²⁺, E°_cell = 0.80 − 0.34 = +0.46 V → feasible.
3. (c) Fe³⁺ + Cu(s): Fe³⁺/Fe²⁺ = +0.77 V, Cu²⁺/Cu = +0.34 V: 2Fe³⁺ + Cu → 2Fe²⁺ + Cu²⁺, E°_cell = 0.77 − 0.34 = +0.43 V → feasible.
4. (d) Ag(s) + Fe³⁺: the possible reaction Ag + Fe³⁺ → Ag⁺ + Fe²⁺ has E°_cell = 0.77 − 0.80 = −0.03 V (negative) → not feasible.
5. (e) Br₂(aq) + Fe²⁺: Br₂/Br⁻ = +1.09 V, Fe³⁺/Fe²⁺ = +0.77 V: Br₂ + 2Fe²⁺ → 2Br⁻ + 2Fe³⁺, E°_cell = 1.09 − 0.77 = +0.32 V → feasible.

Extra depth. Observe that E°_cell does not depend on the stoichiometric coefficients: doubling the equation does not double the E° value (E° is an intensive quantity), though ΔG° = −nFE°_cell does depend on the number n of electrons. The small negative value in (d) (−0.03 V) means the reaction is only marginally unfavourable: with changed concentrations (via the Nernst equation, Class XII) it could occur in the other direction — in the standard state Ag⁺ would oxidise Fe²⁺ slightly. Standard potentials tell only whether a reaction is thermodynamically feasible, not how fast it is.

Conclusion: Feasible: (a), (b), (c), (e). Not feasible: (d).

Concept. In electrolysis of an aqueous solution, at the cathode the species with the highest (most positive) reduction potential is reduced; at the anode the species with the lowest reduction potential (easiest to oxidise) is oxidised — including the electrode itself if it is an active metal (like Ag, Cu) but not if it is inert (Pt). Water can also be reduced (2H₂O + 2e⁻ → H₂ + 2OH⁻, E° = −0.83 V) or oxidised (2H₂O → O₂ + 4H⁺ + 4e⁻, E° = +1.23 V).

1. (i) AgNO₃(aq), silver electrodes: cathode: Ag⁺ + e⁻ → Ag (silver deposits, E° 0.80 V > water). Anode: the silver electrode itself dissolves (Ag → Ag⁺ + e⁻, easier than oxidising water or NO₃⁻). Net: silver is transferred from anode to cathode (silver plating/refining); solution concentration stays constant.
2. (ii) AgNO₃(aq), platinum electrodes: cathode: Ag⁺ + e⁻ → Ag (deposited). Anode: Pt is inert and NO₃⁻ cannot be oxidised, so water is oxidised: 2H₂O → O₂ + 4H⁺ + 4e⁻. Products: Ag at the cathode and O₂ at the anode; the solution turns acidic (HNO₃ forms): 4AgNO₃ + 2H₂O → 4Ag + 4HNO₃ + O₂.
3. (iii) dilute H₂SO₄, platinum electrodes: cathode: 2H⁺ + 2e⁻ → H₂. Anode: SO₄²⁻ is not oxidised in dilute solution, so water: 2H₂O → O₂ + 4H⁺ + 4e⁻. Products: H₂ at the cathode (twice the volume of) O₂ at the anode — electrolysis of water; the acid only conducts.
4. (iv) CuCl₂(aq), platinum electrodes: cathode: Cu²⁺ + 2e⁻ → Cu (E° +0.34 V, far above water’s −0.83 V). Anode: 2Cl⁻ → Cl₂ + 2e⁻ (chloride is discharged in preference to water because of the high overpotential of O₂ evolution even though E°(O₂/H₂O) = 1.23 V is lower than that of Cl₂, 1.36 V). Products: Cu at the cathode and Cl₂ at the anode.

Extra depth. The rule ‘active anode dissolves, inert anode does not’ explains industrial copper refining (impure Cu anode, pure Cu cathode in CuSO₄) and silver plating. In (iv), the textbook answer is Cl₂ because O₂ needs an overpotential on platinum; in very dilute chloride solutions O₂ may come off instead. Faraday’s laws (Class XII) give the amount: 1 mol of electrons (96,500 C) deposits 1 mol of Ag but only ½ mol of Cu.

Conclusion: (i) Ag at cathode, Ag dissolves at anode; (ii) Ag at cathode, O₂ at anode; (iii) H₂ at cathode, O₂ at anode; (iv) Cu at cathode, Cl₂ at anode.

Concept. A metal displaces another from the solution of its salt if it is the stronger reducing agent, i.e. has the lower (more negative) standard electrode potential. Using Table 7.1: 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.

1. Order of reducing power (most negative E° first): Mg (−2.36) > Al (−1.66) > Zn (−0.76) > Fe (−0.44) > Cu (+0.34).
2. Displacing action: Mg displaces Al, Zn, Fe and Cu from their salt solutions; Al displaces Zn, Fe and Cu; Zn displaces Fe and Cu; Fe displaces Cu; Cu displaces none of the others.
3. Answer: in the order in which they displace each other (strongest displacer first): Mg > Al > Zn > Fe > Cu.

Extra depth. This is the same order as the metal reactivity series: Mg > Al > Zn > Fe > Cu, supported by the observations of section 7.2.1 (Zn displaces Cu, Cu displaces Ag). It also predicts which metals liberate hydrogen from dilute acids: those with negative E° — Mg, Al, Zn and Fe do; Cu (+0.34 V) does not. Cu, with the highest E°, is displaced by all others, which is why copper is deposited when iron is dipped in CuSO₄ (blue vitriol) solution.

Conclusion: Mg > Al > Zn > Fe > Cu (decreasing power to displace the others from their salt solutions).

Concept. The reducing power of a metal is its tendency to lose electrons (oxidation). The more negative the standard electrode potential E°(Mⁿ⁺/M), the stronger the metal is as a reducing agent. So arrange the metals in order of decreasing E° to get increasing reducing power. (The printed values are partly illegible in the book’s text; the values above are those of the exercise and agree with Table 7.1, which lists Mg²⁺/Mg as −2.36 V — the order does not change.)

1. List by E°, highest first: Ag (+0.80 V), Hg (+0.79 V), Cr (−0.74 V), Mg (−2.37 V), K (−2.93 V).
2. Reducing power increases as E° decreases: Ag has the least tendency to lose electrons (Ag⁺ is the best oxidant of the group); K has the greatest tendency.
3. Answer: Ag < Hg < Cr < Mg < K (increasing order of reducing power).

Extra depth. Notice that Ag and Hg are close (0.80 vs 0.79 V): both are ‘noble’ metals that do not liberate hydrogen from acids, whereas the three with negative E° do. Potassium, at the bottom, reacts violently with water (2K + 2H₂O → 2KOH + H₂), Mg only with steam, and Cr is protected by an oxide layer despite its negative E° — a reminder that E° gives thermodynamics, not rates. The order of oxidising power of the corresponding ions is the exact reverse: K⁺ < Mg²⁺ < Cr³⁺ < Hg²⁺ < Ag⁺.

Conclusion: Increasing reducing power: Ag < Hg < Cr < Mg < K.

Concept. A galvanic cell converts the energy of a spontaneous redox reaction into electricity. The oxidation half reaction takes place at the anode (negative electrode) and the reduction half reaction at the cathode (positive electrode); a salt bridge completes the circuit. The cell is written with the anode on the left: M | Mⁿ⁺ || N^m+ | N, where | is a phase boundary and || the salt bridge.

1. Cell representation: Zn(s) | Zn²⁺(aq) || Ag⁺(aq) | Ag(s) — a zinc rod in ZnSO₄ solution (left beaker) and a silver rod in AgNO₃ solution (right beaker), connected by a salt bridge and by an external wire (as in the Daniell cell, Fig. 7.3, with Cu replaced by Ag).
2. (i) Negatively charged electrode: the zinc electrode (anode) is negative, because it is the source of electrons (zinc is oxidised); the silver electrode is the positive electrode (cathode).
3. (ii) Carriers of current: in the external circuit (wire) the current is carried by electrons, flowing from Zn to Ag (conventional current flows from Ag to Zn); inside the cell, in the solutions and the salt bridge, by ions — Zn²⁺, Ag⁺, NO₃⁻, SO₄²⁻ and the ions of the salt bridge (K⁺, Cl⁻/NO₃⁻) migrating.
4. (iii) Electrode reactions: anode (oxidation): Zn(s) → Zn²⁺(aq) + 2e⁻; cathode (reduction): Ag⁺(aq) + e⁻ → Ag(s) (×2). Overall: Zn + 2Ag⁺ → Zn²⁺ + 2Ag.
5. E°_cell = E°(cathode) − E°(anode) = 0.80 − (−0.76) = +1.56 V (positive, so the cell reaction is spontaneous).

Extra depth. The cell potential of 1.56 V is larger than the Daniell cell’s 1.10 V because silver is a much weaker reducing agent than copper (0.80 V against 0.34 V) — so Ag⁺ is a stronger oxidant. Flow of electrons and current are opposite in direction. As the cell runs, the Zn rod dissolves, the Ag rod gains silver, and the salt bridge supplies anions to the zinc half-cell and cations to the silver half-cell to maintain electrical neutrality. If the salt bridge were removed, the circuit would be broken and the cell would stop. These ideas are developed in Class XII (Electrochemistry).

Conclusion: Zn(s)|Zn²⁺(aq)||Ag⁺(aq)|Ag(s); Zn is negative; electrons in the wire and ions in solution carry current; anode Zn → Zn²⁺ + 2e⁻, cathode Ag⁺ + e⁻ → Ag.