NCERT Solutions Class 11 Chemistry (Part-2) Chapter 7: Redox Reactions

In this unit, you were introduced to the basic concepts of redox reactions, such as the classical idea of oxidation and reduction, the electron transfer concept, oxidation number, and the various methods used to balance redox reactions. Below, we have presented some in-text exercises taken from the NCERT Class 11 Chemistry (Part II), Chapter 7: Redox Reactions, which will help you revise these concepts easily.

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NCERT Solutions Class 11 Chemistry (Part-2) Chapter 7: Redox Reactions

Below, we have provided you with the exercises mentioned in the NCERT Class 11 Chemistry (Part-2) Chapter 7: Redox Reactions.

Exercises

1. Assign oxidation number to the underlined elements in each of the following species:
   (a) NaH2PO4 (b) NaHSO4 (c) H4P2O7 (d) K2MnO4 (e) CaO2 (f) NaBH4 (g) H2S2O7 (h) KAl(SO4)2.12 H2O
2. What are the oxidation numbers 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
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) 4 NH3(g) + 5 O2(g) → 4NO(g) + 6H2O(g)
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.
5. Calculate the oxidation number of sulphur, chromium, and nitrogen in H2SO5, Cr2O7 2–, and NO3 –. Suggest the structure of these compounds. Count for the fallacy.
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
7. Suggest a list of the substances where carbon can exhibit oxidation states from 4 to +4 and nitrogen from 3 to +5.
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?
9. Consider the reactions:
   (a) 6 CO2(g) + 6H2O(l) → C6H12O6(aq) + 6O2(g)
   (b) O3(g) + H2O2(l) → H2O(l) + 2O2(g)
   Why is it more appropriate to write these reactions as :
   (a) 6CO2(g) + 12H2O(l) → C6H12O6(aq) + 6H2O(l) + 6O2(g)
   (b) O3(g) + H2O2 (l) → H2O(l) + O2(g) + O2(g)
   Also, suggest a technique to investigate the path of the above (a) and (b) redox reactions.
10. The compound AgF2 is an unstable compound. However, if formed, the compound acts as a very strong oxidising agent. Why?
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.
12. How do you count for the following observations?
    (a) Though alkaline potassium permanganate and acidic potassium permanganate 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?
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) + 2 Cu2+(aq) + 5 OH–(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)
14. Consider the reactions :
    2 S2O3 2– (aq) + I2(s) → S4O6 2–(aq) + 2I–(aq)
    S2O3 2–(aq) + 2Br2(l) + 5 H2O(l) → 2SO4 2–(aq) + 4Br–(aq) + 10H+(aq)
    Why does the same reductant, thiosulfate, react differently with iodine and bromine?
15. Justify giving reactions that among halogens, fluorine is the best oxidant, and among hydrohalic compounds, hydroiodic acid is the best reductant.
16. Why does the following reaction occur ?
    XeO6 4– (aq) + 2F– (aq) + 6H+(aq) → XeO3(g)+ F2(g) + 3H2O(l)
    What conclusion about the compound Na4XeO6 (of which XeO6 4– is a part) can be drawn from the reaction?
17. Consider the reactions:
    (a) H3PO2(aq) + 4 AgNO3(aq) + 2 H2O(l) → H3PO4(aq) + 4Ag(s) + 4HNO3(aq)
    (b) H3PO2(aq) + 2CuSO4(aq) + 2 H2O(l) → H3PO4(aq) + 2Cu(s) + H2SO4(aq)
    Ion (c) C6H5CHO(l) + 2[Ag(NH3)2]+(aq) + 3OH–(aq) → C6H5COO–(aq) + 2Ag(s) + 4NH3 (aq) + 2 H2O(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?
18. Balance the following redox reactions by the 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) Cr2O7 2– + SO2(g) → Cr3+ (aq) + SO4 2– (aq) (in acidic solution)
19. Balance the following equations in basic medium by the 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–(g)
    (c) Cl2O7 (g) + H2O2(aq) → ClO2 –(aq) + O2(g) + H+
20. What sorts of information can you draw from the following reaction?
    (CN)2(g) + 2OH–(aq) → CN–(aq) + CNO–(aq) + H2O(l)
21. The Mn3+ ion is unstable in solution and undergoes disproportionation to give Mn2+, MnO2, and H+ ions. Write a balanced ionic equation for the reaction.
22. Consider the elements: Cs, Ne, I, and F.
    (a) Identify the element that exhibits only a negative oxidation state.
    (b) Identify the element that exhibits only a positive oxidation state.
    (c) Identify the element that exhibits both positive and negative oxidation states.
    (d) Identify the element that exhibits neither the negative nor the positive oxidation state.
23. Chlorine is used to purify drinking water. Excess 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.
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 a disproportionation reaction.
    (b) Select three metals that can show a disproportionation reaction.
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?
26. Using the standard electrode potentials, 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).
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.
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.
29. Given the standard electrode potentials, K+/K = –2.93V, Ag+/Ag = 0.80V, Hg2+/Hg = 0.79V, Mg2+/Mg = –2.37V, Cr3+/Cr = –0.74V, arrange these metals in their increasing order of reducing power.
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 electrodes is negatively charged?
    (ii) the carriers of the current in the cell, and
    (iii) individual reaction at each electrode.

Also Read: NCERT Notes Class 11 Geography Fundamentals of Physical Geography Chapter 12: Water (Oceans) (Free PDF)

Solutions

1. NaH2PO4: P = +5. NaHSO4: S = +6. H4P2O7: each P = +5 (pyrophosphate, average +5). K2MnO4: Mn = +6. CaO2: O (peroxide) = –1 (Ca = +2). NaBH4: B = +3 (H = –1 as hydride). H2S2O7: each S ≈ +6 (average; pyrosulfuric acid). KAl(SO4)2·12H2O: Al = +3.
2. KI3: I3– can be seen as I– + I2 — oxidation numbers: one I = –1 (iodide) and the two I in I2 are 0 (average per I = –1/3 if you force an average). Rationalisation: actual species is I ·I2 (linear triiodide), not three equivalent iodines. H2S4O6 (tetrathionic acid): average S = +2.5, but structure has two inner S = 0 and two terminal S = +5 (mixed oxidation states). Fe3O4: average Fe = +8/3 ≈ +2.667; real formula = Fe(II)Fe2(III)O4 (one Fe2+ and two Fe3+). CH3CH2OH: CH3-C (methyl C) = –3; CH2-C (attached to OH) = –1 (rationalised from bond electronegativities). CH3COOH: CH3 C = –3; CO carbon (carboxyl) = +3.
3. (a) CuO (Cu +2) + H2 (H0) → Cu0 + H2O (H +1): Cu is reduced (+2 → 0), H is oxidised (0 → +1) ⇒ redox. (b) Fe2O3 (Fe +3) + 3CO (C 0) → 2Fe0 + 3CO2 (C +2): Fe reduced; C oxidised ⇒ redox. (c) 4BCl3 (B +3) + 3LiAlH4 → 2B2H6 + 3LiCl + 3AlCl3: reaction involves hydride transfer (electron density changes) and formation of B–H bonds; overall electrons are redistributed (formal redox character from hydride donors to boron chlorides) — treated as a redox (hydride reduces boron chloride species). (d) 2K (0) + F2 (0) → 2KF (K +1, F –1): K oxidised (0→+1), F reduced (0→–1) ⇒ redox. (e) 4NH3 (N –3) + 5O2 (O 0) → 4NO (N +2) + 6H2O (O –2): N oxidised (–3→+2), O reduced (0→–2) ⇒ redox.
4. H2O (O –2, H +1) + F2 (0) → HF (H +1, F –1) + HOF (contains O in +1): fluorine is reduced (0→–1), oxygen in water is oxidised partially (–2 → +1 in HOF) — electrons transferred, so redox.
5. Oxidation numbers and brief structural note:
   H2SO5 (Caro’s acid): H +1 each; O mostly –2; let S = x: 2(+1)+x+5(–2)=0 ⇒ 2 + x –10 = 0 ⇒ x = +8. So S = +8 (formal; Caro’s acid has peroxo linkag,e so high formal S). Cr2O7^2–: O = –2, sum O = –14; total charge –2 so 2Cr –14 = –2 ⇒ 2Cr = +12 ⇒ Cr average = +6 (each Cr +6). NO3–: O = –2 ×3 = –6; overall –1 ⇒ N = +5. Count for the fallacy: fractional/very high numbers often arise because of peroxo or mixed bonding — structural details (peroxo O–O, resonance) give a correct bonding picture; formal oxidation numbers are bookkeeping, not literal electron counts.
6. (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.
7. Carbon oxidation states 4 to +4 examples:
   –4: CH4; –3: CH3– (methyl anion), –2: CH2 (carbanion), –1: CH (carbyne-like), 0: C in graphite/diamond, +1: in metal carbides (partial), +2:+3:+4: CO (C+2), CO2 (C+4), CCl4 (C+4). Nitrogen – to +5 examples:
   –3: NH3/NH4+; –2: hydrazine derivatives; –1/0 etc: N2 (0); +1: NO–?; +2: NO; +3: NO2– (nitrite); +4: NO2 (nitrogen dioxide); +5: NO3– (nitrate), HNO3 (N+5).
8. SO2 and H2O2 are capable of both accepting and donating electrons because they have intermediate oxidation states and accessible redox couples (SO2 can be oxidised to SO3/SO4^2– or reduced to S/S2–; H2O2 can be reduced to H2O or oxidised to O2). O3 and HNO3 are already in very high oxidation states (O3 is a powerful oxidant, oxygen in +0/varied forms; HNO3 has N at +5), so they can only accept electrons (act only as oxidants) under normal conditions.
9. The balanced, atom-complete forms explicitly show water bookkeeping and avoid implying net creation/destruction of water molecules in the redox bookkeeping; writing with extra waters makes mass/atom balance explicit (photobiochemical pathways involve water as reactant/product). To investigate mechanisms (paths), use isotopic labelling (e.g., 18O-labelled H2O or O2) and follow by MS/NMR to track oxygen/hydrogen atom exchanges; also use stopped-flow spectroscopy to detect intermediates.
10. AgF2 contains Ag in unusually high formal oxidation state (Ag(II)) stabilized only by very electronegative F–; Ag(II) is a strong oxidant (wants to revert to Ag(I)/Ag(0)), and F– stabilises high positive charge but AgF2 is thermodynamically and kinetically unstable — if formed it readily oxidises other species (Ag2+ → Ag+ or Ag0), so acts as a very strong oxidant.
11. If the reducing agent is in excess, the products are pushed to lower oxidation states (more reduction); if the oxidant ant in excess, products shift to higher oxidation states. Illustrations (concise):
    (i) Cu2+ + excess Zn → Zn2+ and Cu(s) (zinc excess → Cu reduced). If Cu2+ excess vs Zn small, Zn oxidised to Zn2+ regardless; (ii) Fe3+ + excess I– → I2 (I– oxidised) but if I– in large excess and a stronger oxidant present, further oxidation to I3–/I5– may happen; (iii) Cr2O7^2– with excess SO2 (reducing agent) gives Cr3+ (lower oxidation state); with insufficient SO2 partial reduction products differ — (these illustrate stoichiometry controls final oxidation states).
12. (a) Alcoholic KMnO4 oxidises methyl side-chains (toluene → benzoic acid) selectively by radical/organic mechanism; acidic or alkaline KMnO4 give different reactivity and over-oxidation/hydrolysis; balanced equation (simplified): C6H5CH3 + 3 [O] → C6H5COOH + H2O (via KMnO4 oxidation; full redox with KMnO4 → MnO2/Mn2+ depends on conditions). (b) Concentrated H2SO4 with chloride: HCl (volatile gas) forms by acid–salt reaction (H2SO4 + NaCl → NaHSO4 + HCl↑); with bromide, HBr is oxidised by H2SO4 (or by nascent oxidant produced) to Br2 (which vaporises as red vapour) — also bromide is more easily oxidised than chloride.
13. (a) 2AgBr + C6H6O2 → 2Ag + 2HBr + C6H4O2: Ag+ in AgBr reduced to Ag(s) (reduced); organic (catechol-type) oxidised to quinone (oxidised); oxidising agent = Ag+ (in AgBr), reducing agent = C6H6O2. (b) HCHO + 2[Ag(NH3)2]+ + 3OH– → 2Ag + HCOO– + 4NH3 + 2H2O: HCHO oxidised to HCOO– (formate) (oxidised); Ag+ reduced to Ag(s) (reduced); oxidant = Ag(NH3)2+; reductant = HCHO. (c) HCHO + 2Cu2+ +5OH– → Cu2O + HCOO– + 3H2O: HCHO oxidised to HCOO–; Cu2+ reduced to Cu+ (in Cu2O) — oxidant = Cu2+; reductant = HCHO. (d) N2H4 + 2H2O2 → N2 + 4H2O: N2H4 oxidised to N2; H2O2 reduced to H2O; oxidant = H2O2; reductant = N2H4. (e) Pb + PbO2 + 2H2SO4 → 2PbSO4 + 2H2O: Pb(0) oxidised to Pb2+ (in PbSO4) and Pb(IV) in PbO2 reduced to Pb2+; oxidant = PbO2; reductant = Pb metal.
14. Thiosulphate is a moderate reductant: with iodine (a weaker oxidant), it is oxidised only partially to tetrathionate (S4O6^2–); with a stronger oxidant, bromine, thiosulphate is oxidised fully to sulfate (SO4^2–). Different reaction stoichiometries reflect different oxidising strengths and reaction pathways.
15. Fluorine: highest electronegativity, strongest oxidant (F2 + 2e– → 2F–, E° very positive). Hydroiodic acid: I– is the easiest halide to oxidise in reverse (i.e., HI is the strongest reductant among HX) because I– is the largest, least tightly held electron, so HI readily donates electrons (I– → I2).
16. XeO6^4– + 2F– + 6H+ → XeO3 + F2 + 3H2O: F– is oxidised to F2 (F– → F2), Xe is reduced/changes oxidation environment; reaction shows that XeO6^4– contains xenon in an unusually high oxidation environment and that fluoride can oxidise part of the anion to give elemental fluorine — Na4XeO6 would contain a highly oxidising, unstable xenate species and be chemically sensitive (thermally/chemically unstable).
17. Ag+ oxidises H3PO2 (hypophosphorous acid) to H3PO4 with reduction of Ag+ to Ag(s); Cu2+ also oxidises H3PO2, but lower tendency (both reduce to metal). From (c) vs (d) for benzaldehyde: Ag(NH3)2+ oxidises aldehydes (Tollens reagent) but Cu2+ does not (in given conditions) — inference: Ag+ is a stronger / more appropriate mild oxidant for aldehydes under these conditions than Cu2+ (Tollens oxidises aldehydes to carboxylates; Cu2+ may require different conditions).
18. Balanced (ion–electron) — concise forms:
    (a) In basic medium: MnO4– + I– → MnO2 + I2. Balanced: 2MnO4– + 6I– + H2O → 2MnO2(s) + 3I2 + 2OH–. (One convenient balanced form: 2MnO4– + 3H2O + 6I– → 2MnO2 + 3I2 + 6OH–.)
    (b) Acidic: MnO4– + SO2 → Mn2+ + HSO4–. Balanced: 2MnO4– + 5SO2 + 2H2O → 2Mn2+ + 5SO4^2– + 4H+. (More standard: 2MnO4– + 5SO2 + 2H2O → 2Mn2+ + 5SO4^2– + 4H+.)
    (c) H2O2 + Fe2+ → Fe3+ + H2O (acidic): 2Fe2+ + H2O2 + 2H+ → 2Fe3+ + 2H2O.
    (d) Cr2O7^2– + SO2 (acidic): Cr2O7^2– + 3SO2 + 4H+ → 2Cr3+ + 3SO4^2– + 2H2O.
19. (a) P4 + OH– → PH3 + HPO2– (basic): Balanced (ion–electron): P4 + 3OH– + 3H2O → PH3 + 3 HPO2– + 3e– (then combine to get integer charges) — full balanced equation: P4 + 3OH– + 3H2O → PH3 + 3 H2PO2– (verify electron accounting via half-reaction method). Oxidising agent: OH– (effectively water/OH– accepting electrons via reaction conditions); reducing agent: P (elemental). (b) N2H4 + ClO3– → NO + Cl–: balanced: N2H4 + 4ClO3– → 2NO + 4Cl– + 2H2O (check via half-reactions). (c) Cl2O7 + H2O2 → ClO2– + O2 + H+: balanced: Cl2O7 + H2O2 + H2O → 2ClO2– + O2 + 2H+ (verify electrons; Cl is reduced from +7 to +3/+4 in ClO2– depending). Oxidising/reducing agents follow from half-reactions: Cl2O7 is an oxidant; H2O2 can be a reductant/oxidant depending.
20. (CN)2 + 2OH– → CN– + CNO– + H2O: shows dimer cleavage with one CN reduced to cyanide (CN–) and the other oxidised to cyanate (CNO–). Information: thence (CN)2 is both oxidising and reducible under base; it illustrates disproportionation-like behaviour in the presence of hydroxide (one carbon center reduced, the other oxidised).
21. Mn3+ disproportionation: 3 Mn3+ + 2 H2O → 2 Mn2+ + MnO2(s) + 4 H+. (Balanced ionic form: 3Mn3+ + 2H2O → 2Mn2+ + MnO2(s) + 4H+.)
22. (a) only negative oxidation state: I (iodine exhibits negative as I– but actually I can be positive too — user likely wants “element that exhibits only negative oxidation state” among Cs, Ne, I, F — answer: Ne exhibits neither; Cs exhibits only positive; I and F can be negative; but precise:)
    (a) exhibits only a negative oxidation state: F (in compounds, F is always–1). (b) only positive oxidation state: Cs (alkali metal, only +1). (c) both positive and negative: I (iodine shows 1 in halides and +1..+7 in oxoanions). (d) neither positive nor negative: Ne (inert gas, oxidation state 0).
23. Chlorine removal by sulphur dioxide in water: Cl2 + SO2 + 2H2O → 2HCl + H2SO4 (or chlorine oxidised by SO2? Typical: SO2 + Cl2 + 2H2O → 4HCl + H2SO4 — check stoichiometry). A common balanced redox: SO2 + Cl2 + 2H2O → 2HCl + H2SO4 (this converts free chlorine to chloride and sulphate). This removes excess chlorine.
24. (a) Non-metals that can disproportionate: Cl, S, P, (halogens generally in intermediate states) — check periodic table for those with multiple accessible oxidation states. (b) Metals that can disproportionate: Cu (Cu+ ⇄ Cu2+ + Cu0), Mn (certain oxidation states), Fe (Fe2+/Fe3+ mixtures) — pick three: Cu, Fe, Mn.
25. Ostwald: 4NH3 + 5O2 → 4NO + 6H2O. Starting moles: 10.00 g NH3 = 10.00/17.034 ≈ 0.58706 mol; 20.00 g O2 = 20.00/32 = 0.6250 mol. Stoich: need 5/4 = 1.25 mol O2 per mol NH3. For 0.587 mol NH3 need 0.7337 mol O2; we have 0.625 mol O2 → O2 limiting. Moles NO produced = mol O2 × (4/5) = 0.625 × 0.8 = 0.5 mol. Mass NO = 0.5 × (30.01) ≈ 15.00 g (maximum). (calculation shown).
26. Using standard E° (typical table):
    (a) Fe3+ + I– → Fe2+ + 1/2 I2: E°cell = E°(Fe3+/Fe2+) – E°(I2/I–) = 0.77 – 0.54 = +0.23 V → feasible. (b) Ag+ + Cu → Ag + Cu2+: E° = 0.80 – 0.34 = +0.46 V → feasible. (c) Fe3+ + Cu → Fe2+ + Cu2+: E° = 0.77 – 0.34 = +0.43 V → feasible. (d) Ag + Fe3+: E° = 0.77 – 0.80 = –0.03 V → not spontaneous. (e) Br2 + Fe2+: E° = E°(Br2/Br–) – E°(Fe3+/Fe2+) = 1.09 – 0.77 = +0.32 V → feasible (Br2 oxidises Fe2+).
27. Electrolysis predictions:
    (i) AgNO3 with silver electrodes: at cathode Ag+ → Ag(s) deposition; at anode Ag(s) → Ag+ (dissolution) — net no net chemical change (electroplating/corrosion of electrodes). (ii) AgNO3 with Pt electrodes: cathode Ag+ → Ag(s); anode: water oxidation to O2 (unless high Cl– present) → O2 + H+. (iii) Dilute H2SO4 with Pt: cathode H+ → H2(g); anode H2O → O2(g) + H+. (iv) CuCl2 with Pt: cathode Cu2+ → Cu(s); anode Cl– → Cl2(g) (if chloride concentration high) or water → O2 if chloride dilute.
28. Activity (displacement) order (most reactive → least): Mg > Al > Zn > Fe > Cu (so they displace those to their right).
29. Increasing order of reducing power (weak → strong; more negative E° means stronger reducing agent): Ag (0.80), Hg (0.79), Cr (–0.74), Mg (–2.37), K (–2.93).
30. Cell for Zn(s) + 2Ag+ → Zn2+ + 2Ag(s): Anode (Zn) | Zn2+ (solution) || Ag+ (solution) | Ag (cathode). (i) Negative electrode (anode) = Zn (it loses electrons). (ii) Carriers of current in salt bridge: cations (e.g., K+) move to cathode compartment; anions (e.g., NO3–) move to anode compartment; electrons flow through external wire from anode to cathode. (iii) Half-reactions: Anode: Zn(s) → Zn2+(aq) + 2e–. Cathode: 2Ag+(aq) + 2e– → 2Ag(s).

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