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Solutions

Overview

Step-by-step NCERT solutions for Redox Reactions (Chapter 7, NCERT Class 11 Chemistry) — every question and answer worked out in full, not just the final result. You can also read the Redox Reactions textbook chapter.

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What these solutions cover

All 30 questions in Redox Reactions are solved in the PDF. Here's what's inside, exercise by exercise:

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)₂.12H₂O
  2. What are the oxidation number of the underlined elements in each of the following and how do you rationalise your results?
    • (a) KI₃
    • (b) H₂S₄O₆
    • (c) Fe₃O₄
    • (d) CH₃CH₂OH
    • (e) CH₃COOH
  3. Justify that the following reactions are redox reactions:
    • (a) CuO(s) + H₂(g) -> Cu(s) + H₂O(g)
    • (b) Fe₂O₃(s) + 3CO(g) -> 2Fe(s) + 3CO₂(g)
    • (c) 4BCl₃(g) + 3LiAlH₄(s) -> 2B₂H₆(g) + 3LiCl(s) + 3AlCl₃(s)
    • (d) 2K(s) + F₂(g) -> 2K⁺F⁻(s)
    • (e) 4NH₃(g) + 5O₂(g) -> 4NO(g) + 6H₂O(g)
  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, Cr₂O₇²⁻ and NO₃⁻. Suggest 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) 6CO₂(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 redox reactions.
  10. The compound AgF₂ is 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…
  13. Identify the substance oxidised, reduced, oxidising agent and reducing agent for each of the following reactions:
    • (a) 2AgBr(s) + C₆H₆O₂(aq) -> 2Ag(s) + 2HBr(aq) + C₆H₄O₂(aq)
    • (b) HCHO(l) + 2[Ag(NH₃)₂]⁺(aq) + 3OH⁻(aq) -> 2Ag(s) + HCOO⁻(aq) + 4NH₃(aq) + 2H₂O(l)
    • (c) HCHO(l) + 2Cu²⁺(aq) + 5OH⁻(aq) -> Cu₂O(s) + HCOO⁻(aq) + 3H₂O(l)
    • (d) N₂H₄(l) + 2H₂O₂(l) -> N₂(g) + 4H₂O(l)
    • (e) Pb(s) + PbO₂(s) +…
  14. Consider the reactions: 2S₂O₃²⁻(aq) + I₂(s) -> S₄O₆²⁻(aq) + 2I⁻(aq) and S₂O₃²⁻(aq) + 2Br₂(l) + 5H₂O(l) -> 2SO₄²⁻(aq) + 4Br⁻(aq) + 10H⁺(aq). Why does the same reductant, thiosulphate, 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? XeO₆⁴⁻(aq) + 2F⁻(aq) + 6H⁺(aq) -> XeO₃(g) + F₂(g) + 3H₂O(l). What conclusion about the compound Na₄XeO₆ (of which XeO₆⁴⁻ is a part) can be drawn from the reaction?
  17. Consider the reactions:
    • (a) H₃PO₂(aq) + 4AgNO₃(aq) + 2H₂O(l) -> H₃PO₄(aq) + 4Ag(s) + 4HNO₃(aq)
    • (b) H₃PO₂(aq) + 2CuSO₄(aq) + 2H₂O(l) -> H₃PO₄(aq) + 2Cu(s) + H₂SO₄(aq)
    • (c) C₆H₅CHO(l) + 2[Ag(NH₃)₂]⁺(aq) + 3OH⁻(aq) -> C₆H₅COO⁻(aq) + 2Ag(s) + 4NH₃(aq) + 2H₂O(l)
    • (d) C₆H₅CHO(l) + 2Cu²⁺(aq) + 5OH⁻(aq) -> No change observed. What inference do you draw about the behaviour of Ag⁺ and Cu²⁺ from these…
  18. Balance the following redox reactions by ion-electron method:
    • (a) MnO₄⁻(aq) + I⁻(aq) -> MnO₂(s) + I₂(s) (in basic medium)
    • (b) MnO₄⁻(aq) + SO₂(g) -> Mn²⁺(aq) + HSO₄⁻(aq) (in acidic solution)
    • (c) H₂O₂(aq) + Fe²⁺(aq) -> Fe³⁺(aq) + H₂O(l) (in acidic solution)
    • (d) Cr₂O₇²⁻(aq) + SO₂(g) -> Cr³⁺(aq) + SO₄²⁻(aq) (in acidic solution)
  19. Balance the following equations in basic medium by ion-electron method and oxidation number methods and identify the oxidising agent and the reducing agent:
    • (a) P₄(s) + OH⁻(aq) -> PH₃(g) + H₂PO₂⁻(aq)
    • (b) N₂H₄(l) + ClO₃⁻(aq) -> NO(g) + Cl⁻(g)
    • (c) Cl₂O₇(g) + H₂O₂(aq) -> ClO₂⁻(aq) + O₂(g) + H⁺
  20. What sorts of informations 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⁺ ion. Write a balanced ionic equation for the reaction.
  22. Consider the elements: Cs, Ne, I and F.
    • (a) Identify the element that exhibits only negative oxidation state.
    • (b) Identify the element that exhibits only positive oxidation state.
    • (c) Identify the element that exhibits both positive and negative oxidation states.
    • (d) Identify the element which exhibits neither the negative nor the positive oxidation state.
  23. Chlorine is used to purify drinking water. Excess of chlorine is harmful. The excess of chlorine is removed by treating with sulphur dioxide. Present a balanced equation for this redox change taking place in water.
  24. Refer to the periodic table given in your book and now answer the following questions:
    • (a) Select the possible non metals that can show disproportionation reaction.
    • (b) Select three metals that can show disproportionation reaction.
  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 given in Table 7.1, predict if the reaction between the following is feasible:
    • (a) Fe³⁺(aq) and I⁻(aq)
    • (b) Ag⁺(aq) and Cu(s)
    • (c) Fe³⁺(aq) and Cu(s)
    • (d) Ag(s) and Fe³⁺(aq)
    • (e) Br₂(aq) and Fe²⁺(aq).
  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 = -2.93 V, Ag⁺/Ag = 0.80 V, Hg²⁺/Hg = 0.79 V, Mg²⁺/Mg = -2.37 V, Cr³⁺/Cr = -0.74 V. Arrange these metals in their increasing order of reducing power.
  30. Depict the galvanic cell in which the reaction Zn(s) + 2Ag⁺(aq) -> Zn²⁺(aq) + 2Ag(s) takes place. Further show:
    • (i) which of the electrode is negatively charged,
    • (ii) the carriers of the current in the cell, and
    • (iii) individual reaction at each electrode.
Sample

Worked examples

3 questions from Redox Reactions, solved step by step.

Exercises

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.

Assign oxidation numbers to each element:

In H₂O: H = +1, O = -2.

In F₂: F = 0.

In HF: H = +1, F = -1.

In HOF (hypofluorous acid): H = +1; F = -1 (most electronegative element); let O = x: +1 + x + (-1) = 0 ⇒ x = 0.

Changes in oxidation state:

F: 0 (in F₂) → -1 (in HF and HOF): F is reduced.

O: -2 (in H₂O) → 0 (in HOF): O is oxidized.

H: +1 throughout (no change).

Electron balance: One O atom loses 2 electrons (oxidized from -2 to 0); two F atoms each gain 1 electron (reduced from 0 to -1). Total gain = 2e⁻ = total loss. ✓

Since oxidation (of O) and reduction (of F) occur simultaneously, this is a redox reaction.

Answer: O in H₂O is oxidized from -2 to 0 (in HOF); F₂ is reduced from 0 to -1 (in HF and HOF). Since both oxidation and reduction occur simultaneously, this is a redox reaction.

Exercises

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

H₂SO₅ (peroxomonosulphuric acid / Caro's acid):

If all O were -2: 2(+1) + x + 5(-2) = 0 ⇒ x = +8. But S cannot exceed +6 (maximum oxidation state for a period-3 element). This is the fallacy.

Reality: H₂SO₅ contains one -O-O⁻ (peroxide) linkage with two O at -1 each, and three normal O at -2.

Correct: 2(+1) + x + 3(-2) + 2(-1) = 0 ⇒ 2 + x - 6 - 2 = 0 ⇒ x = +6.

Structure: HO-SO₂-O-OH (peroxy acid with one peroxide O-O group).

Cr₂O₇²⁻ (dichromate ion):

2x + 7(-2) = -2 ⇒ 2x - 14 = -2 ⇒ 2x = 12 ⇒ x = +6.

Structure: Two CrO₄ tetrahedra sharing one oxygen vertex (Cr-O-Cr bridge).

NO₃⁻ (nitrate ion):

x + 3(-2) = -1 ⇒ x = +5.

Structure: Planar triangular (trigonal planar) with one N=O and two N-O bonds (or three equivalent N-O bonds by resonance). N is sp² hybridised.

Fallacy for H₂SO₅: Assigning -2 to all oxygen atoms gives S = +8 which is impossible for sulphur (maximum is +6). Therefore H₂SO₅ must possess a peroxide (-O-O⁻) linkage with oxygen in -1 oxidation state to give a chemically valid S = +6.

Answer: S in H₂SO₅ = +6 (has -O-O⁻ peroxide linkage; assuming all O=-2 gives impossible +8); Cr in Cr₂O₇²⁻ = +6; N in NO₃⁻ = +5.

Exercises

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

In AgF₂, silver is in the +2 oxidation state (Ag²⁺).

The normal and stable oxidation state of silver is +1 (Ag⁺).

Ag²⁺ is a highly unstable oxidation state for silver because it involves a [Kr]₄d⁹ electronic configuration, which has a strong tendency to gain one electron to achieve the stable [Kr]₄d¹⁰ configuration of Ag⁺.

Therefore, Ag²⁺ in AgF₂ readily accepts an electron (is reduced to Ag⁺), oxidising whatever substance donates the electron.

This makes AgF₂ an extremely strong oxidising agent.

The instability of the Ag²⁺ state drives the oxidising action: the compound decomposes or reacts to convert Ag²⁺ → Ag⁺ by oxidising other species.

Answer: AgF₂ contains Ag in the unstable +2 oxidation state. Ag²⁺ has a strong tendency to gain one electron and return to the stable +1 state. This high electron affinity of Ag²⁺ makes AgF₂ a very strong oxidising agent.

All 30 questions in Redox Reactions are solved step by step in the Solutions PDF above.

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