Chemistry Class 11 Part II…
🏠 Home Log in
TOPIC 4 OF 13

NCERT Exercises and Solutions: Redox Reactions

🎓 Class 11 Chemistry CBSE Theory Ch 7 – Redox Reactions ⏱ ~8 min
🌐 Language: [gtranslate]

This MCQ module is based on: NCERT Exercises and Solutions: Redox Reactions

This assessment will be based on: NCERT Exercises and Solutions: Redox Reactions

Upload images, PDFs, or Word documents to include their content in assessment generation.

NCERT Exercises and Solutions: Redox Reactions

Chapter 7 Summary — Redox Reactions

Classical view

Oxidation = gain of O or electronegative element, or loss of H or electropositive element. Reduction = opposite.

Electron-transfer view (OIL RIG)

Oxidation Is Loss of electrons; Reduction Is Gain. Redox = one process, two halves that always occur together.

Oxidation number rules

Free element = 0; monatomic ion = its charge; H = +1 (−1 in hydrides); O = −2 (−1 peroxide, +2 OF2, −½ superoxide); F = −1 always; sum = 0 for neutral molecule, = charge for a polyatomic ion.

Types of redox

Combination, Decomposition, Displacement (metal / non-metal), Disproportionation (same element oxidised AND reduced).

Balancing

Oxidation-number method (equate ΔON) or ion-electron method (balance atoms → O with H2O → H with H⁺ or OH⁻ → charge with e⁻ → equalise e⁻ → add).

Galvanic cell & EMF

E°cell = E°cathode − E°anode. Daniell cell Zn|Zn²⁺‖Cu²⁺|Cu gives +1.10 V. Positive EMF ⇒ spontaneous.

Keywords

Redox reaction
Oxidation
Reduction
Oxidising agent
Reducing agent
Oxidation number
Stock notation
Half-reaction
Ion-electron method
Combination
Decomposition
Displacement
Disproportionation
Activity series
Galvanic cell
Daniell cell
Anode / Cathode
Salt bridge
Electrode potential (E°)
EMF
Self-indicator
Redox titration

NCERT Exercises — Solved

7.1 Justify that the following reaction is a redox reaction: 2 Cu2O(s) + Cu2S(s) → 6 Cu(s) + SO2(g).

Assign oxidation numbers: in Cu2O Cu is +1, in Cu2S Cu is +1, S is −2. On the product side, Cu is 0 and S is +4 (SO2). Cu goes +1 → 0 (reduced) and S goes −2 → +4 (oxidised). Since ON changes occur, the reaction is redox.

7.2 Why do species like SO2 and H2O2 act both as oxidising and reducing agents?

Both contain an element in an intermediate oxidation state. In SO2, S is +4 — it can rise to +6 (acting as reducer) or fall to 0/−2 (acting as oxidant). In H2O2, O is −1 — it can rise to 0 (reducer) or fall to −2 (oxidant).

7.3 Identify oxidation and reduction in: (a) 2 H₂S + SO₂ → 3 S + 2 H₂O, (b) CuSO₄ + Zn → Cu + ZnSO₄.

(a) In H₂S, S = −2; in SO₂, S = +4; product S = 0. S (−2) → 0 is oxidation, S (+4) → 0 is reduction. (b) Zn 0 → +2 (oxidised, reducing agent); Cu +2 → 0 (reduced, oxidising agent).

7.4 Write the formulas of the following compounds using Stock notation: (a) Copper(II) sulphide, (b) Iron(III) chloride, (c) Manganese(IV) oxide, (d) Tin(II) chloride.

(a) CuS; (b) FeCl3; (c) MnO2; (d) SnCl2.

7.5 Identify the type of redox reaction: (a) 2 H2O2 → 2 H2O + O2; (b) 2 Al + Fe2O3 → 2 Fe + Al2O3; (c) 2 Na + 2 H2O → 2 NaOH + H2; (d) CaCO3 → CaO + CO2.

(a) Disproportionation (O: −1 → −2 and 0). (b) Metal displacement (thermite). (c) Non-metal displacement (H2 released). (d) Not redox — pure decomposition; no ON changes.

7.6 Find the oxidation number of the underlined element: (a) NaH₂PO₄, (b) NaHSO₄, (c) H₄P₂O₇, (d) K₂MnO₄, (e) CrO₅, (f) H₂S₂O₇.

(a) P: (+1) + 2(+1) + x + 4(−2) = 0 ⇒ x = +5. (b) H in HSO₄⁻ is +1 (standard). (c) P in H₄P₂O₇: 4(+1) + 2x + 7(−2) = 0 ⇒ 2x = +10, x = +5. (d) Mn: 2(+1) + x + 4(−2) = 0 ⇒ x = +6. (e) CrO₅ has two peroxide linkages (4 O at −1) and 1 normal O at −2: x + 4(−1) + (−2) = 0 ⇒ x = +6. (f) S: 2(+1) + 2x + 7(−2) = 0 ⇒ 2x = +12, x = +6.

7.7 Justify whether NO2 → NO3⁻ + NO in basic medium is a disproportionation.

In NO2, N = +4. In NO3⁻, N = +5 (oxidised). In NO, N = +2 (reduced). Same element N both oxidised (+4 → +5) and reduced (+4 → +2) ⇒ disproportionation.

7.8 Balance: NO3⁻ + Cu → NO + Cu2+ in acidic medium.

Ox: Cu → Cu²⁺ + 2e⁻. Red: NO₃⁻ + 4H⁺ + 3e⁻ → NO + 2H₂O. Multiply Ox × 3, Red × 2:
3 Cu + 2 NO₃⁻ + 8 H⁺ → 3 Cu²⁺ + 2 NO + 4 H₂O
Charge check: 0 + (−2) + 8(+1) = +6 = 3(+2) + 0. ✓

7.9 Calculate the oxidation number of S in H2SO5 (peroxomonosulphuric / Caro's acid).

Structure has one peroxide linkage (two O at −1) and three normal O at −2: 2(+1) + x + 3(−2) + 2(−1) = 0 ⇒ x = +6.

7.10 Find the oxidation number of (a) Mn in Mn₃O₄, (b) Fe in Fe0.94O, (c) O in KO2, (d) Cr in K3[Cr(CN)6].

(a) 3x + 4(−2) = 0 ⇒ x = +8/3 (average). (b) 0.94 x + (−2) = 0 ⇒ x ≈ +2.13 (non-stoichiometric wüstite). (c) KO₂ (superoxide): (+1) + 2x = 0 ⇒ x = −½. (d) CN⁻ is −1 each (6 of them): 3(+1) + x + 6(−1) = 0 ⇒ x = +3.

7.11 Classify: (a) 3 Cl2 + 6 NaOH(hot) → 5 NaCl + NaClO3 + 3 H2O; (b) NH4NO3 → N2O + 2 H2O; (c) Ca + 2 H2O → Ca(OH)2 + H2.

(a) Disproportionation of Cl2 (0 → −1 and +5). (b) Intramolecular / thermal disproportionation: N (−3 in NH4+) and N (+5 in NO3−) meet at N (+1 in N2O). (c) Non-metal displacement (H2 released by active metal).

7.12 Identify the reaction type: 2 AgBr(s) →hν→ 2 Ag(s) + Br2(l).

Decomposition and redox: Ag (+1) → 0 (reduced), Br (−1) → 0 (oxidised). This is the principle of black-and-white photographic film.

7.13 Classify: 4 NH3 + 5 O2 → 4 NO + 6 H2O (Ostwald process step).

Redox only (no simple combination/displacement/disproportionation fit). N: −3 → +2 (oxidised). O: 0 → −2 (reduced). NH3 is reducing agent, O2 is oxidising agent.

7.14 Identify oxidising and reducing agents: 2 KClO3 + H2SO4 → 2 HClO3 + K2SO4.

No oxidation number changes — a simple acid–base/double-displacement reaction, not a redox. Hence neither species acts as oxidising or reducing agent.

7.15 Classify: Cu(s) + 2 AgNO3(aq) → Cu(NO3)2(aq) + 2 Ag(s).

Metal-displacement redox. Cu is above Ag in activity series, so Cu displaces Ag. Cu 0 → +2 (oxidised), Ag +1 → 0 (reduced).

7.16 Balance by oxidation-number method: MnO2 + HCl → MnCl2 + Cl2 + H2O.

Mn +4 → +2 (ΔON = −2); Cl −1 → 0 (ΔON = +1 per atom, but Cl2 has 2 atoms ⇒ total +2). Ratios balance 1:1.
MnO₂ + 4 HCl → MnCl₂ + Cl₂ + 2 H₂O
Of the 4 HCl, 2 act as oxidised to Cl2 and 2 form MnCl2.

7.17 Balance: Fe2+ + H2O2 → Fe3+ + H2O (acidic).

Ox: Fe²⁺ → Fe³⁺ + e⁻. Red: H₂O₂ + 2H⁺ + 2e⁻ → 2H₂O. Multiply Ox by 2:
2 Fe²⁺ + H₂O₂ + 2 H⁺ → 2 Fe³⁺ + 2 H₂O

7.18 Balance: Cu + HNO3(dilute) → Cu(NO3)2 + NO + H2O.

Ox: Cu → Cu²⁺ + 2e⁻. Red: NO₃⁻ + 4H⁺ + 3e⁻ → NO + 2H₂O. ×3 and ×2:
3 Cu + 8 HNO₃ → 3 Cu(NO₃)₂ + 2 NO + 4 H₂O
(Of the 8 HNO3, 2 act as oxidiser and 6 as spectator to balance Cu2+.)

7.19 Balance: K2Cr2O7 + FeSO4 + H2SO4 → Cr2(SO4)3 + Fe2(SO4)3 + K2SO4 + H2O.

Ionic core: Cr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O. Molecular form:
K₂Cr₂O₇ + 6 FeSO₄ + 7 H₂SO₄ → Cr₂(SO₄)₃ + 3 Fe₂(SO₄)₃ + K₂SO₄ + 7 H₂O

7.20 Balance: Zn + HNO3(very dilute) → Zn(NO3)2 + NH4NO3 + H2O.

Ox: Zn → Zn²⁺ + 2e⁻. Red: NO₃⁻ + 10H⁺ + 8e⁻ → NH₄⁺ + 3H₂O. Ox × 4, Red × 1:
4 Zn + 10 HNO₃ → 4 Zn(NO₃)₂ + NH₄NO₃ + 3 H₂O

7.21 Balance in acidic medium (half-reaction method): Cr2O72− + SO2 → Cr3+ + SO42−.

Ox: SO₂ + 2H₂O → SO₄²⁻ + 4H⁺ + 2e⁻. Red: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O. Ox × 3:
Cr₂O₇²⁻ + 3 SO₂ + 2 H⁺ → 2 Cr³⁺ + 3 SO₄²⁻ + H₂O

7.22 Balance in basic medium: Cl2 + OH⁻ → Cl⁻ + ClO3⁻ + H2O (hot concentrated alkali).

Disproportionation. Ox: ½Cl₂ + 6OH⁻ → ClO₃⁻ + 3H₂O + 5e⁻ (i.e. Cl₂ + 12OH⁻ → 2ClO₃⁻ + 6H₂O + 10e⁻). Red: Cl₂ + 2e⁻ → 2Cl⁻. Multiply Red × 5:
3 Cl₂ + 6 OH⁻ → 5 Cl⁻ + ClO₃⁻ + 3 H₂O

7.23 Balance in basic medium: N2H4 + ClO3⁻ → NO + Cl⁻.

N: −2 → +2 (loses 4 e⁻ per N; 8 e⁻ per N₂H₄ molecule). Cl: +5 → −1 (gains 6 e⁻). LCM = 24. So take 3 N₂H₄ and 4 ClO₃⁻.
3 N₂H₄ + 4 ClO₃⁻ → 6 NO + 4 Cl⁻ + 6 H₂O
Check: atoms N 6=6, H 12 ⇒ in 6 H₂O that's 12 H ✓, Cl 4=4, O 12=6+6 ✓; charge −4 = −4 ✓.

7.24 Balance in acidic medium: P4 + HNO3 → H3PO4 + NO2 + H2O.

P: 0 → +5 (5 e⁻ per P × 4 = 20 per P4). N: +5 → +4 (1 e⁻ per N). So multiply HNO3 by 20:
P₄ + 20 HNO₃ → 4 H₃PO₄ + 20 NO₂ + 4 H₂O
Check H: 20 LHS = 12 + 0 + 8 = 20 ✓.

7.25 Balance in basic medium: Br2 + OH⁻ → BrO3⁻ + Br⁻ + H2O.

Disproportionation. Ox (per Br): Br⁻ loses 6 e⁻ → BrO3⁻. Red (per Br): gains 1 e⁻. LCM ⇒ 1 Br to BrO3⁻ and 5 Br to Br⁻, i.e. 3 Br2:
3 Br₂ + 6 OH⁻ → BrO₃⁻ + 5 Br⁻ + 3 H₂O

7.26 25.0 mL of 0.02 M KMnO4 is required to titrate 0.25 g of an oxalate in acidic medium. Find the % of C2O42− in the sample.

Stoichiometry: 2 MnO4⁻ reacts with 5 C2O42−. Moles of MnO4⁻ = 0.025 × 0.02 = 5.0 × 10⁻⁴. Moles of C2O42− = (5/2) × 5.0 × 10⁻⁴ = 1.25 × 10⁻³. Mass of C2O42− = 1.25 × 10⁻³ × 88 g·mol⁻¹ = 0.110 g. % = (0.110 / 0.25) × 100 = 44.0%.

7.27 Write the cell reaction and EMF of the cell: Zn | Zn2+(1 M) ‖ Ag+(1 M) | Ag. E°(Zn²⁺/Zn) = −0.76 V; E°(Ag⁺/Ag) = +0.80 V.

Anode (ox): Zn → Zn²⁺ + 2 e⁻. Cathode (red): 2 Ag⁺ + 2 e⁻ → 2 Ag. Overall:
Zn(s) + 2 Ag⁺(aq) → Zn²⁺(aq) + 2 Ag(s)
E°cell = 0.80 − (−0.76) = +1.56 V. Positive ⇒ spontaneous.

7.28 What volume of 0.1 M K2Cr2O7 is needed to oxidise 50 mL of 0.1 M FeSO4 in acidic medium?

Balanced ionic: Cr2O72− + 6 Fe²⁺ + 14 H⁺ → 2 Cr³⁺ + 6 Fe³⁺ + 7 H₂O. Moles Fe²⁺ = 0.050 × 0.1 = 5.0 × 10⁻³. Moles Cr2O72− needed = 5.0 × 10⁻³ / 6 = 8.33 × 10⁻⁴. Volume = 8.33 × 10⁻⁴ / 0.1 = 8.33 × 10⁻³ L = 8.33 mL.

7.29 Calculate the equivalent weight of KMnO4 in acidic medium and in basic / neutral medium.

Molar mass of KMnO4 = 158 g·mol⁻¹. In acidic medium Mn goes +7 → +2 (5 e⁻ change): equivalent weight = 158 / 5 = 31.6 g·eq⁻¹. In neutral / faintly alkaline medium Mn goes +7 → +4 (3 e⁻ change, forms MnO2): equivalent weight = 158 / 3 = 52.67 g·eq⁻¹. In strongly alkaline: 1 e⁻ change ⇒ EW = 158.

7.30 Consider the reactions: (a) H3PO2 + 2 AgNO3 + 2 H2O → 2 Ag + 4 HNO3 + H3PO4; (b) H3PO2 + 4 AgNO3 + 2 H2O → 4 Ag + 4 HNO3 + H3PO4. Why is the equivalent weight of H3PO2 different in the two cases?

Equivalent weight is defined per electron transferred. In (a) each H3PO2 hands over 2 e⁻ (P: +1 → +3 via H3PO3 internally, but see (b) limit): EW = M / 2. In (b) each H3PO2 hands over 4 e⁻ (P: +1 → +5 to form H3PO4): EW = M / 4 = 66/4 = 16.5 g·eq⁻¹. Hence the value depends on the extent of oxidation.

Frequently Asked Questions — NCERT Exercises and Solutions: Redox Reactions

How do you find oxidation number in complex compounds in NCERT exercises?
To find oxidation number in NCERT Class 11 Chemistry Chapter 7 exercise problems: (1) apply known values from the rules; (2) set up an equation with the unknown; (3) solve. For example, in K₂Cr₂O₇: 2(+1) + 2x + 7(−2) = 0 → x = +6 for Cr. In KMnO₄: +1 + x + 4(−2) = 0 → x = +7 for Mn. In Na₂S₂O₃: 2(+1) + 2x + 3(−2) = 0 → x = +2 for S (average). For peroxides like H₂O₂, O is −1. For OF₂, O is +2 (F is more electronegative). Practice with H₂S₂O₈ and K₂Cr₂O₇.
How do you balance redox equations step by step?
Balancing redox equations in NCERT Class 11 Chemistry Chapter 7: by oxidation number method — (1) assign oxidation numbers, (2) find change for each atom, (3) balance electrons by multiplying, (4) balance other atoms by inspection, (5) balance O with H₂O and H with H⁺ or OH⁻. By half-reaction method — (1) split into oxidation and reduction halves, (2) balance atoms (not H, O), (3) balance O with H₂O, (4) balance H with H⁺ (acidic) or OH⁻ (basic), (5) balance charge with electrons, (6) equate electrons and add halves. Always check atoms and charges balance after each step.
How do you identify oxidising and reducing agents in NCERT problems?
To identify oxidising and reducing agents in NCERT Class 11 Chemistry Chapter 7 exercises: (1) calculate oxidation numbers before and after reaction; (2) the species whose element's oxidation number increases is the reducing agent (it has been oxidised); (3) the species whose element's oxidation number decreases is the oxidising agent (it has been reduced). Example: in 2Mg + O₂ → 2MgO, Mg goes from 0 to +2 (oxidised, so Mg is reducing agent); O goes from 0 to −2 (reduced, so O₂ is oxidising agent). Practice with permanganate, dichromate and peroxide reactions.
How is cell potential calculated in NCERT exercises?
To calculate cell potential in NCERT Class 11 Chemistry Chapter 7 exercises: (1) identify cathode (higher reduction potential, undergoes reduction) and anode (lower reduction potential, undergoes oxidation); (2) write half-reactions and look up E° values from the electrochemical series; (3) apply E°_cell = E°_cathode − E°_anode (both as reduction potentials, no sign flipping). Example: for Zn|Zn²⁺||Cu²⁺|Cu, E°_cell = +0.34 − (−0.76) = +1.10 V. Always cite reference standards (SHE, 298 K, 1 M, 1 bar) and use the correct sign convention.
How are cell notations written in NCERT problems?
Cell notation in NCERT Class 11 Chemistry Chapter 7 follows the convention: anode | anode solution (concentration) || cathode solution (concentration) | cathode. The salt bridge is indicated by double vertical lines ||, phase boundaries by single |. Examples: Zn(s) | Zn²⁺(1 M) || Cu²⁺(1 M) | Cu(s); Pt(s) | H₂(1 bar) | H⁺(1 M) || Cu²⁺(1 M) | Cu(s). Anode is always on the left; cathode is always on the right. Inert electrodes (Pt, C) are written explicitly. State activity or concentration in parentheses. Spontaneous cell has positive E°_cell.
What 5-mark questions are common in Chapter 7 board exams?
Common 5-mark CBSE Class 11 Chemistry Chapter 7 board questions: (1) balance a given redox equation by ion-electron method in acidic and basic media; (2) explain classical and electron-transfer concepts of redox with examples of each type (combination, decomposition, displacement, disproportionation); (3) describe construction and working of a Daniell cell, write half-reactions and calculate E°_cell; (4) state the rules for assigning oxidation numbers and apply to given compounds; (5) discuss the standard hydrogen electrode and the electrochemical series. The MyAiSchool exercise set provides model answers and marking schemes.
AI Tutor
Chemistry Class 11 Part II – NCERT (2025-26)
Ready
Hi! 👋 I'm Gaura, your AI Tutor for NCERT Exercises and Solutions: Redox Reactions. Take your time studying the lesson — whenever you have a doubt, just ask me! I'm here to help.