Redox Reactions — NCERT Class 11 Chemistry

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This page provides complete, step-by-step NCERT solutions for CBSE Class 11 Chemistry Chapter 8 — Redox Reactions. Every intext question (grouped by page number) and every chapter-end exercise question is answered with clear explanations, worked-out calculations, and balanced equations. Use the section headings to jump to the question you need help with.

Key Concepts (Quick Recap)

• Oxidation: Loss of electrons; increase in oxidation number.
• Reduction: Gain of electrons; decrease in oxidation number.
• Oxidizing Agent: Species that gets reduced (accepts electrons).
• Reducing Agent: Species that gets oxidized (donates electrons).
• Oxidation Number Rules: Elements = 0; H = +1 (−1 in hydrides); O = −2 (−1 in peroxides, +2 in OF₂); ions = charge; sum of ONs = charge on species.
• Half-Reaction Method: Split into oxidation and reduction half-reactions, balance atoms and charges separately, then combine.
• Oxidation Number Method: Identify ON changes, balance electron transfer, then balance remaining atoms.
• Disproportionation: Same element is simultaneously oxidized and reduced (e.g., Cu⁺ → Cu + Cu²⁺).
• Comproportionation: Reverse of disproportionation — two species of the same element with different ONs combine to form a single product (e.g., Ag²⁺ + Ag → 2Ag⁺).
• Electrode Processes: Anode = oxidation; Cathode = reduction. Electrons flow from anode to cathode in external circuit.

Intext Questions

Page 203–204

Q1. Assign oxidation numbers to each element in: (a) H₂O (b) H₂SO₄ (c) KMnO₄ (d) K₂Cr₂O₇ (e) Fe₂O₃ (f) Na₂O₂

(a) H₂O: H = +1, O = −2 (since 2(+1) + O = 0 → O = −2)

(b) H₂SO₄: H = +1, O = −2, S = +6 (since 2(+1) + S + 4(−2) = 0 → S = +6)

(c) KMnO₄: K = +1, O = −2, Mn = +7 (since +1 + Mn + 4(−2) = 0 → Mn = +7)

(d) K₂Cr₂O₇: K = +1, O = −2, Cr = +6 (since 2(+1) + 2Cr + 7(−2) = 0 → 2Cr = 12 → Cr = +6)

(e) Fe₂O₃: O = −2, Fe = +3 (since 2Fe + 3(−2) = 0 → 2Fe = 6 → Fe = +3)

(f) Na₂O₂ (peroxide): Na = +1, O = −1 (since 2(+1) + 2O = 0 → O = −1; peroxide rule)

Q2. Identify the oxidizing and reducing agents in: CuO(s) + H₂(g) → Cu(s) + H₂O(l)

• Cu: ON changes from +2 (in CuO) to 0 (in Cu) — reduced. CuO is the oxidizing agent.
• H: ON changes from 0 (in H₂) to +1 (in H₂O) — oxidized. H₂ is the reducing agent.

Q3. Justify that the following are redox reactions: (a) CuO(s) + H₂(g) → Cu(s) + H₂O(l) (b) Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g) (c) 2K(s) + F₂(g) → 2KF(s) (d) NH₃(g) + 3Cl₂(g) → NCl₃(g) + 3HCl(g)

(a) Cu: +2 → 0 (reduced); H: 0 → +1 (oxidized). Both oxidation and reduction occur — redox reaction.

(b) Fe: +3 → 0 (reduced); C: +2 → +4 (oxidized). Redox reaction.

(c) K: 0 → +1 (oxidized); F: 0 → −1 (reduced). Redox reaction.

(d) N: −3 → +3 (oxidized, 6 electron loss); Cl: 0 → −1 (reduced, in HCl) and 0 → +1 (oxidized, in NCl₃). Actually, Cl goes from 0 to −1 in HCl (reduction) and N goes from −3 to +3 in NCl₃ (oxidation). Redox reaction.

Page 209–210

Q4. Balance the following redox reaction by the half-reaction method (acidic medium): MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺

Reduction half-reaction (Mn): MnO₄⁻ → Mn²⁺ Balance O: MnO₄⁻ → Mn²⁺ + 4H₂O Balance H: MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O Balance charge: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O ... (i)

Oxidation half-reaction (Fe): Fe²⁺ → Fe³⁺ + e⁻ ... (ii)

Multiply (ii) by 5: 5Fe²⁺ → 5Fe³⁺ + 5e⁻

Add (i) and multiplied (ii): MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺

Balanced equation: MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O

Q5. Balance the following in acidic medium: Cr₂O₇²⁻ + Sn²⁺ → Cr³⁺ + Sn⁴⁺

Reduction half-reaction: Cr₂O₇²⁻ → 2Cr³⁺ Balance O: Cr₂O₇²⁻ → 2Cr³⁺ + 7H₂O Balance H: Cr₂O₇²⁻ + 14H⁺ → 2Cr³⁺ + 7H₂O Balance charge: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O ... (i)

Oxidation half-reaction: Sn²⁺ → Sn⁴⁺ + 2e⁻ ... (ii)

Multiply (ii) by 3: 3Sn²⁺ → 3Sn⁴⁺ + 6e⁻

Add: Cr₂O₇²⁻ + 14H⁺ + 3Sn²⁺ → 2Cr³⁺ + 3Sn⁴⁺ + 7H₂O

Balanced equation: Cr₂O₇²⁻ + 14H⁺ + 3Sn²⁺ → 2Cr³⁺ + 3Sn⁴⁺ + 7H₂O

Q6. Balance in basic medium: MnO₄⁻ + I⁻ → MnO₂ + I₂

Reduction half-reaction: MnO₄⁻ → MnO₂ Balance O: MnO₄⁻ → MnO₂ + 2H₂O (in basic medium, add H₂O to product) Balance H: MnO₄⁻ + 4H⁺ → MnO₂ + 2H₂O → in basic: MnO₄⁻ + 2H₂O → MnO₂ + 4OH⁻ Balance charge: MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻ ... (i)

Oxidation half-reaction: 2I⁻ → I₂ + 2e⁻ ... (ii)

Multiply (i) by 2 and (ii) by 3: 2MnO₄⁻ + 4H₂O + 6e⁻ → 2MnO₂ + 8OH⁻ 6I⁻ → 3I₂ + 6e⁻

Add: 2MnO₄⁻ + 4H₂O + 6I⁻ → 2MnO₂ + 8OH⁻ + 3I₂

Balanced equation: 2MnO₄⁻ + 4H₂O + 6I⁻ → 2MnO₂ + 3I₂ + 8OH⁻

Page 215–216

Example: 2Cu⁺ → Cu²⁺ + Cu Here, Cu⁺ is oxidized to Cu²⁺ (ON: +1 → +2) and reduced to Cu (ON: +1 → 0).

Q8. Identify the oxidizing agent and reducing agent in: 2PbO(s) + C(s) → 2Pb(s) + CO₂(g)

• Pb: +2 → 0 (reduced). PbO is the oxidizing agent.
• C: 0 → +4 (oxidized). C is the reducing agent.

• F: 0 → −1 (in HF) — reduced. F₂ is the oxidizing agent.
• O: −2 (in H₂O) → 0 (in O₂) — oxidized. H₂O is the reducing agent.

Q10. Which of the following are disproportionation reactions? (a) 2Cu⁺ → Cu²⁺ + Cu (b) 3MnO₄²⁻ + 4H⁺ → 2MnO₄⁻ + MnO₂ + 2H₂O (c) 2KClO₃ → 2KCl + 3O₂

(a) Yes. Cu⁺ (+1) → Cu²⁺ (+2, oxidation) and Cu⁺ (+1) → Cu (0, reduction). Same species oxidized and reduced.

(b) Yes. Mn in MnO₄²⁻ has ON = +6. It goes to MnO₄⁻ (Mn = +7, oxidation) and MnO₂ (Mn = +4, reduction). Same element both oxidized and reduced.

(c) No. Cl: +5 → −1 (reduced); O: −2 → 0 (oxidized). Different elements are oxidized and reduced. This is a standard redox reaction, not disproportionation.

Chapter-End Exercise

Q1. Assign oxidation numbers to the underlined atoms in: (a) P⁻₄O₁₀ (b) MⁿₙO₄⁻ (c) N₃⁻H₃ (d) C⁶H₃COOH (e) C⁴H₂₂

Q2. Identify the substance oxidized, reduced, oxidizing agent, and reducing agent in: Fe₂O₃ + 3CO → 2Fe + 3CO₂

• Fe: +3 → 0 — reduced; Fe₂O₃ is the oxidizing agent
• C: +2 → +4 — oxidized; CO is the reducing agent

Q3. Balance the following by oxidation number method: P₄ + OH⁻ → PH₃ + H₂PO₂⁻

Step 1: Assign oxidation numbers: P₄ (ON = 0) → PH₃ (P = −3, reduction by 3 per P) and H₂PO₂⁻ (P = +1, oxidation by 1 per P)

Step 2: Balance electron transfer. Let x P atoms be oxidized and y P atoms be reduced: x(1) = y(3) → x = 3y. Minimum: x = 3, y = 1. Total P atoms = 4 (matches P₄).

Step 3: P₄ + OH⁻ → PH₃ + 3H₂PO₂⁻

Step 4: Balance O: P₄ + 6OH⁻ → PH₃ + 3H₂PO₂⁻ (6 O on both sides)

Step 5: Balance H: P₄ + 6OH⁻ → PH₃ + 3H₂PO₂⁻

Left: 6H (from OH⁻); Right: 3H (from PH₃) + 6H (from 3H₂PO₂⁻) = 9H Add 3H₂O to left: P₄ + 3H₂O + 6OH⁻ → PH₃ + 3H₂PO₂⁻

Check: P: 4=4; O: 3+6=9, 3(2)=6 → need adjustment.

Balanced: P₄ + 3OH⁻ + 3H₂O → PH₃ + 3H₂PO₂⁻

Q4. Balance by half-reaction method (acidic medium): SO₂ + Cr₂O₇²⁻ → Cr³⁺ + SO₄²⁻

Oxidation half-reaction: SO₂ → SO₄²⁻ Balance O: SO₂ + 2H₂O → SO₄²⁻ + 4H⁺ Balance charge: SO₂ + 2H₂O → SO₄²⁻ + 4H⁺ + 2e⁻ ... (i)

Reduction half-reaction: Cr₂O₇²⁻ → 2Cr³⁺ Balance O: Cr₂O₇²⁻ → 2Cr³⁺ + 7H₂O Balance H: Cr₂O₇²⁻ + 14H⁺ → 2Cr³⁺ + 7H₂O Balance charge: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O ... (ii)

Multiply (i) by 3: 3SO₂ + 6H₂O → 3SO₄²⁻ + 12H⁺ + 6e⁻

Add with (ii): 3SO₂ + Cr₂O₇²⁻ + 6H₂O + 14H⁺ → 3SO₄²⁻ + 2Cr³⁺ + 7H₂O + 12H⁺

Simplify H⁺ and H₂O: 3SO₂ + Cr₂O₇²⁻ + 2H⁺ → 3SO₄²⁻ + 2Cr³⁺ + H₂O

Q5. Balance in basic medium: Cr(OH)₃ + IO₃⁻ → CrO₄²⁻ + I⁻

Oxidation half-reaction: Cr(OH)₃ → CrO₄²⁻ Balance O (basic): Cr(OH)₃ + 5OH⁻ → CrO₄²⁻ + 4H₂O Balance charge: Cr(OH)₃ + 5OH⁻ → CrO₄²⁻ + 4H₂O + 3e⁻ ... (i)

Reduction half-reaction: IO₃⁻ → I⁻ IO₃⁻ + 3H₂O → I⁻ + 6OH⁻ Balance charge: IO₃⁻ + 3H₂O + 6e⁻ → I⁻ + 6OH⁻ ... (ii)

Multiply (i) by 2: 2Cr(OH)₃ + 10OH⁻ → 2CrO₄²⁻ + 8H₂O + 6e⁻

Add with (ii): 2Cr(OH)₃ + 10OH⁻ + IO₃⁻ + 3H₂O → 2CrO₄²⁻ + 8H₂O + I⁻ + 6OH⁻

Simplify: 2Cr(OH)₃ + IO₃⁻ + 4OH⁻ → 2CrO₄²⁻ + I⁻ + 5H₂O

Q6. Identify the species undergoing disproportionation: (a) MnO₄⁻ → Mn²⁺ + MnO₄²⁻ (b) Cl₂ → Cl⁻ + ClO⁻

(a) Mn in MnO₄⁻ has ON = +7. Mn²⁺ has ON = +2 (reduction). MnO₄²⁻ has Mn = +6 (also reduction, from +7 to +6). Both changes are reductions — this is not disproportionation.

(b) Cl₂ has ON = 0. Cl⁻ has ON = −1 (reduction). ClO⁻ has Cl = +1 (oxidation). Same element (Cl) is both oxidized and reduced — this is disproportionation.

Q7. Which of the following are redox reactions? (a) CuSO₄ + Zn → ZnSO₄ + Cu (b) AgNO₃ + NaCl → AgCl + NaNO₃ (c) 2H₂ + O₂ → 2H₂O

(a) Redox reaction. Zn: 0 → +2 (oxidized); Cu: +2 → 0 (reduced).

(b) Not a redox reaction. This is a double displacement (precipitation) reaction. No change in oxidation numbers. Ag: +1 throughout; Na: +1 throughout; Cl: −1 throughout; NO₃: −1 throughout.

(c) Redox reaction. H: 0 → +1 (oxidized); O: 0 → −2 (reduced).

Q8. Calculate the oxidation number of S in: (a) H₂SO₄ (b) SO₂ (c) Na₂S₂O₃ (d) H₂S

(a) H₂SO₄: 2(+1) + S + 4(−2) = 0 → S = +6

(b) SO₂: S + 2(−2) = 0 → S = +4

(c) Na₂S₂O₃: 2(+1) + 2S + 3(−2) = 0 → 2S = 4 → average S = +2 (Note: one S is +6 and the other is −2 in thiosulfate)

(d) H₂S: 2(+1) + S = 0 → S = −2

Q9. Why do the following reactions proceed differently? P₄ + 3NaOH + 3H₂O → PH₃ + 3NaH₂PO₂ and P₄ + 5O₂ → 2P₂O₅

Q10. Balance the following redox reaction by the oxidation number method: Cu + HNO₃(dilute) → Cu(NO₃)₂ + NO + H₂O

Oxidation numbers: Cu: 0 → +2 (loss of 2e⁻); N: +5 → +2 (gain of 3e⁻)

Balance electron transfer: LCM of 2 and 3 = 6. So, 3Cu atoms lose 6e⁻ and 2N atoms gain 6e⁻.

3Cu + 2HNO₃ → 3Cu(NO₃)₂ + 2NO + H₂O

But Cu(NO₃)₂ needs 6 NO₃⁻ for 3 Cu, plus 2 NO₃⁻ go to NO. Total HNO₃ = 8.

3Cu + 8HNO₃ → 3Cu(NO₃)₂ + 2NO + 4H₂O

Check: Cu: 3=3; N: 8=6+2; O: 24=18+2+4=24; H: 8=8. Balanced.

Q11. Balance by half-reaction method (acidic): H₂O₂ + Fe²⁺ → Fe³⁺ + H₂O

Oxidation: Fe²⁺ → Fe³⁺ + e⁻ ... (i)

Reduction: H₂O₂ + 2H⁺ + 2e⁻ → 2H₂O ... (ii)

Multiply (i) by 2: 2Fe²⁺ → 2Fe³⁺ + 2e⁻

Add: H₂O₂ + 2Fe²⁺ + 2H⁺ → 2Fe³⁺ + 2H₂O

Example: Ag²⁺ + Ag⁰ → 2Ag⁺ (Ag with ON +2 and Ag with ON 0 give Ag with ON +1)

Q15. Balance the following in basic medium: Al + NO₃⁻ → Al(OH)₄⁻ + NH₃

Oxidation: Al → Al(OH)₄⁻ Balance O and H (basic): Al + 4OH⁻ → Al(OH)₄⁻ + 3e⁻ ... (i)

Reduction: NO₃⁻ → NH₃ NO₃⁻ + 6H₂O + 8e⁻ → NH₃ + 9OH⁻ ... (ii)

Multiply (i) by 8 and (ii) by 3: 8Al + 32OH⁻ → 8Al(OH)₄⁻ + 24e⁻ 3NO₃⁻ + 18H₂O + 24e⁻ → 3NH₃ + 27OH⁻

Add: 8Al + 32OH⁻ + 3NO₃⁻ + 18H₂O → 8Al(OH)₄⁻ + 3NH₃ + 27OH⁻

Simplify OH⁻: 8Al + 3NO₃⁻ + 5OH⁻ + 18H₂O → 8Al(OH)₄⁻ + 3NH₃

Q16. Identify the reducing agent in the reaction: 2Al(s) + Fe₂O₃(s) → Al₂O₃(s) + 2Fe(s)

(a) A reducing agent donates electrons to another species (causes reduction). To donate electrons, it must itself lose electrons (be oxidized). Thus, a good reducing agent is one that is easily oxidized — it readily gives up electrons.

(b) A strong oxidizing agent readily accepts electrons (gets reduced). Species with high reduction potentials have a strong tendency to gain electrons. Therefore, a strong oxidizing agent has a high reduction potential (e.g., F₂, MnO₄⁻, Cr₂O₇²⁻).

Internal Links

• [Chapter 8 Guide — Redox Reactions](/school/cbse/class/11/chemistry/cbse-redox-reactions)
• [All Class 11 Chemistry Solutions](/school/cbse/class/11/chemistry/solutions)
• [Class 11 Chemistry Subject Guide](/school/cbse/class/11/chemistry)