Class 11 Chemistry Chapter 7 Redox Reactions NCERT Solution introduces students to one of the most fundamental concepts in chemistry. This chapter explains how oxidation and reduction occur together in chemical reactions, the use of oxidation numbers, types of redox reactions, and their real-life applications.

Class 11 Chemistry Chapter 7 Redox Reactions

Understanding redox reactions is essential for mastering electrochemistry, metallurgy, and many industrial processes. In this section, you’ll find solved NCERT questions and detailed explanations to help you strengthen your concepts and prepare effectively for exams.

Class 11 Chemistry Chapter 7 Redox Reactions Textbook

Problem 7.1

Identify the species undergoing oxidation and reduction:

(i) H₂S(g) + Cl₂(g) → 2HCl(g) + S(s)

  • H₂S is oxidised (hydrogen is removed from sulfur).
  • Cl₂ is reduced (hydrogen is added to chlorine).

(ii) 3Fe₃O₄(s) + 8Al(s) → 9Fe(s) + 4Al₂O₃(s)

  • Aluminium is oxidised (oxygen is added).
  • Fe₃O₄ is reduced (oxygen is removed).

(iii) 2Na(s) + H₂(g) → 2NaH(s)

  • Sodium is oxidised.
  • Hydrogen is reduced.

Problem 7.2

Justify that the reaction: 2Na(s) + H₂(g) → 2NaH(s) is a redox reaction.

Solution:

  • Na(s) → Na⁺ + e⁻ (oxidation)
  • H₂ + 2e⁻ → 2H⁻ (reduction)
  • So, Na is oxidised and H₂ is reduced.
    Hence, the overall reaction is a redox reaction.

Problem 7.3

Using Stock notation, represent the following compounds:

  • HAuCl₄ → Au(III)
  • Tl₂O → Tl(I)
  • FeO → Fe(II)
  • Fe₂O₃ → Fe(III)
  • CuI → Cu(I)
  • CuO → Cu(II)
  • MnO → Mn(II)
  • MnO₂ → Mn(IV)

Final Stock Notation:
HAu(III)Cl₄, Tl₂(I)O, Fe(II)O, Fe₂(III)O₃, Cu(I)I, Cu(II)O, Mn(II)O, Mn(IV)O₂

Problem 7.4

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

Solution:

  • Cu is reduced from +1 to 0 (in both Cu₂O and Cu₂S).
  • S is oxidised from –2 to +4.
  • So, this is a redox reaction.
  • Oxidant: Cu(I) in Cu₂O
  • Reductant: S in Cu₂S

Problem 7.5

Which of the following does not show disproportionation and why?
ClO⁻, ClO₂⁻, ClO₃⁻, ClO₄⁻

Solution:

  • ClO₄⁻ does not show disproportionation because Cl is in its maximum oxidation state (+7).
    Disproportionation reactions:
  1. 3ClO⁻ → 2Cl⁻ + ClO₃⁻
  2. 6ClO₂⁻ → 4ClO₃⁻ + 2Cl⁻
  3. 4ClO₃⁻ → Cl⁻ + 3ClO₄⁻

Problem 7.6

Classify the following reactions:

(a) N₂ + O₂ → 2NO → Combination redox reaction
(b) 2Pb(NO₃)₂ → 2PbO + 4NO₂ + O₂ → Decomposition redox reaction
(c) NaH + H₂O → NaOH + H₂ → Displacement redox reaction
(d) 2NO₂ + 2OH⁻ → NO₂⁻ + NO₃⁻ + H₂O → Disproportionation redox reaction

Problem 7.7

Why do these reactions proceed differently?

  • Pb₃O₄ + 8HCl → 3PbCl₂ + Cl₂ + 4H₂O
    • PbO is basic, reacts with HCl (acid–base).
    • PbO₂ acts as oxidising agent, oxidises Cl⁻ to Cl₂.
  • Pb₃O₄ + 4HNO₃ → 2Pb(NO₃)₂ + PbO₂ + 2H₂O
    • HNO₃ is itself an oxidising agent, so no redox occurs with PbO₂.
    • Only acid–base reaction between PbO and HNO₃ happens.

Problem 7.8

Write the net ionic equation for the reaction of K₂Cr₂O₇ with Na₂SO₃ in acid solution:

Balanced equation:
Cr₂O₇²⁻ + 3SO₃²⁻ + 8H⁺ → 2Cr³⁺ + 3SO₄²⁻ + 4H₂O

Problem 7.9

Permanganate reacts with bromide in basic medium to form MnO₂ and BrO₃⁻.

Balanced ionic equation:
2MnO₄⁻ + Br⁻ + H₂O → 2MnO₂ + BrO₃⁻ + 2OH⁻

Problem 7.10

MnO₄⁻ oxidises I⁻ to I₂ in basic solution, forming MnO₂.

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

You can access the official NCERT Solutions for Class 10 Mathematics on the NCERT website at the following link:

NCERT Class 10 Mathematics Solutions

📘 Math & Science Solutions by Class

🔹 Class 10

🔹 Class 9

🔹 Class 8

🔹 Class 7

🔹 Class 6

🔹 Class 12

🔹 Class 11

Learn more about circle formulas from NCERT.

Class-wise Solutions

Class 12:

Class 12 Physics – NCERT Solutions

Class 12 Chemistry – NCERT Solutions

Class 11:

Class 10:

Class 9:

Class 8:

Class 7:

Class 6:

Subject-wise Solutions

Physics:

Chemistry:

Biology:

Math:

Science:

NEET BIOLOGY

Understanding the concepts of Class 11 Chemistry Chapter 7 Redox Reactions is essential for mastering chemical processes involving oxidation and reduction. The detailed solutions provided here for Class 11 Chemistry Chapter 7 Redox Reactions will help students grasp the logic behind electron transfer, oxidation number method, and balancing redox equations. By practicing regularly with these answers, students can strengthen their foundation in redox chemistry.

Whether for revision or board exam preparation, Class 11 Chemistry Chapter 7 Redox Reactions solutions serve as a reliable guide. Keep exploring and revising Class 11 Chemistry Chapter 7 Redox Reactions to enhance your problem-solving skills and conceptual clarity.