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NCERT Solutions for Class 11 Chemistry Chapter 7 Redox Reactions

NCERT Solutions for Class 11 Chemistry Chapter 7 Redox Reactions covers classical and electron-transfer concepts of oxidation and reduction, oxidation numbers and rules for calculation, types of redox reactions (combination, decomposition, displacement, disproportionation), balancing redox reactions (oxidation number method and ion-electron/half-reaction method), and redox reactions in electrochemical cells.

Overview of Topics & Subtopics (Click to Jump)

1. Classical & Electron Transfer Concepts of Redox Reactions

Short Note:

• Classical Concept:

  - Oxidation: Addition of oxygen/electronegative element or removal of hydrogen/electropositive element.

  - Reduction: Addition of hydrogen/electropositive element or removal of oxygen/electronegative element.

• Electron Transfer Concept:

  - Oxidation: Loss of electron(s) by a chemical species (Increase in oxidation state).

  - Reduction: Gain of electron(s) by a chemical species (Decrease in oxidation state).

  - Oxidising Agent (Oxidant): Species that accepts electrons and gets reduced.

  - Reducing Agent (Reductant): Species that donates electrons and gets oxidised.

2. Oxidation Number Concept & Stock Notation

Short Note:

• Oxidation Number (O.N.): The residual charge an atom would carry in a compound if all surrounding bonds were completely ionic.

• Rules for Assigning Oxidation Number:

  - Free element / uncombined state = 0 (e.g., O2, H2, Na, S8).

  - Monatomic ions = charge on the ion (e.g., Na+ = +1, Cl- = -1, Al3+ = +3).

  - Fluorine = -1 in all compounds.

  - Oxygen = -2 generally (except -1 in peroxides like H2O2, +2 in OF2).

  - Hydrogen = +1 with non-metals, -1 with metals (metal hydrides like NaH).

  - Sum of oxidation numbers in a neutral molecule = 0; in a polyatomic ion = total charge on the ion.

• Stock Notation: Indicating the oxidation state of a metal in Roman numerals within parentheses, e.g., Gold(III) chloride for AuCl3, Iron(II) oxide for FeO.

3. Types of Redox Reactions & Disproportionation

Short Note:

• Combination Reactions: A + B → C (where at least one reactant is an element), e.g., C(s) + O2(g) → CO2(g).

• Decomposition Reactions: Breakdown of a compound into simpler substances, e.g., 2KClO3(s) → 2KCl(s) + 3O2(g).

• Displacement Reactions: X + YZ → XZ + Y.

  - Metal displacement: CuSO4 + Zn → ZnSO4 + Cu.

  - Non-metal displacement: 2Na + 2H2O → 2NaOH + H2.

• Disproportionation Reactions: Reactions in which the same element simultaneously undergoes oxidation and reduction, e.g., 2H2O2(aq) → 2H2O(l) + O2(g) (O changes from -1 to -2 and 0).

4. Balancing Redox Reactions & Electrochemical Cells

Short Note:

• Methods of Balancing Redox Equations:

  - Oxidation Number Method: Equating total increase in oxidation number with total decrease in oxidation number.

  - Ion-Electron (Half-Reaction) Method: Splitting the overall equation into oxidation and reduction half-reactions, balancing atoms and charges separately, and combining them.

• Redox Titrations: Using self-indicating reagents like KMnO4 or K2Cr2O7 for quantitatively determining reductants.

• Electrochemical Cells: Device converting chemical energy into electrical energy through redox reactions (e.g., Daniell Cell: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)). Anode = Oxidation (-), Cathode = Reduction (+).

Intext & Conceptual Questions

Q1: Assign oxidation numbers to the underlined elements in the following species: (a) KMnO4, (b) K2Cr2O7, (c) H2SO4.
Answer:
(a) KMnO4 (Manganese):
Let oxidation number of Mn = x.
(+1) + x + 4(-2) = 0
1 + x - 8 = 0 ⇒ x = +7.

(b) K2Cr2O7 (Chromium):
Let oxidation number of Cr = x.
2(+1) + 2x + 7(-2) = 0
2 + 2x - 14 = 0 ⇒ 2x = 12 ⇒ x = +6.

(c) H2SO4 (Sulfur):
Let oxidation number of S = x.
2(+1) + x + 4(-2) = 0
2 + x - 8 = 0 ⇒ x = +6.

Textbook Chapter End Exercise Questions

Q1: Justify that the following reaction is a redox reaction: CuO(s) + H2(g) → Cu(s) + H2O(g). Identify the species oxidised, reduced, oxidising agent, and reducing agent.
Answer:
Assigning oxidation numbers to each element in the reaction:
• In CuO: Cu = +2, O = -2
• In H2: H = 0
• In Cu: Cu = 0
• In H2O: H = +1, O = -2

Analysis:
• Oxidation Number of Cu decreases from +2 (in CuO) to 0 (in Cu) → Reduction.
• Oxidation Number of H increases from 0 (in H2) to +1 (in H2O) → Oxidation.

Since both oxidation and reduction occur simultaneously, it is a redox reaction.

• Species Oxidised: H2
• Species Reduced: CuO
• Oxidising Agent: CuO
• Reducing Agent: H2
Q2: Balance the following ionic redox equation in acidic medium using the ion-electron method: Fe2+(aq) + Cr2O72-(aq) → Fe3+(aq) + Cr3+(aq).
Answer:
Step 1: Separate into half-reactions:
• Oxidation half-reaction: Fe2+ → Fe3+
• Reduction half-reaction: Cr2O72- → Cr3+

Step 2: Balance atoms other than O and H:
• Oxidation: Fe2+ → Fe3+
• Reduction: Cr2O72- → 2Cr3+

Step 3: Balance O by adding H2O and H by adding H+ (acidic medium):
• Reduction: Cr2O72- + 14H+ → 2Cr3+ + 7H2O

Step 4: Balance charges by adding electrons (e-):
• Oxidation: Fe2+ → Fe3+ + e-   (Multiply by 6)
• Reduction: Cr2O72- + 14H+ + 6e- → 2Cr3+ + 7H2O

Step 5: Add both half-reactions to cancel electrons:
6Fe2+ + Cr2O72- + 14H+ + 6e- → 6Fe3+ + 2Cr3+ + 7H2O + 6e-

Final Balanced Ionic Equation:
6Fe2+(aq) + Cr2O72-(aq) + 14H+(aq) → 6Fe3+(aq) + 2Cr3+(aq) + 7H2O(l)
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Class 11 › Chemistry › Redox Reactions
Chapter notesRead the chapterPractice MCQsCheck what you rememberChapter test10 questions on one pageOnline testTest 1, 2 and 3, one question at a timePYQPrevious year questionsStudy materialPDF notes
Topics in this chapter: Classical & Electron Transfer Concepts of Redox ReactionsOxidation Number Concept & Stock NotationTypes of Redox Reactions & DisproportionationBalancing Redox Reactions & Electrochemical Cells