Redox Reactions Notes: Class 11 Chemistry Chapter 7

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
2. Oxidation Number Concept & Stock Notation
3. Types of Redox Reactions & Disproportionation
4. Balancing Redox Reactions & Electrochemical Cells

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) + Cr₂O₇^2-(aq) → Fe³⁺(aq) + Cr³⁺(aq).

Answer:
Step 1: Separate into half-reactions:
• Oxidation half-reaction: Fe²⁺ → Fe³⁺
• Reduction half-reaction: Cr₂O₇^2- → Cr³⁺

Step 2: Balance atoms other than O and H:
• Oxidation: Fe²⁺ → Fe³⁺
• Reduction: Cr₂O₇^2- → 2Cr³⁺

Step 3: Balance O by adding H₂O and H by adding H⁺ (acidic medium):
• Reduction: Cr₂O₇^2- + 14H⁺ → 2Cr³⁺ + 7H₂O

Step 4: Balance charges by adding electrons (e^-):
• Oxidation: Fe²⁺ → Fe³⁺ + e^- (Multiply by 6)
• Reduction: Cr₂O₇^2- + 14H⁺ + 6e^- → 2Cr³⁺ + 7H₂O

Step 5: Add both half-reactions to cancel electrons:
6Fe²⁺ + Cr₂O₇^2- + 14H⁺ + 6e^- → 6Fe³⁺ + 2Cr³⁺ + 7H₂O + 6e^-

Final Balanced Ionic Equation:
6Fe²⁺(aq) + Cr₂O₇^2-(aq) + 14H⁺(aq) → 6Fe³⁺(aq) + 2Cr³⁺(aq) + 7H₂O(l)