Class 11 Chemistry Notes Chapter 8 (Chapter 8) – Examplar Problems (English) Book

Detailed Notes with MCQs of Chapter 8: Redox Reactions from your NCERT Exemplar. This chapter is fundamental not just for your Class 11 understanding but also forms the basis for many concepts in electrochemistry and other areas crucial for competitive government exams. Pay close attention to the definitions, rules, and methods we discuss.

Chapter 8: Redox Reactions - Detailed Notes for Government Exam Preparation

1. Introduction to Redox Reactions:

• Redox is a shorthand for Reduction-Oxidation.
• These reactions involve the transfer of electrons between chemical species.
• They are ubiquitous in nature (respiration, photosynthesis, combustion, corrosion) and industry (batteries, metallurgy, chemical synthesis).

2. Classical Concept of Oxidation and Reduction:

• Oxidation:
  • Addition of oxygen or an electronegative element.
  • Removal of hydrogen or an electropositive element.
  • Example: 2Mg(s) + O₂(g) → 2MgO(s) (Mg is oxidized)
• Reduction:
  • Removal of oxygen or an electronegative element.
  • Addition of hydrogen or an electropositive element.
  • Example: CuO(s) + H₂(g) → Cu(s) + H₂O(l) (CuO is reduced)

3. Electronic Concept of Oxidation and Reduction (More Fundamental):

• Oxidation: Loss of electron(s) by a species. Results in an increase in oxidation number.
  • Example: Na → Na⁺ + e⁻
• Reduction: Gain of electron(s) by a species. Results in a decrease in oxidation number.
  • Example: Cl₂ + 2e⁻ → 2Cl⁻
• Redox Reaction: A reaction where both oxidation and reduction occur simultaneously. Electrons lost by one species are gained by another.
  • Example: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)
    • Zn → Zn²⁺ + 2e⁻ (Oxidation)
    • Cu²⁺ + 2e⁻ → Cu (Reduction)

4. Oxidizing and Reducing Agents:

• Oxidizing Agent (Oxidant): The species that accepts electrons and gets reduced itself. It causes the oxidation of another species. (Examples: O₂, Cl₂, MnO₄⁻, Cr₂O₇²⁻)
• Reducing Agent (Reductant): The species that donates electrons and gets oxidized itself. It causes the reduction of another species. (Examples: H₂, Na, Fe²⁺, C)

5. Oxidation Number (or Oxidation State):

• It represents the hypothetical charge an atom would have if all bonds to atoms of different elements were 100% ionic.

• It helps track electron shifts in redox reactions.

• Rules for Assigning Oxidation Number (O.N.):

  • Rule 1: O.N. of an atom in its elemental state (e.g., Na, O₂, P₄, S₈) is zero.
  • Rule 2: O.N. of a monoatomic ion is equal to its charge (e.g., Na⁺ = +1, Cl⁻ = -1, Mg²⁺ = +2).
  • Rule 3: O.N. of Oxygen is usually -2.
    • Exceptions: Peroxides (e.g., H₂O₂, Na₂O₂) where O.N. is -1. Superoxides (e.g., KO₂) where O.N. is -1/2. Oxygen fluorides (e.g., OF₂) where O.N. is +2 (as F is more electronegative).
  • Rule 4: O.N. of Hydrogen is usually +1 when bonded to non-metals and -1 when bonded to metals (metal hydrides, e.g., NaH, CaH₂).
  • Rule 5: O.N. of Fluorine (most electronegative element) is always -1 in its compounds. Other halogens (Cl, Br, I) usually have O.N. = -1, except when bonded to oxygen or a more electronegative halogen (e.g., in ClO₄⁻, O.N. of Cl is +7).
  • Rule 6: The algebraic sum of O.N. of all atoms in a neutral molecule is zero.
  • Rule 7: The algebraic sum of O.N. of all atoms in a polyatomic ion equals the charge on the ion.
  • Rule 8: In compounds, Group 1 metals always have O.N. = +1, Group 2 metals always have O.N. = +2.

• Stock Notation: Representing the O.N. of a metal in a compound using Roman numerals in parentheses after the metal's symbol or name (e.g., Aurous chloride = Au(I)Cl, Stannic chloride = Sn(IV)Cl₄).

6. Types of Redox Reactions:

• a) Combination Reactions: Two or more substances combine to form a single product. Can be redox if at least one element changes O.N.
  • C(s) [0] + O₂(g) [0] → CO₂(g) [+4, -2] (Redox)
  • CaO(s) [+2, -2] + CO₂(g) [+4, -2] → CaCO₃(s) [+2, +4, -2] (Not Redox)
• b) Decomposition Reactions: A compound breaks down into two or more simpler substances. Can be redox if elements change O.N.
  • 2H₂O(l) [+1, -2] → 2H₂(g) [0] + O₂(g) [0] (Redox)
  • CaCO₃(s) [+2, +4, -2] → CaO(s) [+2, -2] + CO₂(g) [+4, -2] (Not Redox)
• c) Displacement Reactions: An ion (or atom) in a compound is replaced by an ion (or atom) of another element. Always redox.
  • i) Metal Displacement: A more reactive metal displaces a less reactive metal from its salt solution.
    • Zn(s) [0] + CuSO₄(aq) [+2] → ZnSO₄(aq) [+2] + Cu(s) [0]
  • ii) Non-metal Displacement: Usually hydrogen displacement or halogen displacement.
    • Zn(s) [0] + 2HCl(aq) [+1] → ZnCl₂(aq) [+2] + H₂(g) [0] (Hydrogen displacement)
    • Cl₂(g) [0] + 2KBr(aq) [-1] → 2KCl(aq) [-1] + Br₂(aq) [0] (Halogen displacement)
• d) Disproportionation Reactions: A reaction in which the same element in a particular oxidation state is simultaneously oxidized and reduced. The element must exist in at least three oxidation states.
  • P₄(s) [0] + 3OH⁻(aq) + 3H₂O(l) → PH₃(g) [-3] + 3H₂PO₂⁻(aq) [+1] (P is reduced to -3 and oxidized to +1)
  • 2H₂O₂(aq) [-1] → 2H₂O(l) [-2] + O₂(g) [0] (O is reduced to -2 and oxidized to 0)
• e) Comproportionation Reactions: (Reverse of disproportionation) A reaction where two species containing the same element in different oxidation states react to form a product where the element is in an intermediate oxidation state.
  • Ag²⁺(aq) [+2] + Ag(s) [0] → 2Ag⁺(aq) [+1]

7. Balancing Redox Reactions:

Essential skill for stoichiometry involving redox reactions. Two main methods:

• a) Oxidation Number Method:

  • Step 1: Write the skeletal equation.
  • Step 2: Assign O.N. to all atoms and identify atoms undergoing change in O.N.
  • Step 3: Calculate the total increase and decrease in O.N. per formula unit/ion.
  • Step 4: Equalize the total increase and decrease in O.N. by multiplying the species involved with suitable integers.
  • Step 5: Balance atoms other than O and H.
  • Step 6: Balance O and H.
    • Acidic Medium: Balance O by adding H₂O, then balance H by adding H⁺.
    • Basic Medium: Balance O by adding H₂O. Balance H by adding H₂O to the side deficient in H and adding an equal number of OH⁻ ions to the opposite side. (Alternatively, balance as if in acidic medium, then add OH⁻ ions equal to the number of H⁺ ions to both sides; combine H⁺ and OH⁻ to form H₂O).
  • Step 7: Verify the balanced equation (atoms and charges).

• b) Half-Reaction (Ion-Electron) Method:

  • Step 1: Write the skeletal ionic equation.
  • Step 2: Split the reaction into two half-reactions: oxidation and reduction.
  • Step 3: Balance atoms other than O and H in each half-reaction individually.
  • Step 4: Balance O and H.
    • Acidic Medium: Balance O by adding H₂O, then balance H by adding H⁺.
    • Basic Medium: Balance O by adding H₂O, then balance H by adding OH⁻ (add H₂O to the side needing H, add OH⁻ to the other side).
  • Step 5: Balance the charge in each half-reaction by adding electrons (e⁻) to the appropriate side.
  • Step 6: Equalize the number of electrons in both half-reactions by multiplying them with suitable integers.
  • Step 7: Add the two balanced half-reactions; cancel electrons and any common species on opposite sides.
  • Step 8: Verify the balanced equation (atoms and charges).

8. Redox Reactions and Electrode Processes:

• Electrochemical Cell (Galvanic/Voltaic Cell): A device that converts chemical energy from a spontaneous redox reaction into electrical energy. (e.g., Daniell cell: Zn|Zn²⁺||Cu²⁺|Cu)
  • Oxidation occurs at the Anode (-ve terminal in galvanic cell).
  • Reduction occurs at the Cathode (+ve terminal in galvanic cell).
• Electrode Potential: The potential difference developed between an electrode and its electrolyte. It's a measure of the tendency of an electrode to lose or gain electrons.
• Standard Electrode Potential (E°): Electrode potential measured under standard conditions (298 K, 1 M concentration for solutions, 1 bar pressure for gases). Measured relative to the Standard Hydrogen Electrode (SHE), whose E° is defined as 0.00 V.
• Electrochemical Series: Arrangement of elements (or half-reactions) in order of their increasing standard reduction potentials (E°).
  • Applications:
    • Comparing relative oxidizing/reducing power (Higher E° = stronger oxidant, Lower E° = stronger reductant).
    • Predicting spontaneity of redox reactions (E°cell = E°cathode - E°anode; If E°cell > 0, reaction is spontaneous).
    • Predicting whether a metal can displace H₂ from acid (Metals with negative E° can).

9. Redox Titrations:

• Titrations based on redox reactions. Used to determine the concentration of an analyte.
• Often use indicators that change color at the endpoint, or sometimes one of the reactants/products acts as a self-indicator (e.g., KMnO₄).

Multiple Choice Questions (MCQs):

1. The oxidation number of Cr in K₂Cr₂O₇ is:
   a) +3
   b) +6
   c) +7
   d) +4

2. Which of the following reactions is an example of a disproportionation reaction?
   a) Zn + CuSO₄ → ZnSO₄ + Cu
   b) 2KClO₃ → 2KCl + 3O₂
   c) 3Cl₂ + 6NaOH → 5NaCl + NaClO₃ + 3H₂O
   d) NH₄NO₂ → N₂ + 2H₂O

3. In the reaction: MnO₂ + 4HCl → MnCl₂ + Cl₂ + 2H₂O, the reducing agent is:
   a) MnO₂
   b) HCl
   c) MnCl₂
   d) Cl₂

4. What is the change in the oxidation number of carbon in the following reaction?
   CH₄(g) + 4Cl₂(g) → CCl₄(l) + 4HCl(g)
   a) 0 to +4
   b) -4 to +4
   c) +4 to -4
   d) 0 to -4

5. When Cl₂ gas reacts with hot and concentrated sodium hydroxide solution, the oxidation number of chlorine changes from:
   a) Zero to +1 and Zero to -1
   b) Zero to +5 and Zero to -1
   c) Zero to +3 and Zero to -1
   d) Zero to -1 and Zero to +7

6. The standard electrode potentials (E°) for Al³⁺/Al, Ag⁺/Ag, K⁺/K and Cr³⁺/Cr are -1.66 V, +0.80 V, -2.93 V and -0.74 V, respectively. The correct decreasing order of reducing power is:
   a) Ag > Cr > Al > K
   b) K > Al > Cr > Ag
   c) K > Al > Ag > Cr
   d) Al > K > Ag > Cr

7. Which of the following is NOT a redox reaction?
   a) CaCO₃ → CaO + CO₂
   b) 2H₂ + O₂ → 2H₂O
   c) Na + H₂O → NaOH + ½ H₂
   d) CuSO₄ + Zn → ZnSO₄ + Cu

8. The oxidation state of sulfur in the anions SO₃²⁻, S₂O₄²⁻, and S₂O₆²⁻ follows the order:
   a) S₂O₄²⁻ < SO₃²⁻ < S₂O₆²⁻
   b) SO₃²⁻ < S₂O₄²⁻ < S₂O₆²⁻
   c) S₂O₄²⁻ < S₂O₆²⁻ < SO₃²⁻
   d) S₂O₆²⁻ < S₂O₄²⁻ < SO₃²⁻

9. In balancing the reaction Cr₂O₇²⁻ + SO₃²⁻ → Cr³⁺ + SO₄²⁻ in acidic medium, the coefficient of H⁺ is:
   a) 4
   b) 6
   c) 8
   d) 10

10. Which substance serves as the reducing agent in the following reaction?
    14H⁺ + Cr₂O₇²⁻ + 3Ni → 2Cr³⁺ + 7H₂O + 3Ni²⁺
    a) H⁺
    b) Cr₂O₇²⁻
    c) Ni
    d) H₂O

Answer Key for MCQs:

1. b (+6)
2. c (Cl goes from 0 to -1 and +5)
3. b (HCl; Cl goes from -1 to 0, getting oxidized)
4. b (-4 in CH₄ to +4 in CCl₄)
5. b (Zero in Cl₂ to -1 in NaCl and +5 in NaClO₃)
6. b (Lower E° means stronger reducing power: K < Al < Cr < Ag)
7. a (No change in O.N. for Ca, C, O)
8. a (O.N. of S: S₂O₄²⁻ = +3, SO₃²⁻ = +4, S₂O₆²⁻ = +5)
9. c (Balanced equation: Cr₂O₇²⁻ + 3SO₃²⁻ + 8H⁺ → 2Cr³⁺ + 3SO₄²⁻ + 4H₂O)
10. c (Ni goes from 0 to +2, getting oxidized)

Make sure you practice assigning oxidation numbers quickly and accurately, and master both methods of balancing redox equations. Understanding the electrochemical series is also vital for predicting reaction feasibility. Go through the Exemplar problems thoroughly, as they often highlight common pitfalls and important concepts.