Class 11 Notes

Class 11 Chemistry Notes Chapter 1 (Redox Reactions) – Chemistry Part-II Book

Philoid

Philoid

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Chemistry Part-II
Alright students, let's dive into a crucial chapter for your competitive exams: Redox Reactions. Understanding this chapter is fundamental not just for direct questions but also for grasping concepts in electrochemistry and other related topics. We will cover the core ideas systematically.

Chapter 1: Redox Reactions (NCERT Class 11, Chemistry Part-II)

1. Introduction to Redox Reactions

  • Redox is short for Reduction-Oxidation.
  • Redox reactions involve the transfer of electrons between chemical species.
  • One species loses electrons (Oxidation), and another species gains electrons (Reduction). These processes always occur simultaneously.

2. Classical Concept of Oxidation and Reduction

  • Oxidation:
    • Addition of oxygen (e.g., 2Mg + O₂ → 2MgO)
    • Addition of an electronegative element (e.g., Mg + Cl₂ → MgCl₂)
    • Removal of hydrogen (e.g., H₂S + Cl₂ → 2HCl + S)
    • Removal of an electropositive element (e.g., 2K₄[Fe(CN)₆] + H₂O₂ → 2K₃[Fe(CN)₆] + 2KOH)
  • Reduction:
    • Removal of oxygen (e.g., CuO + H₂ → Cu + H₂O)
    • Removal of an electronegative element (e.g., 2FeCl₃ + H₂ → 2FeCl₂ + 2HCl)
    • Addition of hydrogen (e.g., Cl₂ + H₂ → 2HCl)
    • Addition of an electropositive element (e.g., HgCl₂ + SnCl₂ → Hg₂Cl₂ + SnCl₄)
  • Limitation: This concept is not comprehensive as many reactions not involving O or H are also redox reactions.

3. Electronic Concept of Oxidation and Reduction (Modern Concept)

  • 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⁻
  • Oxidizing Agent (Oxidant): The species that accepts electron(s) and gets reduced itself.
  • Reducing Agent (Reductant): The species that donates electron(s) and gets oxidized itself.
  • Overall Redox Reaction: Example: 2Na + Cl₂ → 2Na⁺Cl⁻ (Here, Na is oxidized and acts as the reducing agent; Cl₂ is reduced and acts as the oxidizing agent).

4. Oxidation Number (O.N.) or Oxidation State (O.S.)

  • It is the charge (real or imaginary) assigned to an atom in a species according to a set of rules, assuming electron pairs in covalent bonds belong entirely to the more electronegative atom.

  • Rules for Assigning Oxidation Number:

    1. Free Elements: O.N. of an atom in its elemental state (e.g., Na, O₂, P₄, S₈) is zero.
    2. Monoatomic Ions: O.N. equals the charge on the ion (e.g., Na⁺ = +1, Cl⁻ = -1, Mg²⁺ = +2).
    3. Oxygen: Usually -2. Exceptions:
      • Peroxides (e.g., H₂O₂, Na₂O₂): O.N. is -1.
      • Superoxides (e.g., KO₂, RbO₂): O.N. is -1/2.
      • Compounds with Fluorine (e.g., OF₂): O.N. is +2 (as F is more electronegative). O₂F₂: O.N. is +1.
    4. Hydrogen: Usually +1 when bonded to non-metals (e.g., HCl, H₂O). O.N. is -1 when bonded to metals (metal hydrides, e.g., NaH, CaH₂).
    5. Fluorine: Always -1 in its compounds (most electronegative element).
    6. Other Halogens (Cl, Br, I): Usually -1, except when bonded to oxygen or a more electronegative halogen (e.g., in ClO₄⁻, Cl is +7; in ICl, Cl is -1, I is +1).
    7. Sum of O.N.s:
      • For a neutral molecule, the sum of O.N.s of all atoms is zero.
      • For a polyatomic ion, the sum of O.N.s of all atoms equals the charge on the ion.
    8. Alkali Metals (Group 1): Always +1 in their compounds.
    9. Alkaline Earth Metals (Group 2): Always +2 in their compounds.

  • Calculating O.N. (Examples):

    • KMnO₄: Let O.N. of Mn be x. (+1) + x + 4(-2) = 0 => x = +7
    • K₂Cr₂O₇: Let O.N. of Cr be y. 2(+1) + 2y + 7(-2) = 0 => 2y = +12 => y = +6
    • H₂SO₄: Let O.N. of S be z. 2(+1) + z + 4(-2) = 0 => z = +6
    • CrO₅ (Butterfly structure): Contains one O(-2) and four O(-1) peroxide linkages. Let O.N. of Cr be w. w + 1(-2) + 4(-1) = 0 => w = +6.
    • SO₄²⁻: Let O.N. of S be p. p + 4(-2) = -2 => p = +6

  • Redox Reactions in terms of O.N.:

    • Oxidation: Increase in Oxidation Number.
    • Reduction: Decrease in Oxidation Number.

5. Types of Redox Reactions

  • a) Combination Reactions: Two or more substances combine to form a single substance. Can be redox if at least one element changes O.N.
    • Example: C(s) [O.N.=0] + O₂(g) [O.N.=0] → CO₂(g) [C=+4, O=-2] (C oxidized, O reduced)
  • b) Decomposition Reactions: A compound breaks down into two or more simpler substances. Can be redox if elements change O.N.
    • Example: 2KClO₃(s) [K=+1, Cl=+5, O=-2] → 2KCl(s) [K=+1, Cl=-1] + 3O₂(g) [O=0] (Cl reduced, O oxidized)
  • c) Displacement Reactions: An ion (or atom) in a compound is replaced by an ion (or atom) of another element.
    • Metal Displacement: A more reactive metal displaces a less reactive metal from its salt solution.
      • Example: Zn(s) [0] + CuSO₄(aq) [Cu=+2] → ZnSO₄(aq) [Zn=+2] + Cu(s) [0] (Zn oxidized, Cu reduced)
    • Non-metal Displacement: Usually involves hydrogen displacement or halogen displacement.
      • Example (H displacement): Zn(s) [0] + 2HCl(aq) [H=+1] → ZnCl₂(aq) [Zn=+2] + H₂(g) [H=0] (Zn oxidized, H reduced)
      • Example (Halogen displacement): Cl₂(g) [0] + 2KBr(aq) [Br=-1] → 2KCl(aq) [Cl=-1] + Br₂(aq) [Br=0] (Br oxidized, Cl reduced)
  • 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.
    • Example 1: P₄(s) [0] + 3OH⁻(aq) + 3H₂O(l) → PH₃(g) [P=-3] + 3H₂PO₂⁻(aq) [P=+1] (P is reduced to -3 and oxidized to +1)
    • Example 2: 2H₂O₂(aq) [O=-1] → 2H₂O(l) [O=-2] + O₂(g) [O=0] (Oxygen is reduced to -2 and oxidized to 0)
    • Example 3: Cl₂(g) [0] + 2OH⁻(aq) → Cl⁻(aq) [-1] + ClO⁻(aq) [+1] + H₂O(l) (Chlorine is reduced to -1 and oxidized to +1)

6. Balancing Redox Reactions

  • Ensures that the number of atoms of each element and the net charge on both sides of the equation are equal.
  • Two Methods:
    • a) Oxidation Number Method:
      1. Write the skeletal equation.
      2. Assign O.N.s to all atoms and identify atoms undergoing change in O.N.
      3. Calculate the total increase and decrease in O.N.
      4. Multiply the species involved by suitable integers to equalize the total increase and decrease in O.N.
      5. Balance atoms other than O and H.
      6. Balance O atoms by adding H₂O molecules to the side deficient in O.
      7. Balance H atoms by adding H⁺ ions (in acidic medium) or OH⁻ ions (in basic medium - see specific steps below).
      8. Verify the equation for atom and charge balance.
      • For Basic Medium: After balancing H with H⁺ (as if in acidic medium), add as many OH⁻ ions to both sides of the equation as there are H⁺ ions. Combine H⁺ and OH⁻ on the same side to form H₂O and simplify.
    • b) Half-Reaction (Ion-Electron) Method:
      1. Write the skeletal ionic equation.
      2. Split the reaction into two half-reactions: Oxidation and Reduction.
      3. Balance atoms other than O and H in each half-reaction.
      4. Balance O atoms by adding H₂O.
      5. Balance H atoms by adding H⁺ (for acidic medium).
      6. Balance the charge by adding electrons (e⁻) to the appropriate side.
      7. Multiply the half-reactions by suitable integers so that the number of electrons lost in oxidation equals the number of electrons gained in reduction.
      8. Add the two balanced half-reactions and cancel out common species (including electrons).
      9. Verify the final equation for atom and charge balance.
      • For Basic Medium: Follow steps 1-4. For step 5, balance H by adding H₂O to the side deficient in H and an equal number of OH⁻ to the opposite side. Then proceed with steps 6-9. (Alternatively, balance as in acidic medium till step 8, then add OH⁻ to both sides to neutralize H⁺, forming H₂O).

7. Redox Reactions as the Basis for Titrations

  • Redox titrations are used to determine the concentration of an analyte (oxidizing or reducing agent) by titrating it against a standard solution of a reducing or oxidizing agent.
  • The endpoint is often detected using:
    • Self-Indicator: One of the reactants acts as an indicator (e.g., KMnO₄ is purple, turns colorless when reduced to Mn²⁺).
    • Redox Indicator: A substance that changes color when the potential of the solution reaches a specific value near the equivalence point.
  • Common Titrants: Potassium permanganate (KMnO₄), Potassium dichromate (K₂Cr₂O₇).
  • Stoichiometry based on balanced redox reactions is crucial for calculations.

8. Electrode Processes and Electrode Potential (Brief Overview)

  • Redox reactions are the basis of electrochemical cells (batteries, electrolysis).
  • When a metal strip is dipped in its salt solution, a potential difference develops at the interface - this is electrode potential.
  • Standard Electrode Potential (E°): Electrode potential measured under standard conditions (1 M concentration, 1 atm pressure for gases, 298 K). Measured relative to the Standard Hydrogen Electrode (SHE), whose potential is defined as 0.00 V.
  • A more positive E° indicates a greater tendency for the species to be reduced (stronger oxidizing agent).
  • A more negative E° indicates a greater tendency for the species to be oxidized (stronger reducing agent).
  • The Electrochemical Series lists elements/species in order of their standard electrode potentials.

Key Takeaways for Exams:

  • Master the rules for assigning Oxidation Numbers.
  • Be able to identify oxidizing and reducing agents in a reaction.
  • Recognize different types of redox reactions, especially disproportionation.
  • Practice balancing redox equations thoroughly using both methods (Oxidation Number and Half-Reaction) in acidic and basic media.
  • Understand the basic principle of redox titrations.

Multiple Choice Questions (MCQs)

  1. What is the oxidation number of Manganese (Mn) in KMnO₄?
    (a) +2
    (b) +4
    (c) +6
    (d) +7

  2. In the reaction: 2Na(s) + Cl₂(g) → 2NaCl(s), which species acts as the reducing agent?
    (a) Na
    (b) Cl₂
    (c) NaCl
    (d) Na⁺

  3. Which of the following reactions is an example of a disproportionation reaction?
    (a) Zn + CuSO₄ → ZnSO₄ + Cu
    (b) 2H₂O₂ → 2H₂O + O₂
    (c) C + O₂ → CO₂
    (d) NH₄NO₃ → N₂O + 2H₂O

  4. The oxidation number of Oxygen in H₂O₂ (Hydrogen Peroxide) is:
    (a) -2
    (b) 0
    (c) -1
    (d) +1

  5. Oxidation involves:
    (a) Gain of electrons
    (b) Decrease in oxidation number
    (c) Loss of electrons
    (d) Addition of hydrogen

  6. In the reaction: CuO + H₂ → Cu + H₂O, which species is oxidized?
    (a) CuO
    (b) H₂
    (c) Cu
    (d) H₂O

  7. When balancing a redox reaction in acidic medium using the half-reaction method, Oxygen atoms are balanced by adding:
    (a) OH⁻ ions
    (b) H₂O molecules
    (c) H⁺ ions
    (d) Electrons

  8. The oxidation state of Cr in K₂Cr₂O₇ is:
    (a) +3
    (b) +4
    (c) +6
    (d) +7

  9. Which of the following is the strongest oxidizing agent based on standard electrode potentials (values given)?
    (a) Zn²⁺ (E° = -0.76 V)
    (b) Cu²⁺ (E° = +0.34 V)
    (c) Ag⁺ (E° = +0.80 V)
    (d) F₂ (E° = +2.87 V)

  10. In the balanced equation: a Cr₂O₇²⁻ + b Fe²⁺ + c H⁺ → d Cr³⁺ + e Fe³⁺ + f H₂O, what is the coefficient 'b' (for Fe²⁺)?
    (a) 1
    (b) 3
    (c) 6
    (d) 14

Answer Key:

  1. (d) +7
  2. (a) Na (Sodium loses electrons, gets oxidized, hence is the reducing agent)
  3. (b) 2H₂O₂ → 2H₂O + O₂ (Oxygen in O₂⁻¹ is reduced to O⁻² and oxidized to O⁰)
  4. (c) -1 (It's a peroxide)
  5. (c) Loss of electrons (or increase in O.N.)
  6. (b) H₂ (Hydrogen's O.N. changes from 0 in H₂ to +1 in H₂O)
  7. (b) H₂O molecules
  8. (c) +6
  9. (d) F₂ (Highest positive E° value indicates the greatest tendency to get reduced)
  10. (c) 6 (The balanced equation is Cr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O)

Remember to practice balancing equations extensively, as it's a frequently tested skill. Good luck with your preparation!

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