Unit 6: Oxidation and Reduction

Chemistry – Class 11

This note delves into unit 6: oxidation and reduction, offering clear explanations and practical insights. It is designed to help students grasp core ideas through structured content. Whether preparing for exams or seeking conceptual clarity, this resource provides valuable support. Enhance your understanding with simplified notes and organized materials tailored to learners.

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Unit 6: Oxidation and Reduction (5 Teaching Hours)

1. General and Electronic Concept of Oxidation and Reduction

  • Oxidation: The process in which an atom, ion, or molecule loses electrons, resulting in an increase in oxidation state.
    • Example: Zn→Zn2++2e−\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-
  • Reduction: The process in which an atom, ion, or molecule gains electrons, resulting in a decrease in oxidation state.
    • Example: Cu2++2e−→Cu\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}
  • Redox Reaction: A reaction where oxidation and reduction occur simultaneously.
    • Example: Zn+Cu2+→Zn2++Cu\text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu}

2. Oxidation Number and Rules for Assigning Oxidation Number

  • Oxidation Number (Oxidation State): A number assigned to an element in a compound that represents the number of electrons lost or gained by the atom.
  • Rules for Assigning Oxidation Numbers:
    1. The oxidation number of an atom in its elemental form (e.g., O2\text{O}_2, N2\text{N}_2) is 0.
    2. For monoatomic ions, the oxidation number is equal to the charge on the ion (e.g., Na+=+1\text{Na}^+ = +1, Cl−=−1\text{Cl}^- = -1).
    3. Hydrogen has an oxidation number of +1 when bonded to non-metals and -1 when bonded to metals.
    4. Oxygen has an oxidation number of -2 in most compounds, except in peroxides (where it is -1) and when bonded to fluorine (where it is +2).
    5. The sum of oxidation numbers in a neutral compound is 0, and in a polyatomic ion, it equals the charge of the ion.

3. Balancing Redox Reactions

  • By Oxidation Number Method:
    • Identify the oxidation and reduction half-reactions.
    • Balance the changes in oxidation numbers by adjusting coefficients.
    • Ensure that the total increase and decrease in oxidation numbers are equal.
  • By Ion-Electron (Half Reaction) Method:
    1. Separate the redox reaction into two half-reactions (oxidation and reduction).
    2. Balance all elements except hydrogen and oxygen.
    3. Balance oxygen atoms by adding water (H2O\text{H}_2\text{O}) molecules.
    4. Balance hydrogen atoms by adding hydrogen ions (H+\text{H}^+) in acidic medium or hydroxide ions (OH−\text{OH}^-) in basic medium.
    5. Balance charges by adding electrons (e−e^-).
    6. Combine the half-reactions and ensure the total number of electrons is the same in both half-reactions.

4. Electrolysis

  • Electrolysis: The process by which electrical energy is used to drive a non-spontaneous chemical reaction.
    • Qualitative Aspect: Electrolysis occurs in an electrolytic cell where the positive ions move to the cathode (reduction occurs), and the negative ions move to the anode (oxidation occurs).
      • Example: Electrolysis of NaCl(aq)\text{NaCl(aq)} produces H2\text{H}_2 gas at the cathode and Cl2\text{Cl}_2 gas at the anode.
    • Quantitative Aspect: Governed by Faraday’s Laws of Electrolysis:
      1. First Law: The mass of a substance deposited or liberated at an electrode during electrolysis is directly proportional to the quantity of electricity (charge) passed through the electrolyte. m=Z×Qwhere Q=I×tm = Z \times Q \quad \text{where} \, Q = I \times t mm = mass of substance, ZZ = electrochemical equivalent, QQ = charge (in coulombs), II = current, and tt = time.
      2. Second Law: When the same quantity of electricity is passed through different electrolytes, the mass of the substances produced is proportional to their equivalent weights. m1m2=E1E2where E1 and E2 are equivalent weights.\frac{m_1}{m_2} = \frac{E_1}{E_2} \quad \text{where} \, E_1 \, \text{and} \, E_2 \, \text{are equivalent weights.}
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