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NOTES

ASSIGNMENT

1. Define oxidation number and how it differs from valency.
2. What are the basic steps followed in the oxidation number method of redox balancing?
3. How does the medium (acidic/basic) affect the balancing of redox reactions?
4. Assign oxidation numbers to all elements in the compound: K₂Cr₂O₇
5. Identify whether the following is a redox reaction and justify:
   CuO(s) + H₂(g) → Cu(s) + H₂O(g)

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Section B: Short Answer Questions (2 Marks Each)

6. Balance the following redox reaction using the oxidation number method:
   Fe²⁺ + Cr₂O₇²⁻ + H⁺ → Fe³⁺ + Cr³⁺ + H₂O
7. In the reaction: MnO₄⁻ + SO₃²⁻ → Mn²⁺ + SO₄²⁻ (acidic medium),
   identify the oxidation number changes and name the oxidising and reducing agents.
8. Balance the following redox reaction using the oxidation number method (basic medium):
   MnO₄⁻ + I⁻ → MnO₂ + I₂
9. Explain why oxidation and reduction must occur together in a redox reaction.
10. Assign oxidation numbers to sulphur in the following:
    a) H₂S
    b) H₂SO₄
    c) SO₃
    d) Na₂S₂O₃

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Section C: Long Answer/Problem-Based Questions (3 Marks Each)

11. Balance the redox reaction using the oxidation number method (acidic medium):
    Cr₂O₇²⁻ + SO₃²⁻ → Cr³⁺ + SO₄²⁻
12. Balance the reaction in basic medium using the oxidation number method:
    MnO₄⁻ + Br⁻ → MnO₂ + BrO₃⁻
13. A reaction is written as:
    Cl₂ + NaOH → NaCl + NaClO + H₂O
    a) Assign oxidation numbers.
    b) Identify oxidised and reduced species.
    c) Balance the reaction using oxidation number method (basic medium).
14. Explain the role of H⁺ and OH⁻ ions in balancing redox equations using oxidation number method. Provide one example for each case.
15. What are the key differences between the half-reaction method and the oxidation number method of redox balancing? State with example.

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KEY

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Section A: Very Short Answer Questions

1. Oxidation number is the charge assigned to an atom in a compound assuming that electrons are completely transferred.
   It differs from valency, which is the combining capacity of an atom.
2. Basic steps:
   • Assign oxidation numbers.
   • Identify oxidized and reduced elements.
   • Balance change in oxidation numbers.
   • Balance atoms other than H and O.
   • Balance O with H₂O, H with H⁺ (acidic) or OH⁻ (basic).
   • Ensure charge and atoms are balanced.
3. In acidic medium, use H⁺ and H₂O to balance H and O.
   In basic medium, use OH⁻ and H₂O.
4. K₂Cr₂O₇:
   K = +1, O = –2
   Total charge = 2(+1) + 2(x) + 7(–2) = 0 → x = +6
   Cr = +6
5. Yes, it is a redox reaction:
   Cu: +2 → 0 (reduction),
   H: 0 → +1 (oxidation)

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Section B: Short Answer Questions

6. Balance (acidic medium):
   Fe²⁺ + Cr₂O₇²⁻ + H⁺ → Fe³⁺ + Cr³⁺ + H₂O
   Final balanced equation:
   6Fe²⁺ + Cr₂O₇²⁻ + 14H⁺ → 6Fe³⁺ + 2Cr³⁺ + 7H₂O
7. Mn: +7 → +2 (reduction)
   S: +4 → +6 (oxidation)
   Oxidising agent: MnO₄⁻
   Reducing agent: SO₃²⁻
8. Balanced (basic medium):
   2MnO₄⁻ + I⁻ + 2H₂O → 2MnO₂ + I₂ + 4OH⁻
9. Because electron loss (oxidation) must be accompanied by electron gain (reduction), to conserve charge.
10. a) H₂S: S = –2
    b) H₂SO₄: S = +6
    c) SO₃: S = +6
    d) Na₂S₂O₃: One S = +6, other = –2 (average +2)

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Section C: Long Answer/Problem-Based Questions

11. Cr₂O₇²⁻ + SO₃²⁻ → Cr³⁺ + SO₄²⁻ (acidic medium)
    Balanced:
    Cr₂O₇²⁻ + 3SO₃²⁻ + 8H⁺ → 2Cr³⁺ + 3SO₄²⁻ + 4H₂O
12. MnO₄⁻ + Br⁻ → MnO₂ + BrO₃⁻ (basic medium)
    Balanced:
    2MnO₄⁻ + Br⁻ + 2H₂O → 2MnO₂ + BrO₃⁻ + 4OH⁻

a) Cl₂: 0; Cl⁻: –1; ClO⁻: +1
b) Cl₂ is both oxidised and reduced (disproportionation)
c) Balanced:
Cl₂ + 2NaOH → NaCl + NaClO + H₂O

Acidic example:
MnO₄⁻ + Fe²⁺ + H⁺ → Mn²⁺ + Fe³⁺ + H₂O
H⁺ used to balance H atoms
Basic example:
MnO₄⁻ + SO₃²⁻ + OH⁻ → MnO₂ + SO₄²⁻ + H₂O
OH⁻ balances H atoms

Feature	Half-Reaction Method	Oxidation Number Method
Basis	Splitting into oxidation/reduction	Changes in oxidation number
Medium consideration	More suited to acidic/basic media	Works well with simple redox
Electron balancing	Explicit electron transfer shown	Based on change in oxidation states
Example:
Fe²⁺ + Cr₂O₇²⁻ + H⁺ → Fe³⁺ + Cr³⁺ + H₂O (can be done by both methods)

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