Redox Reactions — Previous Year Questions (Class 11 Chemistry)
Redox reactions involve electron transfer between species. Master oxidation states, balancing, and identifying oxidizing/reducing agents.
TL;DR: Redox reactions involve electron transfer between species. Master oxidation states, balancing, and identifying oxidizing/reducing agents.
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Redox reactions involve electron transfer between species. Master oxidation states, balancing, and identifying oxidizing/reducing agents.
Redox Reactions — Previous Year Questions with Solutions
Q (2023, 3 marks): Assign oxidation states to all elements in the compounds K₂Cr₂O₇, MnO₄⁻, and K₃[Fe(CN)₆].
Answer: K₂Cr₂O₇ (potassium dichromate):
K: +1, O: -2
For Cr: 2(+1) + 2x + 7(-2) = 0
+2 + 2x - 14 = 0
2x = +12, x = +6
Oxidation states: K(+1), Cr(+6), O(-2)
MnO₄⁻ (permanganate ion):
O: -2, charge = -1
x + 4(-2) = -1
x - 8 = -1
x = +7
Oxidation states: Mn(+7), O(-2)
K₃[Fe(CN)₆] (potassium ferrocyanide):
K: +1
In [Fe(CN)₆]³⁻: CN⁻ is a ligand (no change in formal oxidation state)
N in CN: -3 (as nitrile carbon), C: +2
For Fe: Let Fe be x
3(+1) + x + 6(-1) = 0 (for K₃Fe(CN)₆ neutral)
x + 3 - 6 = 0
x = +3
Oxidation states: K(+1), Fe(+3), C(+2), N(-3)
Final Answer: K₂Cr₂O₇ [K(+1), Cr(+6), O(-2)]; MnO₄⁻ [Mn(+7), O(-2)]; K₃Fe(CN)₆ [K(+1), Fe(+3), C(+2), N(-3)]
Q (2022, 5 marks): Balance the redox reaction: MnO₄⁻ + C₂O₄²⁻ → Mn²⁺ + CO₂ in acidic medium using the oxidation number method.
Answer: Step 1: Identify oxidation state changes
Mn in MnO₄⁻: +7 → Mn²⁺: +2 (reduction, gain 5 electrons)
C in C₂O₄²⁻: +3 → C in CO₂: +4 (oxidation, loss 1 electron per C, 2 electrons per C₂O₄²⁻)
Step 2: Multiply to equalize electrons
Mn: gains 5 electrons per Mn
C₂O₄²⁻: loses 2 electrons per C₂O₄²⁻
LCM = 10
2 MnO₄⁻ (gain 10 electrons) : 5 C₂O₄²⁻ (lose 10 electrons)
Step 3: Balance oxygen and hydrogen
2 MnO₄⁻ + 5 C₂O₄²⁻ → 2 Mn²⁺ + 10 CO₂
Oxygen: Left = 8 + 20 = 28; Right = 20
Add water to balance: 28 - 20 = 8 O, so add 8 H₂O on right
2 MnO₄⁻ + 5 C₂O₄²⁻ → 2 Mn²⁺ + 10 CO₂ + 8 H₂O
Now count H: Right = 16 H
Add H⁺ on left:
2 MnO₄⁻ + 5 C₂O₄²⁻ + 16 H⁺ → 2 Mn²⁺ + 10 CO₂ + 8 H₂O
Step 4: Verify charges
Left: 2(-1) + 5(-2) + 16(+1) = -2 - 10 + 16 = +4
Right: 2(+2) = +4 ✓
Final Answer: 2 MnO₄⁻ + 5 C₂O₄²⁻ + 16 H⁺ → 2 Mn²⁺ + 10 CO₂ + 8 H₂O
Q (2023, 3 marks): Balance the reaction: Fe²⁺ + MnO₄⁻ → Fe³⁺ + Mn²⁺ in acidic medium and identify the oxidizing and reducing agents.
Answer: Step 1: Oxidation state changes
Fe²⁺ → Fe³⁺: +2 → +3 (oxidation, loss 1 electron)
Mn in MnO₄⁻ → Mn²⁺: +7 → +2 (reduction, gain 5 electrons)
Step 2: Equalize electrons
LCM(1,5) = 5
5 Fe²⁺ + 1 MnO₄⁻ → 5 Fe³⁺ + Mn²⁺
Step 3: Balance oxygen and hydrogen
5 Fe²⁺ + MnO₄⁻ → 5 Fe³⁺ + Mn²⁺
Oxygen imbalance: 4 O on left, 0 on right
Add 4 H₂O on right, then add 8 H⁺ on left:
5 Fe²⁺ + MnO₄⁻ + 8 H⁺ → 5 Fe³⁺ + Mn²⁺ + 4 H₂O
Step 4: Verify
Left charge: 5(+2) + (-1) + 8(+1) = 10 - 1 + 8 = +17
Right charge: 5(+3) + (+2) = 15 + 2 = +17 ✓
Identification:
Reducing agent (oxidized): Fe²⁺ (loses electrons)
Oxidizing agent (reduced): MnO₄⁻ (gains electrons)
Final Answer: 5 Fe²⁺ + MnO₄⁻ + 8 H⁺ → 5 Fe³⁺ + Mn²⁺ + 4 H₂O; Reducing agent: Fe²⁺, Oxidizing agent: MnO₄⁻
Q (2021, 3 marks): Define oxidation and reduction. Explain why the decomposition of H₂O₂ is not a redox reaction, while combustion of methane is.
Answer: Oxidation: Loss of electrons or increase in oxidation state.
Reduction: Gain of electrons or decrease in oxidation state.
H₂O₂ decomposition: 2 H₂O₂ → 2 H₂O + O₂
Oxidation states: H(+1), O(-1) in H₂O₂
After decomposition: H(+1), O(-2) in H₂O; O(0) in O₂
Some O: -1 → -2 (reduction), some O: -1 → 0 (oxidation)
This is a redox reaction (disproportionation, where the same element is both oxidized and reduced).
CH₄ + 2 O₂ → CO₂ + 2 H₂O
C: -4 → +4 (oxidation, loses 8 electrons)
O: 0 → -2 (reduction, gains electrons)
This is clearly a redox reaction.
Note: The decomposition of H₂O₂ IS actually a redox reaction because oxygen changes from -1 to both -2 and 0.
Final Answer: Both H₂O₂ decomposition and CH₄ combustion are redox reactions; H₂O₂ shows disproportionation with same element changing to two different oxidation states
Q (2022, 2 marks): Calculate the equivalent weight of KMnO₄ when it acts as an oxidizing agent in acidic medium where it is reduced to Mn²⁺.
Answer: In acidic medium: MnO₄⁻ + 8 H⁺ + 5 e⁻ → Mn²⁺ + 4 H₂O
Mn: +7 → +2 (gain 5 electrons)
Number of electrons transferred per molecule of KMnO₄ = 5
Molar mass of KMnO₄ = 39 + 55 + 64 = 158 g/mol
Equivalent weight = Molar mass / Number of electrons
Equivalent weight = 158 / 5 = 31.6 g/equivalent
Final Answer: Equivalent weight = 31.6 g/equivalent
Q (2023, 3 marks): Explain the concept of disproportionation reaction and give two examples.
Answer: Disproportionation reaction: A redox reaction where the same element is oxidized and reduced simultaneously in the same compound, producing two different products with higher and lower oxidation states.
Example 1: Decomposition of H₂O₂
2 H₂O₂ → 2 H₂O + O₂
O in H₂O₂: -1
O in H₂O: -2 (reduction)
O in O₂: 0 (oxidation)
Oxygen is both oxidized and reduced.
Example 2: Disproportionation of Cl₂ in water
Cl₂ + H₂O → HCl + HClO
Cl: 0 → -1 (reduction to form Cl⁻ in HCl)
Cl: 0 → +1 (oxidation to form ClO⁻ in HClO)
Chlorine is both oxidized and reduced.
Other examples: P₄ in alkali, S₈ in alkali, Cu⁺ in aqueous solution
Final Answer: Disproportionation reaction involves simultaneous oxidation and reduction of the same element; examples: H₂O₂ decomposition and Cl₂ in water
Frequently Asked Questions
How do you determine the oxidation state of an element in a compound?
Rules (in order of priority): (1) Oxidation state of an element in its elemental form is 0. (2) Oxidation state of a monatomic ion equals the charge on the ion. (3) Oxygen is -2 (except in peroxides where it's -1, and in OF₂ where it's +2). (4) Hydrogen is +1 (except in metal hydrides where it's -1). (5) Group 1 metals are +1; Group 2 metals are +2. (6) Sum of oxidation states in a neutral compound is 0; in a polyatomic ion, it equals the charge.
Why is the half-reaction method preferred over the oxidation number method for balancing redox equations?
The half-reaction method separates oxidation and reduction into two half-reactions, making it easier to balance complex equations. It clearly shows: (1) Which species is oxidized/reduced, (2) Electron transfer, (3) Balancing of oxygen and hydrogen independently. It works equally well in acidic or basic media and is systematic, reducing errors.
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