JEE Chemistry Redox Reactions MCQs Set 08

Answer: (d) S₈ < S₂O₃²⁻ < S₄O₆²⁻ < S₂O₈²⁻

Explanation:
1. S₈: 0
2. S₂O₃²⁻: 2x − 6 = −2, so x = +2
3. S₄O₆²⁻: 4x − 12 = −2, so x = +2.5
4. S₂O₈²⁻ has a peroxy link, so sulphur is +6
5. The order is 0 < +2 < +2.5 < +6. Answer: (d)

Teacher's Note:
1. Elemental sulphur has oxidation number zero.
2. In peroxodisulphate two oxygen atoms are -1, so sulphur stays +6.

Answer: (d) SO₂ and SO₃ (sulphur)

Explanation:
1. NO₂ and N₂O₄: +4 and +4, difference 0
2. P₂O₅ and P₄O₁₀: +5 and +5, difference 0
3. N₂O (+1) and NO (+2): difference 1
4. SO₂ (+4) and SO₃ (+6): difference 2, the greatest. Answer: (d)

Teacher's Note:
1. Compute each oxidation number from the formula.
2. Different molecular formulas for the same empirical formula give the same oxidation number.

Answer: (d) \( \mathrm{Ag_2O + H_2O_2 \to 2Ag + H_2O + O_2} \)

Explanation:
1. In (a), (b) and (c), H₂O₂ is reduced (O from −1 to −2) and acts as an oxidising agent
2. In (d), O in H₂O₂ goes from −1 to 0 (O₂), so H₂O₂ is oxidised and acts as the reducing agent while Ag₂O is reduced to Ag. Answer: (d)

Teacher's Note:
1. A reducing agent is itself oxidised.
2. H2O2 forms O2 only when it loses electrons.

Answer: (d) \( \mathrm{Cu_2S + 2FeO \to 2Cu + 2Fe + SO_2} \)

Explanation:
1. (a) is neutralisation, (b) is precipitation and (c) is the action of an acid on a carbonate, so no oxidation numbers change in them
2. In (d), Cu goes from +1 to 0 and Fe from +2 to 0 while S goes from −2 to +4, so it is a redox reaction. Answer: (d)

Teacher's Note:
1. Look for a change in oxidation number to identify a redox reaction.
2. Neutralisation, precipitation and carbonate decomposition by acid are not redox reactions.

Answer: (c) 2, 5 and 16

Explanation:
1. Mn gains 5 electrons per MnO₄⁻ and each C₂O₄²⁻ loses 2 electrons
2. Equating electrons: 5x = 2y, so x : y = 2 : 5
3. Charge balance: \( -x - 2y + z = 2x \), so \( z = 3x + 2y = 6 + 10 = 16 \)
4. So x = 2, y = 5, z = 16. Answer: (c)

Teacher's Note:
1. Balance electrons first, then charge, and finally hydrogen.
2. Check: hydrogen 16 on the left and 2 times 8 on the right.

Answer: (c) +3

Explanation:
1. In N₂H₄ nitrogen is −2
2. Two N atoms lose 10 electrons, so each loses 5
3. New oxidation state = −2 + 5 = +3. Answer: (c)

Teacher's Note:
1. Divide the total electrons lost by the number of nitrogen atoms.
2. Loss of electrons raises the oxidation number.

Answer: (b) Highest = OF₂, lowest = H₂O₂

Explanation:
1. OF₂: +2
2. H₂O₂: −1
3. KO₂ (superoxide): −1/2
4. O₂F₂: +1
5. Highest is OF₂ (+2) and lowest is H₂O₂ (−1). Answer: (b)

Teacher's Note:
1. Fluorine is more electronegative than oxygen, so oxygen is positive in fluorides.
2. Superoxide oxygen carries -1/2.