CBSE Class 11 Chemistry Redox Reactions MCQs Set 05

Answer: (b) H₂O₂

Explanation:
1. In H₂O₂ oxygen is in the −1 state, between its lowest (−2) and highest (0) states.
2. So it can be reduced to H₂O (oxidising agent) or oxidised to O₂ (reducing agent).
3. In KMnO₄ and K₂Cr₂O₇ the metal is already in its highest state (+7 and +6), so they only oxidise.

Teacher's Note:
1. An element in an intermediate oxidation state can act both ways.
2. Remember H₂O₂, SO₂ and NO₂⁻ as the standard examples.

Answer: (b) SO₂

Explanation:
1. Sulphur in SO₂ is at +4, between its lowest (−2) and highest (+6) states, so SO₂ can be oxidised to SO₄²⁻ or reduced to S.
2. In H₂SO₄ sulphur is at +6 (highest), so it can only be reduced.
3. In H₂S sulphur is at −2 (lowest), so it can only be oxidised; in HNO₃ nitrogen is at +5 (highest).

Teacher's Note:
1. First write the oxidation number of the central atom.
2. Highest state: only oxidising; lowest state: only reducing; in between: both.

Answer: (a) CaCO₃ → CaO + CO₂

Explanation:
1. In CaCO₃ → CaO + CO₂, Ca stays +2, C stays +4 and O stays −2, so no oxidation number changes.
2. In the others, H₂ and O₂ (0 to +1 and −2), Na (0 to +1) and Mn (+3 to +2) change oxidation number.

Teacher's Note:
1. A redox reaction always shows a change in some oxidation number.
2. Simple decomposition of a carbonate is not redox.

Answer: (b) Ni

Explanation:
1. Nickel goes from 0 to +2, so it loses electrons and is oxidised.
2. The substance that is oxidised is the reducing agent, so Ni is the reducing agent.
3. Cr₂O₇²⁻ is reduced (Cr from +6 to +3), so it is the oxidising agent.

Teacher's Note:
1. The reducing agent is the one that is oxidised.
2. Check each oxidation number before and after the arrow.

Answer: (b) 2H₂S + SO₂ → 3S + 2H₂O

Explanation:
1. In 2H₂S + SO₂ → 3S + 2H₂O, sulphur in SO₂ goes from +4 to 0, so SO₂ is reduced and acts as an oxidising agent.
2. In (d) sulphur goes from +4 to +6, so SO₂ acts as a reducing agent.
3. In (a) and (c) no oxidation number changes.

Teacher's Note:
1. An oxidising agent gets reduced.
2. Look for the reaction in which the S in SO₂ goes down in oxidation number.

Answer: (c) BaO₂ + H₂SO₄ → BaSO₄ + H₂O₂

Explanation:
1. In BaO₂ + H₂SO₄ → BaSO₄ + H₂O₂, Ba stays +2, S stays +6, H stays +1 and O stays −1 (peroxide).
2. In (a) chlorine changes from +5 to +7 and −1; in (b) sulphur goes from +4 and −2 to 0; in (d) oxygen goes from 0 to −1.

Teacher's Note:
1. Peroxide oxygen is −1 on both sides of this reaction.
2. This is an acid–base type reaction that forms H₂O₂.

Answer: (b) 2NO + 3I₂ + 4H₂O → 2NO₃⁻ + 6I⁻ + 8H⁺

Explanation:
1. In (b) nitrogen goes from +2 in NO to +5 in NO₃⁻, so NO is oxidised and is the reducing agent.
2. In (c) and (d) nitrogen goes from +2 to +1 (reduced), so NO is an oxidising agent there.
3. In (a) NO is a product.

Teacher's Note:
1. The reducing agent is oxidised, so its oxidation number rises.
2. +2 to +5 is an increase.

Answer: (d) +8

Explanation:
1. Let the oxidation number of Os be x.
2. x + 4(−2) = 0, so x = +8.

Teacher's Note:
1. Oxygen is −2 in oxides.
2. The sum of oxidation numbers in a neutral molecule is zero.

Answer: (d) −3

Explanation:
1. Treat the ammonium ion: x + 4(+1) = +1, so x = −3.
2. The sulphate ion is SO₄²⁻, so two NH₄⁺ ions balance it.

Teacher's Note:
1. Work with the ion NH₄⁺ and not the whole formula.
2. H is +1 here.

Answer: (b) I₃⁻

Explanation:
1. In I₃⁻, 3x = −1, so the average oxidation number is −1/3.
2. In IF₇, IF₅ and IF₃ iodine is +7, +5 and +3, all whole numbers.

Teacher's Note:
1. A fractional value is an average over several atoms of the element.
2. Use the charge on the ion.

Answer: (c) +5

Explanation:
1. Let the oxidation number of P be x.
2. 2(+2) + 2x + 7(−2) = 0, so 2x = 10 and x = +5.

Teacher's Note:
1. Mg is +2 and O is −2.
2. Set the total to zero for a neutral compound.

Answer: (c) NH₂OH

Explanation:
1. In NH₂OH, x + 3(+1) + (−2) = 0, so x = −1.
2. N is +1 in N₂O, +3 in NO₂⁻ and −2 in N₂H₄.

Teacher's Note:
1. Use H = +1 and O = −2.
2. Calculate each option if you are not sure.

Answer: (d) None of these

Explanation:
1. In (a) N is +4 on both sides.
2. In (b) N is −3 on both sides (NH₃ and NH₄⁺).
3. In (c) N is +5 on both sides (N₂O₅ and HNO₃).
4. So the oxidation number of nitrogen does not change in any of them.

Teacher's Note:
1. Dimerisation, protonation and the reaction of an acidic oxide with water are not redox changes.

Answer: (c) +4 to +6

Explanation:
1. Dichromate is an oxidising agent, so it oxidises SO₂.
2. SO₂ is oxidised to sulphate, SO₄²⁻.
3. Sulphur goes from +4 to +6.

Teacher's Note:
1. The oxidising agent raises the oxidation number of the other substance.
2. Cr⁶⁺ is reduced to Cr³⁺ in this reaction.

Answer: (d) Mg₃N₂: −3

Explanation:
1. In Mg₃N₂, 3(+2) + 2x = 0, so x = −3.
2. In [Co(NH₃)₅Cl]Cl₂ the N is in NH₃, so it is −3 and not 0.
3. In NH₂OH it is −1, and in (N₂H₅)₂SO₄ (N₂H₅⁺: 2x + 5 = +1) it is −2.

Teacher's Note:
1. Check every option; three of them are wrong.
2. In a complex, find the oxidation state of N from the ligand NH₃.

Answer: (a) 0

Explanation:
1. CO is a neutral ligand, so its oxidation number is 0.
2. The complex is neutral, so Cr is 0.

Teacher's Note:
1. Neutral ligands such as CO and NH₃ do not change the oxidation state.

Answer: (d) SO₂ and SO₃

Explanation:
1. NO₂ and N₂O₄: N is +4 in both (difference 0).
2. P₂O₅ and P₄O₁₀: P is +5 in both (difference 0).
3. N₂O (+1) and NO (+2): difference 1.
4. SO₂ (+4) and SO₃ (+6): difference 2, which is the greatest.

Teacher's Note:
1. Work out all four pairs before choosing.

Answer: (c) K₂MnO₄

Explanation:
1. In MnO₂, Mn is +4.
2. In Mn₃O₄, Mn is +8/3 (average).
3. In K₂MnO₄, 2(+1) + x + 4(−2) = 0, so Mn is +6.
4. In MnSO₄, Mn is +2.

Teacher's Note:
1. Calculate all four values and compare.

Answer: (a) 0, +1 and −2

Explanation:
1. S₈ is an element, so S is 0.
2. In S₂F₂, 2x + 2(−1) = 0, so S is +1.
3. In H₂S, 2(+1) + x = 0, so S is −2.

Teacher's Note:
1. Fluorine is always −1.
2. An element in its free state is 0.

Answer: (b) +6

Explanation:
1. Losing electrons raises the oxidation number.
2. Each electron lost adds +1, so +3 + 3 = +6.

Teacher's Note:
1. Loss of electrons means oxidation, so the number goes up.