CBSE Class 11 Chemistry Redox Reactions Worksheet Set 03

Redox Reactions Worksheet

Question 1. Calculate the oxidation number of sulphur in (a) Na₂S₂O₃ and (b) Na₂S₄O₆.
Answer:
(a) Let the oxidation number of S be x. (+1) × 2 + (x) × 2 + (−2) × 3 = 0, so x = +2.
(b) (+1) × 2 + (x) × 4 + (−2) × 6 = 0, so x = +2.5.
In simple words: Add up the known oxidation numbers, then solve for x so that the total is zero.

Teacher's Note:
1. Na is +1 and O is −2 in these salts.
2. The answers are averages, because the S atoms are not all in the same state.

Question 2. Na₂S₂O₃ has two sulphur atoms and Na₂S₄O₆ has four. Explain why the oxidation number of S in Na₂S₄O₆ is a fraction (+2.5), although no single sulphur atom has this oxidation number.
Answer:
The value +2.5 is the average oxidation number of the four sulphur atoms.
The structure of the S₄O₆²⁻ ion shows that the four atoms are in different states: two terminal S atoms are at +5 and the two middle S atoms are at 0. The average is (5 + 0 + 0 + 5) ÷ 4 = +2.5.
So a fractional oxidation number should be taken with care; the true picture comes from the structure.
In simple words: A fraction is an average of atoms in different states.

Teacher's Note:
1. Fractional oxidation numbers appear when the same element is present in different states in one species.
2. Always write 'average oxidation number' in the answer.

Question 3. Find the average and the individual oxidation numbers of (i) Fe in Fe₃O₄ and (ii) Pb in Pb₃O₄, which are mixed oxides.
Answer:
(i) Fe₃O₄ is a mixture of FeO and Fe₂O₃ in the ratio 1 : 1, so the individual oxidation numbers are +2 and +3. Average = [1(+2) + 2(+3)] ÷ 3 = +8/3.
(ii) Pb₃O₄ is a mixture of PbO and PbO₂ in the ratio 2 : 1, so the individual oxidation numbers are +2 and +4. Average = [2(+2) + 1(+4)] ÷ 3 = +8/3.
In simple words: Split the mixed oxide into its two simple oxides, then average.

Teacher's Note:
1. Both oxides give an average of +8/3.
2. Fe₃O₄ = FeO·Fe₂O₃ and Pb₃O₄ = 2PbO·PbO₂.

Question 4. State the oxidation number of oxygen in (i) H₂O₂, (ii) KO₂, (iii) KO₃, (iv) OF₂ and (v) O₂F₂.
Answer:
(i) H₂O₂ (a peroxide): −1
(ii) KO₂ (a superoxide): −1/2
(iii) KO₃ (an ozonide): −1/3
(iv) OF₂: +2
(v) O₂F₂: +1
In simple words: Oxygen is −2 in most compounds, but these five are the usual exceptions.

Teacher's Note:
1. Fluorine is more electronegative than oxygen, so oxygen is positive in OF₂ and O₂F₂.
2. Learn the five values as one list.

Question 5. Define an oxidising agent and a reducing agent. Give two examples of each, and name two substances that can act as both.
Answer:
An oxidising agent (oxidant) oxidises other substances and is itself reduced. The oxidation number of one of its elements decreases, or it gains electrons. Examples: KMnO₄, K₂Cr₂O₇ (also HNO₃ and conc. H₂SO₄).
A reducing agent (reductant) reduces other substances and is itself oxidised. The oxidation number of one of its elements increases, or it loses electrons. Examples: KI, Na₂S₂O₃.
Substances that can act as both include H₂O₂ and NO₂⁻.
In simple words: The oxidising agent is reduced and the reducing agent is oxidised.

Teacher's Note:
1. An element in its highest oxidation state can only oxidise.
2. An element in its lowest state can only reduce; an intermediate state can do both.

Question 6. How can you tell from its oxidation number whether a substance can act as an oxidising agent, a reducing agent or both?
Answer:
Find the oxidation state of the central atom.
If it is the maximum oxidation state of the element, the substance is an oxidising agent only.
If it is the minimum oxidation state, the substance is a reducing agent only.
If it is an intermediate state, the substance can act as both and may also disproportionate.
In simple words: Highest state: oxidises. Lowest state: reduces. In between: both.

Teacher's Note:
1. Use this rule to answer 'can it act as both?' questions quickly.

Question 7. Is the reaction 2KClO₃ → 2KCl + 3O₂ a disproportionation reaction? Give a reason.
Answer:
No. In this reaction chlorine is reduced (+5 to −1) and oxygen is oxidised (−2 to 0).
Different elements are oxidised and reduced, so it only looks like a disproportionation reaction. KClO₃ acts as both oxidant and reductant, but not through the same element.
In simple words: Disproportionation needs the same element to be oxidised and reduced.

Teacher's Note:
1. Name the elements whose oxidation numbers change.
2. If they are two different elements, it is not disproportionation.

Question 8. Identify the disproportionation reaction and write the oxidation-number changes: 4KClO₃ → 3KClO₄ + KCl
Answer:
Chlorine in KClO₃ is at +5. It goes up to +7 in KClO₄ (oxidation) and down to −1 in KCl (reduction).
The same element, in the same compound, is oxidised and reduced, so this is a disproportionation reaction.
In simple words: One element, one compound, two products with a higher and a lower oxidation number.

Teacher's Note:
1. The element must start in an intermediate state.
2. Chlorine can be −1, +5 and +7 here, so +5 is intermediate.

Question 9. Show that Cl₂ reacts with cold dilute NaOH in a disproportionation reaction: Cl₂ + 2OH⁻ → ClO⁻ + Cl⁻ + H₂O
Answer:
Chlorine in Cl₂ is at 0. In ClO⁻ it is +1 (oxidised) and in Cl⁻ it is −1 (reduced).
The same element is oxidised and reduced at the same time, so the reaction is a disproportionation reaction.
Other examples: 2H₂O₂ → 2H₂O + O₂ (oxygen −1 to −2 and 0), and S₈ + 12OH⁻ → 4S²⁻ + 2S₂O₃²⁻ + 6H₂O.
In simple words: Cl₂ at 0 splits into Cl at +1 and Cl at −1.

Teacher's Note:
1. Write the oxidation number above each atom.
2. Fluorine never disproportionates, because it has no positive oxidation state.

Question 10. What is a comproportionation reaction? Show with NH₄NO₂ → N₂ + 2H₂O.
Answer:
A comproportionation reaction is the reverse of disproportionation: an element in two different oxidation states is converted into a single oxidation state.
In NH₄NO₂ nitrogen is at −3 (in NH₄⁺) and +3 (in NO₂⁻). Both change to 0 in N₂.
Another example is 5I⁻ + IO₃⁻ + 6H⁺ → 3I₂ + 3H₂O.
In simple words: Two different states of the same element meet in one state.

Teacher's Note:
1. Compare it with disproportionation, where one state splits into two.

Question 11. Balance the following redox reaction by the ion-electron method in acidic medium: Fe²⁺ + MnO₄⁻ → Fe³⁺ + Mn²⁺. Write the equation in molecular form for FeSO₄ + KMnO₄ + H₂SO₄.
Answer:
Oxidation half: Fe²⁺ → Fe³⁺ + e⁻ (×5)
Reduction half: 5e⁻ + 8H⁺ + MnO₄⁻ → Mn²⁺ + 4H₂O (×1)
Adding: 5Fe²⁺ + 8H⁺ + MnO₄⁻ → 5Fe³⁺ + Mn²⁺ + 4H₂O
Molecular form: 10FeSO₄ + 2KMnO₄ + 8H₂SO₄ → 5Fe₂(SO₄)₃ + 2MnSO₄ + K₂SO₄ + 8H₂O
In simple words: Make the electrons lost equal to the electrons gained, then add.

Teacher's Note:
1. Balance O with H₂O and H with H⁺ in acidic medium.
2. Check atoms and charge on both sides at the end.

Question 12. Balance in basic medium by the ion-electron method: ClO⁻ + CrO₂⁻ + OH⁻ → Cl⁻ + CrO₄²⁻ + H₂O
Answer:
Reduction: ClO⁻ → Cl⁻, giving 2e⁻ + H₂O + ClO⁻ → Cl⁻ + 2OH⁻ (×3)
Oxidation: CrO₂⁻ → CrO₄²⁻, giving 4OH⁻ + CrO₂⁻ → CrO₄²⁻ + 2H₂O + 3e⁻ (×2)
Adding: 3ClO⁻ + 2CrO₂⁻ + 2OH⁻ → 3Cl⁻ + 2CrO₄²⁻ + H₂O
In simple words: Balance as in acid first, then add OH⁻ on both sides to cancel the H⁺.

Teacher's Note:
1. In basic medium, add as many OH⁻ on both sides as there are H⁺.
2. Combine H⁺ and OH⁻ into H₂O and cancel.

Question 13. Balance the equation (acidic medium): H₂O₂ + MnO₄⁻ → Mn²⁺ + O₂
Answer:
Oxidation: H₂O₂ → O₂ + 2H⁺ + 2e⁻ (×5)
Reduction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O (×2)
Adding: 6H⁺ + 5H₂O₂ + 2MnO₄⁻ → 2Mn²⁺ + 5O₂ + 8H₂O
In simple words: Here H₂O₂ acts as the reducing agent.

Teacher's Note:
1. The 5 and 2 come from making 10 electrons on each side.

Question 14. Balance: Zn + HNO₃ (dilute) → Zn(NO₃)₂ + H₂O + NH₄NO₃
Answer:
Zinc goes from 0 to +2 (loses 2 e⁻); nitrogen in the NH₄⁺ product goes from +5 to −3 (gains 8 e⁻).
So the ratio of Zn to the N that is reduced is 4 : 1, giving 4Zn + 10HNO₃ → 4Zn(NO₃)₂ + 3H₂O + NH₄NO₃.
In simple words: Four zinc atoms supply 8 electrons, which reduce one N from +5 to −3.

Teacher's Note:
1. Of the 10 HNO₃, one is reduced and the rest form nitrate ions.
2. Check N, H and O to confirm the balance.

Question 15. In a redox reaction the total increase in oxidation number equals the total decrease. Verify this for 10FeSO₄ + 2KMnO₄ + 8H₂SO₄ → 5Fe₂(SO₄)₃ + 2MnSO₄ + K₂SO₄ + 8H₂O
Answer:
Fe goes from +2 to +3: 10 atoms × (+1) = a total increase of 10.
Mn goes from +7 to +2: 2 atoms × (−5) = a total decrease of 10.
The two are equal, so the equation is correctly balanced for electron transfer.
In simple words: 10 units gained = 10 units lost.

Teacher's Note:
1. Use this check to confirm the coefficients of oxidant and reductant.