Chemistry · Redox reactions · chapter 7 of 9

Class 11 Chemistry Chapter 7: Redox Reactions case study question with answers

One case-based question of the kind CBSE-pattern school exams ask: a passage, then 3 parts worth 4 marks in all, with the answer and the marking scheme for each. Written from the NCERT chapter.

Case study · 4 marks

To find the copper in a brass sample, Nisha dissolves it, removes the excess acid, and adds excess potassium iodide solution to the Cu²⁺ solution: 2Cu²⁺(aq) + 4I⁻(aq) → Cu₂I₂(s) + I₂(aq). The iodine set free stays in solution and is titrated with 0.100 M sodium thiosulphate: I₂(aq) + 2S₂O₃²⁻(aq) → 2I⁻(aq) + S₄O₆²⁻(aq). When the brown colour fades to pale yellow she adds a little starch solution and continues the titration drop by drop. The end point is reached after 18.0 mL of thiosulphate in all.

Part 1 · 1 mark

At the end point, the colour of the solution in the flask changes from

(a) colourless to pale pink(b) pale yellow to violet(c) intense blue to colourless(d) colourless to intense blue

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Answer: (c) intense blue to colourless

Starch gives an intense blue with iodine; the blue disappears as soon as the last iodine is used up by thiosulphate. Pale pink belongs to permanganate titrations.

Part 2 · 1 mark

The amount of Cu²⁺ in the sample is

(a) 0.90 mmol(b) 1.80 mmol(c) 3.60 mmol(d) 18.0 mmol

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Answer: (b) 1.80 mmol

Thiosulphate used = 0.100 × 18.0 = 1.80 mmol, which reacts with 0.90 mmol I₂. Two Cu²⁺ give one I₂, so Cu²⁺ = 2 × 0.90 = 1.80 mmol: Cu²⁺ and S₂O₃²⁻ are 1 : 1. 0.90 mmol stops at iodine; 3.60 mmol doubles twice.

Part 3 · 2 marks

Using oxidation numbers, identify the species oxidised and the species reduced in each of the two reactions.

Answer and marking scheme

First reaction: Cu +2 → +1 in Cu₂I₂, so Cu²⁺ is reduced; I −1 → 0, so I⁻ is oxidised. Second reaction: I 0 → −1, so I₂ is reduced; S +2 in S₂O₃²⁻ → +2.5 (average) in S₄O₆²⁻, so thiosulphate is oxidised.

first reaction: Cu²⁺ reduced (+2 → +1), I⁻ oxidised (−1 → 0) (1)
second reaction: I₂ reduced (0 → −1), S₂O₃²⁻ oxidised (+2 → +2.5) (1)

Iodine is nearly insoluble in water. Explain why the iodine set free stays in solution here, and why the thiosulphate reaction suits this titration.

Answer and marking scheme

The excess KI holds the iodine in solution as KI₃ (I₂ + I⁻ → I₃⁻). Iodine reacts very specifically with thiosulphate, and with starch it gives an intense blue that vanishes the moment the iodine is used up, so the end point is sharp.

excess KI keeps iodine dissolved as KI₃ (1)
specific reaction with thiosulphate; starch blue vanishes sharply at the end point (1)

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