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UPSC Civil Services (Mains)

Chemistry optional

  • 2 papers
  • 250 marks each
  • 3 h per paper
  • 5 of 8 questions, Q1 and Q5 compulsory

Paper I covers physical and inorganic chemistry.

UPSC Civil Services (Mains), Chemistry optional · 2018 · Paper I

Q1(e). Spontaneity of four redox reactions from E°

  • 10 marks
  • Answered
Question

Observe the following half cell reactions:
(i) I₂ + 2e⁻ → 2I⁻; E°=+0.54 V
(ii) 2Br⁻ → Br₂ + 2e⁻; E°=-1.09 V
(iii) Fe → Fe²⁺ + 2e⁻; E°=+0.44 V
Using the above Standard Electrode Potential Values, explain the spontaneity or non-spontaneity for the following reactions:
(A) Br₂ + 2I⁻ → 2Br⁻ + I₂ (B) Fe + Br₂ → Fe²⁺ + 2Br⁻
(C) Fe + I₂ → Fe²⁺ + 2I⁻ (D) I₂ + 2Br⁻ → 2I⁻ + Br₂

Answer in brief

A, B and C are spontaneous (E°_(cell)=+0.55, +1.53 and +0.98 V); D is non-spontaneous (E°_(cell)=-0.55 V).

Key points

  • For each reaction, the species reduced is at the cathode and the species oxidised at the anode: E°_(cell)=E°_(cathode)-E°_(anode) (both as reduction potentials), and Δ G°=-nFE°_(cell) with n=2.
  • (A) Br₂ is the stronger oxidant (higher E°); it takes electrons from I⁻. Bromine water turns a KI solution brown with liberated I₂.
  • (B) and (C) Fe has a negative E°, so it is a good reducing agent: both halogens oxidise it to Fe²⁺. B has the larger driving force (Δ G°=-295 kJ) because Br₂ is the stronger oxidant.
  • (D) is the reverse of A: E°_(cell)<0, Δ G°>0. I₂ cannot oxidise Br⁻; the reverse reaction (A) goes instead.

The full model answer

Written out in full, as it should go in the answer booklet. The solution page with its diagrams follows it.

Reading: (ii) and (iii) are written as oxidations, so their values are oxidation potentials. As reduction potentials: E°(Br₂/Br⁻)=+1.09 V and E°(Fe²⁺/Fe)=-0.44 V.

  1. For each reaction, the species reduced is at the cathode and the species oxidised at the anode: E°_(cell)=E°_(cathode)-E°_(anode) (both as reduction potentials), and Δ G°=-nFE°_(cell) with n=2.
ReactionReducedOxidisedE°_(cell) / VΔ G° / kJResult
(A) Br₂ + 2I⁻Br₂ (+1.09)I⁻ (+0.54)1.09-0.54=+0.55-106
(B) Fe + Br₂Br₂ (+1.09)Fe (−0.44)1.09+0.44=+1.53-295
(C) Fe + I₂I₂ (+0.54)Fe (−0.44)0.54+0.44=+0.98-189
(D) I₂ + 2Br⁻I₂ (+0.54)Br⁻ (+1.09)0.54-1.09=-0.55+106non-spontaneous
  1. (A) Br₂ is the stronger oxidant (higher E°); it takes electrons from I⁻. Bromine water turns a KI solution brown with liberated I₂.
  2. (B) and (C) Fe has a negative E°, so it is a good reducing agent: both halogens oxidise it to Fe²⁺. B has the larger driving force (Δ G°=-295 kJ) because Br₂ is the stronger oxidant.
  3. (D) is the reverse of A: E°_(cell)<0, Δ G°>0. I₂ cannot oxidise Br⁻; the reverse reaction (A) goes instead.
  4. General rule: in the electrochemical series, the oxidised form of a couple higher in the series oxidises the reduced form of any couple below it. E° is intensive, so it is not multiplied by the coefficients.

Application. the same order (F₂ > Cl₂ > Br₂ > I₂) is used in the laboratory test for halides and in the industrial recovery of bromine from sea water, where Cl₂ displaces Br₂ from bromide.

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Model answer to Q1(e), 2018 Paper I: solution page 6
Solution page 6 of 55
Read the whole 2018 solution2018 question paper

All parts of Question 1