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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 · 2019 · Paper I

Q7(d). A cell for AgBr(s) → Ag⁺ + Br⁻

  • 10 marks
  • Answered
Question

Devise an electrochemical cell in which the following reaction takes place:
AgBr(s) → Ag⁺ + Br⁻

Answer in brief

Ag(s) | Ag+(aq) | Br-(aq) | AgBr(s) | Ag(s). E°_(cell)=0.0713-0.799=-0.728 V; Kₛₚ=5.0×10⁻¹³.

Key points

  • split the reaction into two half-reactions, both involving Ag, so that their sum is the dissolution of AgBr. Write the reduction on the right and the oxidation on the left.
  • Right electrode (silver–silver bromide, an electrode of the second kind): AgBr(s) + e⁻ → Ag(s) + Br-(aq)
  • Left electrode (silver dipping in Ag⁺): Ag(s) → Ag+(aq) + e⁻
  • Ag cancels and AgBr(s) → Ag+(aq) + Br-(aq). The two solutions are connected by a salt bridge (KNO₃; KCl would precipitate AgCl).

The full model answer

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

  1. Method: split the reaction into two half-reactions, both involving Ag, so that their sum is the dissolution of AgBr. Write the reduction on the right and the oxidation on the left.
  2. Right electrode (silver–silver bromide, an electrode of the second kind): AgBr(s) + e⁻ → Ag(s) + Br-(aq)
  3. Left electrode (silver dipping in Ag⁺): Ag(s) → Ag+(aq) + e⁻
  4. Sum: Ag cancels and AgBr(s) → Ag+(aq) + Br-(aq). The two solutions are connected by a salt bridge (KNO₃; KCl would precipitate AgCl).
  5. EMF: E°_(cell)=E°_(right)-E°_(left)=0.0713-0.799, E°_(cell)=-0.728 V
  6. Meaning: Δ G°=-FE°=+70.2 kJ mol⁻¹: dissolution is non-spontaneous under standard conditions. The cell as written runs only when driven (or runs backwards, precipitating AgBr); its EMF gives the solubility product: ln Kₛₚ=(FE°)/(RT), Kₛₚ=5.0×10⁻¹³, solubility 7.0×10⁻⁷ mol dm⁻³.

Application. such cells measure the solubility products of sparingly soluble salts too small to find by analysis, and the Ag/AgBr and Ag/AgCl electrodes serve as reference and halide-sensing electrodes.

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Model answer to Q7(d), 2019 Paper I: solution page 34
Solution page 34 of 40
Read the whole 2019 solution2019 question paper

All parts of Question 7