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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 · 2026 · Paper I

Q1(f). Concentration cell of two silver electrodes

  • 5 marks
  • Answered
Question

One silver rod is immersed in a silver nitrate solution of the concentration C₁(AgNO₃)=5×10⁻⁴ kmol m⁻³ and a second silver rod in the concentration C₂(AgNO₃)=1×10⁻² kmol m⁻³. The electrolyte solutions are connected by salt bridge. Calculate the difference Δ E of the redox potential for the electrochemical cell at 25°C (under equilibrium conditions).

Answer in brief

Δ E=RT/Fln 20=0.0770 V (77.0 mV); the dilute side is the anode.

Key points

  • E=E^°_(Ag+/Ag)+RT/Fln[Ag⁺] (Nernst equation).
  • The same couple at both ends, so E^° cancels: Δ E=RT/Fln(C₂)/(C₁)
  • RT/F=(8.314×298.15)/(96485)=0.02569 V; ln1×10⁻²5×10⁻⁴=20=2.996
  • Δ E=0.02569×2.996=0.0770 V=77.0 mV

The full model answer

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

Given: 1 kmol m⁻³ = 1 mol dm⁻³, so C₁=5×10⁻⁴ M and C₂=1×10⁻² M; T=298.15 K. Activities are taken equal to concentrations, and the salt bridge removes the junction potential.

  1. Each electrode: E=E^°_(Ag+/Ag)+RT/Fln[Ag⁺] (Nernst equation).
  2. The same couple at both ends, so E^° cancels: Δ E=RT/Fln(C₂)/(C₁)
  3. RT/F=(8.314×298.15)/(96485)=0.02569 V; ln1×10⁻²5×10⁻⁴=20=2.996
  4. Δ E=0.02569×2.996=0.0770 V=77.0 mV

Reading the answer: net change Ag⁺ (C₂) → Ag⁺ (C₁), Δ G=-FΔ E=-7.43 kJ mol⁻¹; Ag dissolves at the dilute side (anode). “Equilibrium conditions” = the reversible, zero-current potential.

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Model answer to Q1(f), 2026 Paper I: solution page 5
Solution page 5 of 55
Read the whole 2026 solution2026 question paper

All parts of Question 1