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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 · 2020 · Paper I

Q4(c). Does a KCl salt bridge eliminate the liquid junction?

  • 10 marks
  • Answered
Question

To eliminate liquid-junction potential, KCl-salt bridge is used, generally. Does it eliminate liquid-junction? If not, how salt-bridge eliminates liquid junction potential?

Answer in brief

No. The bridge replaces one junction by two (solution 1 | KCl and KCl | solution 2). It reduces the junction potential to a few millivolts because in concentrated KCl the current across each junction is carried mainly by K⁺ and Cl⁻, which move almost equally fast, and the two junction potentials are opposite in sign and nearly cancel.

Key points

  • where two different solutions meet, ions diffuse across the boundary. If cation and anion move at different speeds, the faster one runs ahead and a double layer of charge forms; the resulting potential slows the faster ion.…
  • 1 electrolyte at two concentrations: E_J=(t₊-t₋)RT/Fln(c₁)/(c₂). It vanishes when t₊=t₋.
  • The salt bridge does not eliminate the junction: it creates two junctions, solution 1 | KCl and KCl | solution 2. Liquid still meets liquid at each end.
  • the bridge holds concentrated (3.5 M or saturated) KCl. At each junction KCl is in large excess, so K⁺ and Cl⁻ carry almost all of the current, and their transport numbers are almost equal (t₊≈0.49, t₋≈0.51; λ° = 73.5 and…

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. Origin of a liquid-junction potential: where two different solutions meet, ions diffuse across the boundary. If cation and anion move at different speeds, the faster one runs ahead and a double layer of charge forms; the resulting potential slows the faster ion. For HCl(0.1 M) | HCl(0.01 M), H⁺ runs ahead of Cl⁻.
  2. For a 1:1 electrolyte at two concentrations: E_J=(t₊-t₋)RT/Fln(c₁)/(c₂). It vanishes when t₊=t₋.
  3. The salt bridge does not eliminate the junction: it creates two junctions, solution 1 | KCl and KCl | solution 2. Liquid still meets liquid at each end.
  4. Why E_J becomes small: the bridge holds concentrated (3.5 M or saturated) KCl. At each junction KCl is in large excess, so K⁺ and Cl⁻ carry almost all of the current, and their transport numbers are almost equal (t₊≈0.49, t₋≈0.51; λ° = 73.5 and 76.3 S cm² mol⁻¹ at 25°C). Each junction potential is therefore small: against 0.1 N HCl it falls from 27.0 mV with a 0.1 N KCl bridge to 1.1 mV with 3.5 N KCl.
  5. Opposite signs: the two junctions face opposite ways (electrolyte → KCl, KCl → electrolyte), so their small potentials are of opposite sign and largely cancel. The residual is usually 1 to 2 mV: the junction potential is minimised, not eliminated.
  6. Limits: KCl cannot be used where Cl⁻ reacts (with Ag⁺, Hg₂²⁺, Pb²⁺); NH₄NO₃ or KNO₃, also with t₊≈ t₋, is used instead. With strongly acid or alkaline solutions (H⁺, OH⁻ very mobile) the residual is larger.

In cell notation the double bar | means a liquid junction whose potential has been made negligible by a salt bridge.

Application. every pH measurement with a glass electrode against a calomel or Ag/AgCl reference uses a KCl junction; the small residual junction potential limits the accuracy of pH readings to a few hundredths of a unit.

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Model answer to Q4(c), 2020 Paper I: solution page 19
Solution page 19 of 45