Q4(c). Does a KCl salt bridge eliminate the liquid junction?
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.
- 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⁻.
- For a 1:1 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.
- 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.
- 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.
- 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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All parts of Question 4
- 4(a) LCAO; bonding, non-bonding and antibonding orbitals of H₂10 marks
- 4(b) Capillary action, contact angle and γ=12hρ gr10 marks
- 4(c) Does a KCl salt bridge eliminate the liquid junction?10 marks
- 4(d) Diamagnetic [Ni(CN)₄]²⁻ and paramagnetic [NiCl₄]²⁻10 marks
- 4(e) Why d–d transitions are forbidden, weak, and yet observed10 marks