Paper: CSIR-NET December 2019 · Subject: Physical Chemistry · Chapter: Electrochemistry · Topic: Electrochemistry – General · Marks: 2 · Difficulty: Medium
For the electrochemical cell $\mathrm{Ag}|\mathrm{AgCl}| \mathrm{MCl}(0.01 \mathrm{M})|\mathrm{MCl}(0.02 \mathrm{M})| \mathrm{AgCl} \mid \mathrm{Ag}$, the junction potential is the highest when $\mathrm{M}^{+}$is
(a)$\mathrm{H}^{+}$
(b)$\mathrm{Li}^{+}$
(c)$\mathrm{Na}^{+}$
(d)$\mathrm{K}^{+}$
Answer
Answer: A ✓ checked by 4AB · confidence high
Explanation
The liquid-junction potential grows with the difference between cation and anion mobilities. H⁺ (λ ≈ 350) differs most from Cl⁻ (λ ≈ 76); K⁺ (≈ 73) differs least. So M⁺ = H⁺ gives the largest junction potential.