An electrochemical cell consists of two metallic electrodes (anode and cathode) dipping in electrolytic solutions. Oxidation takes place at the anode and reduction takes place at the cathode. These cells are of two types:
1) Galvanic cells, in which the chemical energy of a spontaneous redox reaction is converted into electrical energy. Eocell = Eocathode - Eoanode, where Eocell is the standard potential of the cell, and ΔrGo = -nFEocell, where ΔrGo is the standard Gibbs energy change of the cell reaction.
2) Electrolytic cells, in which electrical energy is used to carry out a non-spontaneous redox reaction. The amount of substance produced at an electrode depends on the quantity of electricity passed, Q = I × t (where I is the current in ampere and t is the time in seconds). One faraday is the charge carried by 1 mole of electrons, that is 96500 C.
The conductivity (κ) of an electrolytic solution depends on the nature and the concentration of the electrolyte. According to Kohlrausch’s law of independent migration of ions, the limiting molar conductivity of an electrolyte is the sum of the limiting molar conductivities of its cation and anion, for example Λom(NaCl) = λoNa+ + λoCl–, where λo is the limiting molar conductivity of an ion. The law can be used to calculate the limiting molar conductivity of weak electrolytes.
Λ°m for NaCl, HCl and CH3COONa are 126.4, 425.9 and 91.0 S cm2 mol–1 respectively. Calculate Λ°m for CH3COOH
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 390.5 S cm2 mol-1
By Kohlrausch’s law of independent migration of ions, the limiting molar conductivity of an electrolyte is the sum of the contributions of its cation and anion. So Λ°m of a weak electrolyte such as acetic acid (\( \ce{CH3COOH} \), written HAc) can be found from strong electrolytes that together supply the same ions (NaAc = \( \ce{CH3COONa} \)):
\( \displaystyle \Lambda^{\circ}_m(\mathrm{HAc})=\Lambda^{\circ}_m(\mathrm{HCl})+\Lambda^{\circ}_m(\mathrm{NaAc})-\Lambda^{\circ}_m(\mathrm{NaCl}) \)
This works because the \( \ce{Na^{+}} \) and \( \ce{Cl^{-}} \) contributions cancel:
\( \displaystyle =[\lambda^{\circ}_{H^+}+\lambda^{\circ}_{Cl^-}]+[\lambda^{\circ}_{Na^+}+\lambda^{\circ}_{Ac^-}]-[\lambda^{\circ}_{Na^+}+\lambda^{\circ}_{Cl^-}] \)
\( \displaystyle =\lambda^{\circ}_{H^+}+\lambda^{\circ}_{Ac^-} \)
Substituting the values:
\( \displaystyle \Lambda^{\circ}_m(\mathrm{HAc})=425.9+91.0-126.4 \)
\( \displaystyle =390.5\ \mathrm{S\,cm^2\,mol^{-1}} \)
Note 1: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.
NCERT (2026-27 edition):
Variation of Conductivity and Molar Conductivity with Concentration – …
Note 2: The same/similar question has been asked in different years by NTA as per the details given below:
| 2026 | 2025 | 2024 | 2023 | 2022 |
|---|---|---|---|---|
| 11-05 S1 | 03-06 S2 28-05 S1 22-05 S1 15-05 S1 | – | – | 20-08 S1 10-08 S1 |