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.
Λom for NH4OH is equal to:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Λom NH4Cl + Λom NaOH - Λom NaCl
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. For a weak electrolyte such as \( \ce{NH4OH} \) it cannot be measured directly, so it is obtained by combining strong electrolytes that contain the same ions:
\( \displaystyle \Lambda^\circ_m(\ce{NH4OH}) = \lambda^\circ_{\ce{NH4+}} + \lambda^\circ_{\ce{OH-}} \)
\( \displaystyle = (\lambda^\circ_{\ce{NH4+}} + \lambda^\circ_{\ce{Cl-}}) + (\lambda^\circ_{\ce{Na+}} + \lambda^\circ_{\ce{OH-}}) - (\lambda^\circ_{\ce{Na+}} + \lambda^\circ_{\ce{Cl-}}) \)
\( \displaystyle = \Lambda^\circ_m(\ce{NH4Cl}) + \Lambda^\circ_m(\ce{NaOH}) - \Lambda^\circ_m(\ce{NaCl}) \)
Why the other options are incorrect
Option (1) — Incorrect. It contains Λ°m(NH4OH) itself on the right-hand side, so it cannot be used to calculate it.
Option (3) — Incorrect. The NaCl and NaOH terms are swapped; this leaves \( \ce{NH4^{+}} \) + \( \ce{2Cl^{-}} \) − \( \ce{OH^{-}} \).
Option (4) — Incorrect. This gives \( \ce{2Na^{+}} \) + \( \ce{OH^{-}} \) − \( \ce{NH4^{+}} \), not \( \ce{NH4^{+}} \) + \( \ce{OH^{-}} \).
NCERT (2026-27 edition):
Applications of Kohlrausch Law – “Example 2.8: Λ°m for NaCl, HCl and NaAc are 126.4, 425.9 and 91.0 S cm2 mol–1 respectively. Calculate Λ° for HAc.
\( \Lambda^\circ_m(\text{HAc}) = \lambda^\circ_{\ce{H+}} + \lambda^\circ_{\ce{Ac-}} = \lambda^\circ_{\ce{H+}} + \lambda^\circ_{\ce{Cl-}} + \lambda^\circ_{\ce{Ac-}} + \lambda^\circ_{\ce{Na+}} - \lambda^\circ_{\ce{Cl-}} - \lambda^\circ_{\ce{Na+}} \)
\( = \Lambda^\circ_m(\ce{HCl}) + \Lambda^\circ_m(\text{NaAc}) - \Lambda^\circ_m(\ce{NaCl}) \)
\( = (425.9 + 91.0 - 126.4)\ \mathrm{S\,cm^2\,mol^{-1}} = 390.5\ \mathrm{S\,cm^2\,mol^{-1}} \)”