Which of the following relationships are not correct ?
(A) \( \displaystyle E^{\circ}_{\mathrm{cell}}=\frac{0.0591}{nF}\log K_c \)
(B) \( \lambda_m^{\circ}\,\mathrm{Na_2SO_4}=\lambda_m^{\circ}(\mathrm{Na^+})+\lambda_m^{\circ}(\mathrm{SO_4^{2-}}) \)
(C) \( \displaystyle \text{Cell constant}=\frac{\text{Conductivity}}{\text{Conductance}} \)
(D) \( \Delta G^{\circ}=nFE^{\circ}_{\mathrm{Cell}} \)
(E) \( \mathrm{Zn_{(s)}|Zn^{2+}_{(aq)}||Cu^{2+}_{(aq)}|Cu} \)
Choose the correct answer from the options given below :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A), (B), (D) only
The correct forms of the electrochemical relations used here are:
\( \displaystyle E^{\circ}_{\mathrm{cell}} = \frac{2.303\,RT}{nF}\log K_c = \frac{0.059}{n}\log K_c\ \ (\text{at } 298\ \mathrm{K}) \)
\( \displaystyle \Lambda^{\circ}_m = \nu_+\lambda^{\circ}_+ + \nu_-\lambda^{\circ}_-\ \ (\text{Kohlrausch law}) \)
\( \displaystyle G^* = \frac{l}{A} = \frac{\kappa}{G} = \frac{\text{conductivity}}{\text{conductance}} \)
\( \displaystyle \Delta_r G^{\circ} = -nFE^{\circ}_{\mathrm{cell}} \)
(A) \( \displaystyle E^{\circ}_{\mathrm{cell}}=\frac{0.0591}{nF}\log K_c \) — Incorrect. F is already included in the 0.059 V factor (0.059 = 2.303RT/F at 298 K); the correct relation is E°cell = (0.059/n) log Kc.
(B) \( \lambda_m^{\circ}\,\mathrm{Na_2SO_4}=\lambda_m^{\circ}(\mathrm{Na^+})+\lambda_m^{\circ}(\mathrm{SO_4^{2-}}) \) — Incorrect. \( \ce{Na2SO4} \) → \( \ce{2Na^{+}} \) + \( \ce{SO4^{2-}} \), so ν+ = 2 and Λ°m(Na2SO4) = 2λ°(\( \ce{Na^{+}} \)) + λ°(\( \ce{SO4^{2-}} \)).
(C) Cell constant = Conductivity / Conductance — Correct. Since κ = G × (l/A), the cell constant l/A = κ/G.
(D) ΔG° = nFE°Cell — Incorrect. The minus sign is missing: ΔrG° = −nFE°cell, so that a positive E°cell gives a negative ΔG° (spontaneous reaction).
(E) Zn(s)|Zn2+(aq)||Cu2+(aq)|Cu — Correct. This is the standard representation of the Daniell cell: anode on the left, cathode on the right, salt bridge shown by ||.
NCERT (2026-27 edition):
Variation of Conductivity and Molar Conductivity with Concentration – “… On the basis of the above observations he enunciated Kohlrausch law of independent migration of ions. The law states that limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte. Thus, if \( \lambda^{\circ}_{\mathrm{Na^+}} \) and \( \lambda^{\circ}_{\mathrm{Cl^-}} \) are limiting molar conductivity of the sodium and chloride ions respectively, then the limiting molar conductivity for sodium chloride is given by the equation:
\( \Lambda^{\circ}_m(\mathrm{NaCl}) = \lambda^{\circ}_{\mathrm{Na^+}} + \lambda^{\circ}_{\mathrm{Cl^-}} \) (2.24)
In general, if an electrolyte on dissociation gives ν+ cations and ν− anions then its limiting molar conductivity is given by:
\( \Lambda^{\circ}_m = \nu_+\lambda^{\circ}_+ + \nu_-\lambda^{\circ}_- \) (2.25)”
Electrochemical Cell and Gibbs Energy of the Reaction – “Electrical work done in one second is equal to electrical potential multiplied by total charge passed. If we want to obtain maximum work from a galvanic cell then charge has to be passed reversibly. The reversible work done by a galvanic cell is equal to decrease in its Gibbs energy and therefore, if the emf of the cell is E and nF is the amount of charge passed and ΔrG is the Gibbs energy of the reaction, then
\( \Delta_r G = -nFE_{\mathrm{(cell)}} \) (2.15)”