Match List-I with List-II
| List-I Expression | List-II Known as |
| (A) \( \Delta_r G^{\circ}=\mathrm{n}FE^{\circ}_{\mathrm{(cell)}} \) | (I) Cell constant |
| (B) \( \displaystyle G^{*}=\frac{l}{A}=\mathrm{R}\,k \) | (II) Standard Gibbs energy |
| (C) \( \displaystyle \Lambda_m=\frac{k}{c} \) | (III) Conductivity |
| (D) \( \displaystyle \kappa=\frac{1}{\rho} \) | (IV) Molar conductivity |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) - (II), (B) - (I), (C) - (IV), (D) - (III)
Each expression is one of the basic relationships of electrochemistry. The Gibbs energy links a cell to its emf. The other three describe how a solution conducts electricity: resistance R = ρl/A, conductivity κ = 1/ρ, and molar conductivity is the conductivity per unit concentration.
(A) ΔrG° = nFE°(cell) — This matches (II) Standard Gibbs energy. The standard Gibbs energy of the cell reaction is linked to the standard emf of the cell, where n is the number of electrons transferred and F is the Faraday constant.
\( \displaystyle \Delta_r G^{\circ}=-nFE^{\circ}_{\mathrm{cell}} \)
(B) G* = l/A = R k — This matches (I) Cell constant. The cell constant is the distance between the electrodes (l) divided by their area of cross section (A). Since R = ρl/A and ρ = 1/κ, it is found by measuring the resistance of a solution of known conductivity.
\( \displaystyle G^{*}=\frac{l}{A}=R\,\kappa \)
(C) Λm = k/c — This matches (IV) Molar conductivity. It is the conductivity of the solution divided by its molar concentration.
\( \displaystyle \Lambda_m=\frac{\kappa}{c} \)
(D) κ = 1/ρ — This matches (III) Conductivity. Conductivity is the reciprocal of resistivity; its SI unit is S m−1.
Note: It seems that there is a mistake in the original NTA paper, which prints ΔrG° = nFE°(cell) in (A) when it should read ΔrG° = −nFE°(cell), and k for conductivity in (B) and (C) when it should read κ.