The figure shows two simple designs of a vessel fitted with the elect…

The correct answer is Option (4) → To measure the resistance, and so the conductivity, of the solution

Each design in the figure has two identical platinized Pt electrodes (platinum coated with platinum black) dipping in the same solution and joined to connecting wires, with no salt bridge and no gas supply. This vessel is a conductivity cell. The solution between the electrodes (area of cross section A, distance l apart) forms a column whose resistance is measured, and the conductivity of the solution is found from it:

    \( R = \rho\,\frac{l}{A} = \frac{l}{\kappa A} \)

Why the other options are incorrect

Option (1) To get an emf from the two electrodes acting as anode and cathode — Incorrect. Both electrodes are the same platinized Pt in the same solution, so they do not form two different half-cells and give no cell emf.

Option (2) To serve as a standard hydrogen electrode for measuring electrode potentials — Incorrect. The standard hydrogen electrode has pure hydrogen gas at 1 bar bubbled over a platinum electrode in a 1 M H⁺ solution; the vessels shown have two electrodes and no hydrogen supply.

Option (3) To produce electricity from hydrogen and oxygen fed to the electrodes — Incorrect. In the H₂–O₂ fuel cell the gases are bubbled through porous carbon electrodes into concentrated NaOH; the vessels shown have no gas inlets at the electrodes.

NCERT (2026-27 edition):
2.4.1 Measurement of the Conductivity of Ionic Solutions – “…for measuring the resistance of an ionic solution we face two problems. …Secondly, a solution cannot be connected to the bridge like a metallic wire or other solid conductor. …The second problem is solved by using a specially designed vessel called conductivity cell. It is available in several designs and two simple ones are shown in Fig. 2.4.”

“Basically it consists of two platinum electrodes coated with platinum black (finely divided metallic Pt is deposited on the electrodes electrochemically). These have area of cross section equal to ‘A’ and are separated by distance ‘l’. Therefore, solution confined between these electrodes is a column of length l and area of cross section A. The resistance of such a column of solution is then given by the equation:”

    \( R = \rho\,\frac{l}{A} = \frac{l}{\kappa A} \qquad (2.17) \)

Fig. 2.4: Two different types of conductivity cells.