In a conductivity cell, the two platinum electrodes, each of area 10 cm are fixed 1.5 cm apart. The — Electrochemistry Chemistry Question
Question
In a conductivity cell, the two platinum electrodes, each of area 10 cm$^2$ are fixed 1.5 cm apart. The cell contained 0.05 N solution of a salt. If the two electrodes are just half dipped into the solution which has a resistance of 50 $\Omega$, the equivalent conductance of the salt solution, in $\Omega^{–1}\text{ cm}^2\text{ eq}^{–1}$, is
💡 Solution & Explanation
Because the electrodes are half dipped, the effective cross-sectional area is $A = 10 / 2 = 5\text{ cm}^2$. The distance between them is $l = 1.5\text{ cm}$. Cell constant $G^* = l / A = 1.5 / 5 = 0.3\text{ cm}^{–1}$. Specific conductance $\kappa = G^* / R = 0.3 / 50 = 0.006\ \Omega^{–1}\text{cm}^{–1}$. Equivalent conductance $\Lambda_{eq} = (\kappa \times 1000) / N = (0.006 \times 1000) / 0.05 = 6 / 0.05 = 120\ \Omega^{–1}\text{cm}^2\text{eq}^{–1}$. Therefore, correct answer is A.