A dilute solution of KCl was placed between two platinum electrodes, 12 cm apart, across which a pot — Electrochemistry Chemistry Question
Question
A dilute solution of KCl was placed between two platinum electrodes, 12 cm apart, across which a potential of 1.93 volts was applied. How far (in cm) would the $K^+$ ion move in 20 h at 25°C? Ionic conductance of $K^+$ ion at infinite dilution at 25°C is $7.5 \times 10^{-3}$ mho $m^2$ $mol^{-1}$.
💡 Solution & Explanation
Electric field $E = V / l = 1.93 / 12 = 0.1608$ V/cm. Ionic conductance $\lambda = 7.5 \times 10^{-3} \text{ S m}^2 \text{ mol}^{-1} = 75 \text{ S cm}^2 \text{ mol}^{-1}$. Ionic mobility $u = \lambda / F = 75 / 96500 = 7.77 \times 10^{-4} \text{ cm}^2 \text{ V}^{-1} \text{ s}^{-1}$. Drift speed $v = u \times E = 7.77 \times 10^{-4} \times 0.1608 = 1.25 \times 10^{-4}$ cm/s. Time $t = 20 \text{ h} = 72000$ s. Distance moved $d = v \times t = 1.25 \times 10^{-4} \times 72000 = 9.0$ cm. Therefore, correct answer is 9.