Question

The mobility of a Li+ ion in aqueous solution is 4.01 × 10⁻⁸ m² s⁻¹ V⁻¹ at 25 °C. The potential difference between two electrodes placed in the solution is 24.0 V. If the electrodes are 5.0 mm apart, what is the drift speed of the ion?

Answer 1

The drift speed of a Li+ ion in an aqueous solution with mobility 4.01 × 10⁻⁸ m² s⁻¹ V⁻¹ and a potential difference of 24.0 V between electrodes 5.0 mm apart is calculated to be 1.9248 × 10⁻⁴ m/s.

Step-by-step explanation:

To calculate the drift speed of a Li+ ion in an aqueous solution, given its mobility and the potential difference and distance between two electrodes, we can use the formula:

v_d = μ × E

where:

Now, we can calculate the drift speed:

v_d = (4.01 × 10⁻⁸ m² s⁻¹ V⁻¹) × (4800 V/m) = 1.9248 × 10⁻⁴ m/s

Therefore, the drift speed of the Li+ ion is 1.9248 × 10⁻⁴ m/s.