Question

a dil solution of KCLwas placed between 2 platinum electrodes12cm apart across which 1.93V is appliedHow far would K+ ions move in 20 hours at 298Kif ionic conductance and inifnite dilution for K+ is 7.5 10^-2

• A. 0.0090 cm

Answer 1

Answer: (A)

The K⁺ ions move approximately 0.0090 cm in 20 hours.

Answer 2

Solution:

1. Ionic mobility calculation:

   Given ionic conductance (at infinite dilution) for K⁺, λ₍K⁺₎ = 7.5×10⁻² (in S·cm²·mol⁻¹). Ionic mobility is: μ = λ₍K⁺₎ / F, with F = 96500 C/mol. Thus, μ = (7.5×10⁻²) / (96500) ≈ 7.77×10⁻⁷ cm²/(V·s).

2. Potential gradient:

   Electrodes are 12 cm apart with an applied voltage V = 1.93 V, so the field is: E = V/d = 1.93/12 ≈ 0.1608 V/cm.

3. Drift velocity of K⁺ ions:

   v = μ × E = (7.77×10⁻⁷ cm²/(V·s)) × (0.1608 V/cm) ≈ 1.25×10⁻⁷ cm/s.

4. Distance moved in 20 hours:

   Time = 20 hours = 20×3600 = 72000 seconds. Distance = v × time = (1.25×10⁻⁷ cm/s) × 72000 s ≈ 0.0090 cm.

Thus, the K⁺ ions move approximately 0.0090 cm in 20 hours.