Question

What are the drift speeds of Li+, Na+, and K+ in water when a potential difference of 10 V is applied across a 1.00 cm conductivity cell? How long would it take an ion to move from one electrode to the other? In conductivity measurements it is normal to use alternating current: what are the displacements of the ions in centimeters during a half cycle of 1.0 kHz applied potential.

Answer 1

(i) What are the drift speeds of Li+, Na+, and K+ in water when a potential difference of 10 V
is applied across a 1.00 cm conductivity cell?
Answer:
Potential difference, Δφ = 10 V
distance, L = 1cm
Electric field, E = Δφ/L = 10/1 = 10.0 V cm-1

Ionic mobilities in water at 298 K
Ion mobilities, u(Li+) = 4.01E-4 cm2 s-1 V-1
Ion mobilities, u(Na+) = 5.19E-4 cm2 s-1 V-1
Ion mobilities, u(K+) = 7.62E-4 cm2 s-1 V-1

Drift speed, S = u*E
Drift speed, S(Li+) = 4.01E-4 cm2 s-1 V-1*10.0 V cm-1 = 4.01E-3 cm s-1
Drift speed, S(Na+) = 5.19E-4 cm2 s-1 V-1*10.0 V cm-1= 5.19E-3 cm s-1
Drift speed, S(K+) = 7.62E-4 cm2 s-1 V-1*10.0 V cm-1 = 7.26E-3 cm s-1

(ii) How long would it take an ion to move from one electrode to the other?
time, t = L/S = 1/S s
time, t(Li+) = 1/4.01E-3 cm s-1   = 249.4 s
time, t(Na+) = 1/5.19E-3 cm s-1 = 192.7 s
0 = time, t(K+) = 1/7.26E-3 cm s-1   = 137.7 s

(iii) In conductivity measurements it is normal to use alternating current: what are the displacements of the ions
in centimeters during a half cycle of 1.0 kHz applied potential.
frequency, f = 1 hz
time taken to move half cycle =1/(2*f)=1/2 s
distatnce moved in half cycle, d = Int[S*dt](t=0 to 1/(2*f)) = u*E0*Int[sin(2*pi*f*t)dt](t=0 to 1/(2*f)) = u*E0/(pi*f)

d = u*E0/(pi*f) = u*10/(pi*1) = 3.183*u cm
d(Li+) = 3.183*u(Li+) = 3.183*4.01E-4 = 12.76E-4 cm    = 1.28 mm
d(Na+) = 3.183*u(Na+) = 3.183*5.19E-4 = 16.52E-4 cm = 1.65 mm
d(K+) = 3.183*u(K+) = 3.183*7.62E-4 = 24.25E-4 cm = 2.425 mm