The m^2/sv had me for a minute, that is for a voltage, instead of a
gradient, so there is a v/m term which accounts for the square.

So if we convert to cm instead of meters we end up with this:

6.4X10-8 m^2, or 6.4X10-4 cm^2 per second per volt.  So if you have 100
volts across 1 cm, the silver ions will move at a rate of .064 mm per
second.

Marshall

Dan Nave wrote:

> How about this?
> From:
>
> http://www.tannerm.com/IonSpeeds.htm
>
> Ion speeds and conductivity-the Einstein and Stokes-Einstein equations
>
> Movement of an ion through a solution under the influence of a
> potential gradient (electric field) is the result of the acceleration of
> the charged particle in the field and the opposing forces of assymetry
> and the electrophoretic effect. Thus the ion moves at a constant rate
> determined by a balance of these forces. This leads to the definition of
> mobility ui of the ion i.
>
> Ion mobilities at 298 K in aqueous solution
>
> Ion u0(m2s-1V-1)
>
> H3O+     36.3 x 10-8
> OH-        20.5 x 10-8
> Li+         4.0 x 10-8
> Na+       5.2 x 10-8
> K+         7.6 x 10-8
> Ag+       6.4 x 10-8
> Mg2+     5.5 x 10-8
> Zn2+     5.5 x 10-8
> Cl-         7.9 x 10-8
> Br-         8.1 x 10-8
> NO3-     7.4 x 10-8
> SO42-    8.3 x 10-8
>
> >>> Marshall Dudley <[email protected]> 1/12/2006 9:32:37 AM >>>
> If you know the ion's mobility it can be figured rather easily from
> the
> voltage gradient:
>
> http://www.du.edu/~jcalvert/phys/elechem.htm and
> http://www.ce-mag.com/archive/1999/mayjune/mrstatic.html
>
> The velocity of an ion is v = KE, where K is the mobility of the ion in
> cm/s
> per V/cm.
>
> So if you can find the mobility of a silver ion, it would be very easy
> to
> figure.  I searched a little but was unable to find any mobiity
> numbers, but
> they are probably somewhere if you want to look long enough.
>
> Marshall
>
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