Wayne said:
I've conducted a simple experiment on a glass sample that has two
clean copper electrodes biased with 20v DC. Between the two
electrodes I have placed a drop of de-ionised water with Universal
Indicator added.
Over a period of time the electrodes changes colour, as expected,
with the cathode going purple (OH-) and the anode going red (H+).
After a longer period of time the colours start to spread out
toward the opposing electrodes, pH gradient. However, the amount
the purple (OH-) spreads is apx three times greater that that of
the red (H+).
I have repeated this 10 times in random positions and I get the
same results.
I would have thought that the H+ ions would have more mobility
than the OH- ions and thus I would have expected the opposite to
happen.
Can anyone help and through some light on this phenomena, or have
I missed something?
WayneL
Hi Wayne,
Raw hydrogen ions cannot exist naked in solution. They combine with
a nearby hydroxyl molecule to form H3O. You are actually generating
copper ions at the anode, not hydrogen ions. The process is
1. At the anode, a copper atom gives up two electrons to become an
ion:
Cu(s) - 2e --> Cu(++)
2. At the cathode, water dissociates and hydrogen ions accept
electrons to form hydrogen gas which escapes:
2H2O --> 2H(+) + 2OH(-)
2H(+) + 2e --> H2(g)
So for every copper ion, two hydroxyl ions are produced.
This is why telephone circuits use negative polarity. If the voltage
was positive, small leakage currents would cause the wire to
disappear through electrolysis.
You can calculate the amount of copper ions in solution by knowing
the current and the time interval. The calculations are highly prone
to error due to the numerous
unit conversions needed, but a dos
program written by Roger Schafley called Mercury will do the
conversions for you. Here is an example calculation I use for a much
larger electrolysis cell:
--------------------------------------------------------------------
; Colloidal Copper Generator
; Calculations Bob Lee's method
; Note change
; k = 0.5* 63.5 / 96485 ; Coulombs required per gram of copper
; Roger Schafly's Mercury is available at
;
http://www.mindspring.com/~schlafly/eureka.htm
;
http://archives.math.utk.edu/software/msdos/calculus/mrcry209/.html
;
http://archives.math.utk.edu/software/msdos/calculus/mrcry209/mrcry209.zip
Cou = I * sec ; total number of Coulombs
esec = I / 1.60217733e-19; electrons per second
gm = k * I * sec ; Faraday's equation
isin = esec / sqin ; ions per sq. in. per sec
isnm = isin / 6.45e14 ; ions per square nanometer per sec
k = 0.5* 63.5 / 96485 ; Coulombs required per gram of copper
lt = 3.785 * gal ; convert gallons to litres
lt = ml / 1000 ; convert millilitres to litres
mg = gm * 1000 ; convert grams to milligrams
ml = 29.57 * oz ; convert ounce to milliliters
phr = ppm / hrs ; ppm per hour
ppm = mg / lt ; 1 ppm is 1 milligram per litre
sec = hrs * 3600 + mnt * 60 ; convert hours to seconds
uAin = 1e6 * I / sqin ; current density in uA per sq in
hrs = 1
I = 3.111e-3 ; current 02 3.3k 3.111 mA
ml = 1450 ; volume of dw
mnt = 0 ; minutes
sqin = 9.5 ; wetted area
--------------------------------------------------------------------
Here is the solution:
--------------------------------------------------------------------
Cou = 11.1996
gal = 0.38309
gm = 0.00368
hrs = 1.00000
I = 0.00311
k = 0.00032
lt = 1.45000
mg = 3.68541
ml = 1450.00
mnt = 0.00000
oz = 49.0361
phr = 2.54166
ppm = 2.54166
sec = 3600.00
sqin = 9.50000
uAin = 327.473
--------------------------------------------------------------------
In this example, 0.00368 grams of copper were released giving a
concentration of 2.54166 ppm (parts per million). It turns out the
maximum you can achieve is about 3ppm before the copper starts
plating out on the cathode.
As far as the change in color, I posted the results of two
experiments using silver ions that show how to make these ions
visible:
CS> Making Ions Visible
http://escribe.com/health/thesilverlist/m61491.html
Re: CS> Making Ions Visible
http://escribe.com/health/thesilverlist/m61527.html
In my example, the silver ions seemed to have the same mobility as
the hydroxyl ions, since the color bands appeared to meet in the
middle in the first experiment.
Since the mobility is a function of the size of the ion and the
applied voltage, I would expect the copper to move much slower than
the hydroxyl. This appears to be the reverse of what you observe.
Perhaps someone at sci.chem can offer an explanation.
Mike Monett
