> Re: CS>Silver Generation with ULVDC
> From: Ode Coyote
> Date: Mon, 12 May 2003 05:01:53
> The term "ion cloud" has the quotation marks because that's the
> common language used.
> Actually, it's a particle cloud.
> Ions are too small to see...too small to reflect the visible
> wavelength of light that you CAN see.
> Plate out would be like a silver spot on the container due to the
> 'particle cloud' coming too close to the side r bottom as it
> follows a conductive track from electrode to electrode. Some oxide
> plate out will generally be apparent as a black spot too.
> Sufficient stirring eliminates that effect. Excessive stirring
> seems to make it worse but more evenly distributed. I think that
> it, like the yellow deposit is also an inertial thing but from a
> different angle.
[...]
> Ken
Hi Ken,
You are right. The cloud has to be particles. But how do they form?
The ion has to grab an electron from somewhere, which will
eventually come from the cathode. But the cathode and anode current
must be equal, so some other process has to take place to equalize
the current in the series circuit.
When the ions arrive at the cathode, they can't all grab an electron
and jump onto the cathode. Hydrogen is produced at the cathode,
which requires electrons. So the ions have to wait their turn, and
they form an invisible cloud around the cathode.
(This is similar to the space charge of electrons around the cathode
in a vacuum tube, so all the old hams should easily see this
concept.)
With the conventional process, the voltage across the cell may start
at 30V, then decrease after the regulator goes into current
limiting.
The high voltage gradient increases the ion velocity, so they arrive
at the cathode sooner.
(For the ion velocity equation, see Ivan's excellent post on
electrolysis at:)
http://escribe.com/health/thesilverlist/m46719.html
When the ions arrive, the high voltage gradient keeps them close to
the cathode, which means the ion cloud will be quite dense. This
increases the probability that the ions which have found an electron
will be close to other atoms, and Brownian motion will bring them
close enough so the Van der Walls force can start the agglomeration.
The dense cloud also means the particles will grow larger.
So the idea that a high current rips large particles off the anode
may be incorrect. The high current means a higher voltage across the
cell, which means a denser ion cloud at the cathode. The particles
are not produced at the anode, but come from the ion cloud around
the cathode.
So we have been barking up the wrong tree, and I am the worst
offender. (See my 130VDC article at:)
http://www3.sympatico.ca/add.automation/misc/130vdc.htm
A high voltage is not needed to get appreciable conduction with
typical distilled water.
The initial conductance of distilled water is not constant with
applied voltage. It increases as applied voltage increases, which is
why everyone uses the highest voltage they can get without
destroying their current regulator.
With the ULV process, the initial voltage across the electrodes is
only 2V to 4V. This is an order of magnitude less than conventional
systems, but the initial current may still be 180uA. This is in the
same ballpark as I used to get with 27 Volts from 3 nines!
With the low voltage, the ion velocity is an order of magnitude
less, so the ions take longer to reach the cathode. This means more
time for ion production before we start running into problems with
the ion cloud forming at the cathode.
When the cloud starts to form, the lower voltage gradient means the
cloud density is much less than before. This means the ions that
have found an electron have less probability of interacting, so
fewer particles are produced. This means less silver is wasted as
the black residue when we wipe the rods.
Because the cloud is less dense, the particles that do form are
smaller, so they don't have as much tendency to fall to the bottom
and form a black smudge. And the sides of the glass stay clear
instead of turning black.
And the solution doesn't turn yellow and plate out. Stirring is not
needed to get high quality cs.
Here is a run I started this morning:
Mon May 12, 2003, 09:04:56 am 4.374V 180uA
Mon May 12, 2003, 12:05:01 pm 3.024V 240uA
Mon May 12, 2003, 12:44:32 pm 2.811V 250uA
Mon May 12, 2003, 01:15:55 pm 2.517V 263uA
I will let it go another couple of hours, but you can see the
voltage across the cell is quite low. When I stop the run, it will
be among the strongest cs I have ever made, and it will not turn
yellow and plate out.
Any comments?
Best Regards,
Mike Monett
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