At 03:44 PM 7/3/2012, Finlay MacNab wrote:
It should be noted that in an electrolyte the current results from a
chemical reaction at the anode and cathode (in this case the
generation of hydrogen and oxygen) there are no free charge carriers
in the solution itself. The cations and anions are bound together
by electrostatic attraction and exist inside cloud quasi organized
solvent molecules. Electrolyte ions do organize on the surface of
electrodes to screen the electric field at low potentials (most of
the voltage drop in an electrochemistry experiment happens within
the first nanometer of the electrode surface). At the high fields
quoted in the linked paper, I cannot imagine how the electrolyte
could screen the applied field. It seems reasonable to me that an
electric field could exist inside the cell, since electrolytes do
not have free charges that can migrate to the surface of the dielectric.
Electrolytes do not conduct electrons, they accept electrons and
donate electrons. There are no charges flowing through the
solution, just reactions at the electrode surface.
Now I must get back to my electrodeposition experiment.
An ounce of experiment is worth a pound of theory. Or even a ton.
Now, I'd love to be wrong here. However, I remain unconvinced, and
obviously so does Rich. The objection is an obvious one, so one might
think there would be a definitive answer somewhere. I see, however,
that Mr. MacNab may have confused himself with his own knowledge. The
situation has nothing to do with "free charges that can migrate to
the surface" of anything. The mode of conduction is irrelevant.
An electric field *does* exist in the cell. It is complex, and varies
from location to location. If the statement about the "first
nanometer" is true, we could be looking at a field strength there of
more than 10^7 V/cm. Much higher than the field from the "high
voltage" supply. But just for a nanometer.
Here is the problem. Electric fields are measured relative to some
potential. There is only one electric field at any given location.
How do we know what the electric field is at a location? Well, we can
use a voltage probe. That won't tell us the field, we will need to
use two probes for that, which will give us the potential difference
between the two locations.
We can use a bridge to measure potential difference without any need
for current to flow through the probe, complicating things.
So if we stick two probes into the electrolyte, on either side of the
cell, when we have this 6 KV sitting across the cell, what voltage
will we need to place across the probes, so that the current through
them is zero?
In the electrochemical cell, I'll predict this. The voltage will be
very low, probably less even than the voltage between the anode and
cathode, if Mr. MacNab's statement about the voltage drop is true
(and I have no reason to doubt it).
Imagine, though, that it would instead be thousands of volts. This is
at zero current. But thousands of volts across two probes --
electrodes -- in a conductive electrolyte? If you had the available
current, the thing would blow up!
(In fact, here, the high voltage power supply is from a TV set, there
is only low available current.)
So the voltage across the probes would be very low, perhaps
millivolts. If the probes were in contact with the cathode and anode,
respectively, it might be a few volts, whatever the electrolysis voltage is.
There is no "screening" of the field. There is just the shorting of a
portion of the field, by the electrolyte. The current from the HV
supply is very low, it might be picoamps. [I estimate it below]
If we plot the field with a series of measurements, we'd find that
there is about 3 KV across a cell wall, about 1/16 inch thick.
Acrylic plastic, probably. Then there is a very low voltage across
the electrolyte. then there would be another 3 KV across the opposite
wall, giving us a total drop across the cell of 6 KV.
The cell wall is about 1.6 mm thick, and with 3 KV across it -- which
could easily be measured -- that's 19 KV/cm. Looking for the
electrical properties of acrylic, I found that it has a bulk
resistance of 1.6 X 10^16 ohm-cm . We might be looking at about 16
cm^2 for each plate. I get on the order of 1.6 x 10^14 ohms per piece
of acryclic. Current for 3 KV would be about 5 x 10^10 amps, or 500
pA. That is the leakage current through the acrylic.
Breakdown voltage for acrylic is 17 KV/mm. That's probably a minimum
guarantee. 170 KV/cm. Actual breakdown would not normally occur until
substantially higher voltage. (I've tested actual breakdown voltage,
it was, under the situations I was testing, over double the
specification or more.) If the acrylic does break down, all bets are
off. The current though the acrylic would go way up, but the supply,
though, won't supply much current. The current from the electrolysis
supply will probably still be greater. I.e., the "high voltage"
supply wouldn't be high voltage any more.
Other than with breakdown, the field is completely undetectable in
the electrolyte, given that the actual DC current in the electrolyte
is in the range of 1 - 500 mA, perhaps. The electric field from the
electrolysis voltage will be much higher.
This is a DC field, remember. Things would get quite dicey if the
voltage is varying.
Ah, but what if there was breakdown? Could that explain the effect?
Well, the problem is that the voltage wouldn't be 6 KV any more. Did
they measure the voltage? And all this would do would be to add a bit
of current between two new "electrodes." And you could get the same
effect with just putting two electrodes in the cell. Might work. What
would that do? Who knows?
There cannot be a high voltage DC field across an electrolyte without
a corresponding current, and put a few hundred volts across an
electrolyte -- there are some youtube videos showing this -- you get
some spectacular effects. Not just a little shift in plating morphology!
