On Nov 3, 2009, at 2:13 AM, Michel Jullian wrote:
2009/11/3 Abd ul-Rahman Lomax
...
>>> It should be noted that the SPAWAR video is showing the back side
>>> of the cathode, not the side facing the anode.
>>
>> See Figure 1 of:
>>
>> http://www.lenr-canr.org/acrobat/SzpakSpolarizedd.pdf
>>
>> It shows the camera on the outside, next the Mylar film, next the
>> *unsealed* surface of the cathode. It appears to me there is
>> electrolyte between the Mylar and the surface. My experience with
>> anode spots is they form well on both sides.
>
> "unsealed surface"? The mylar film is a window into the cell.
Next to it is
> the cathode. And on the other side of the cell, apparently, not
shown, is
> the anode. However, the cathode was not a foil, as I wrote, but a
"Ni
> screen." It's still viewed from the "back," but it's more complex
than I had
> thought. Somehow I thought this was a foil cathode.
Yes, it's Ni screen, it's viewed from the back through a Mylar
window and a thin layer of intervening electrolyte, and Horace is
definitely right that the observed hot spots occur _on the back side_:
http://www.lenr-canr.org/acrobat/SzpakSlenrresear.pdf :
Yes, those photos are of spots on the back side. However, I
seriously doubt there are not similar spots on the front side too.
The back side cathode-electrolyte interface is not much different
from the front side. That said, one of the problems here is not
knowing the potentials and currents at which the cell photgraphed is
running. This would probably be a lot easier to discuss in
specifics. If SPAWAR is running with a fixed current supply and they
plate out all the PdCl electrolyte, then the voltage across the cell
could be very high.
<<The steep temperature gradients of the hot spots, Figure 1c,
indicate that the heat sources are of high intensity and located
very close to the contact surface.
Figure 1. ...(b) View of the obverse side of the cathode showing the
distribution of hot spots ranging from <29°C (purple) to >49°C
(white). (c) Temperature gradients on the back side of the cathode.>>
So, what is the direction of the deuterium flux through the back
surface of a Pd cathode close to a plastic window do you think?
Out, because electrolyte paths to get there are more resistive.
I don't think this is right. I think the interface will be about
the same on both sides. The potential drop through the mesh has
little practical effect. If you look at ordinary cathodes in
electrolysis, even big ones, they evolve hydrogen on both sides.
Have you ever seen a cathode evolve gas on only one side?
This confirms my suggestion, earlier in this thread, that the hot
spots occur where the deuterium desorbs from Pd deuteride to D2
gas (not solvated D Horace), which chemically is an endothermic
reaction.
These assertions are highly debatable. It appears you are assuming
the front side looks any different.
Am I the only one to see how significant this is?
It might be significant if it were true. My point is I don't think
it is true in the sense you are implying it is. I don't think the
interface layers differ much in potential on either side of that
cathode, and certainly don't think one side is desorbing more than
the other. Except at cracks or any location where there is no
electrolyte, desorption must occur in reverse through the interface,
which is equilibrium with respect to hydrogen flow. The chemical
concentrations are in equilibrium as well, meaning there is no
enthalpy of reaction because there is no net reaction, except when
there have been major changes in potential such that the Pd loading
is in change and loading or de-loading is in progress. I expect the
electrode filmed is pretty much in equilibrium, when it comes to
hydrogen concentrations of all kinds, in the Pd (i.e. the loading
percentage), and in the electrolyte across the interface, and in
terms of flow rates cross the interface, which net to zero. There
is thus no enthalpy of reaction involved, especially on the front
side vs back side.
I'll certainly grant the above is clearly not true at cracks, where H
+ H -> H2 recombination occurs not in any equilibrium, resulting in a
net change in enthalpy when the Gibbs free energy of the palladium
hydride is considered, and which is complicated because it is a
function of loading phase and temperature.
Best regards,
Horace Heffner
http://www.mtaonline.net/~hheffner/
