On Oct 30, 2009, at 8:29 AM, Michel Jullian wrote:
2009/10/30 Horace Heffner <[email protected]>:
...
No it isn't. A hot cathode means the global reaction at the
cathode is
exothermic, nothing else. Palladium hydride formation is exothermic
(it releases around 11 kJ/mol heat for a loading factor of 0.8),
so a
hot cathode is to be expected, correct me someone if I am wrong.
Michel
I was under the impression the palladium was mostly plated out
before SPAWAR
jacked up the voltage and current to get the live results.
I answered that (Abd made the same objection), don't you agree that
steady hydrogen absorption spots in steady state should be hotter due
to steady hydride formation, and conversely desorption spots should be
cooler? I visualize heat flowing through the lattice in the opposite
direction to the steady state deuterium flows, but this may be just my
imagination!
I don't think it is that simple. There is no electrolyte interface in
the cracks where the hydrogen tends to diffuse. Any surface of the
cathode exposed to the electrolyte will have the interface potential
drop there. For this reason the hydrogen is driven to cracks or
places where there is a metal-gas interface. In these places
hydrogen recombines at the surface:
H + H -> H2 + 406.6 kJ/mol Gibbs free energy
Hydrogen (H) has a Gibbs free energy of +203.3 kJ/mol.
Hydrogen will move through palladium from gas phase to gas phase with
just a small pressure differential. Some energy is lost in the
diffusion kinetically, i.e. to resistance, which raises the Pd
temperature. I don't think there can be a major temperature
difference between the two sides of the Pd in this case, because so
little energy is required to push the gas through the pd. The forces
doing the disassociating are largely balanced by the forces doing the
re-associating on the other side.
Hydrogen loaded into Pd by electrolysis can and does flow through the
Pd and out into enormous pressures, e.g. 100 atmospheres. This is
one way some commercial electrolysers gain efficiency over others, by
having the electrolysis perform the compression as well. It has been
suggested by more than just myself the possibility there is free
energy from this kind of electrolysis through to the back side of Pd
to gas, if the output pressure is high enough.
Electrolysis itself can utilize ambient heat for the dissociation of
water:
2 H2O <-> H3O+(aq) + OH-(aq)
I think solvated hydrogen at stp has a Gibbs free energy of -252 kJ/
mol, so thus takes about 203.3 kJ/mol - 252 kJ/mol = 48.7 kJ/mol, or
0.504 eV / atom minimum to push hydrogen through the Pd. This
voltage is reduced by higher temperatures though. The hydride
formation energy can be left out of the balance because it is
reversed at the opposing end, netting in effect to zero.
Ambient heat has little effect on dissociation at room temperature,
but at high pressures and temperatures a large percentage of the
energy of disassociation comes from heat, i.e. a significant change
to the Gibbs free energy of the hydrogen occurs. This principle is
used to raise the efficiency of commercial electrolysers. Hot Elly,
for example gets about 30% of the electrolysis energy from heat and
can run in steam mode at 1.07 volts total.
At any rate, as I see it, pumping energy from the electrolysis
potential is required to load solvated hydrogen into the Pd, there is
some pressure drop throughout the Pd, and not much of a temperature
change occurs at any gas release surface. The gas release surface is
not different from the surface at which H2 has is merely pumped
though Pd. Even though the gas is known to disassociate into ionic
bonds within the Pd lattice, not much energy is released at the metal-
gas surfaces.
That's my understanding anyway. Could be all wrong!
In any case much
of the i*W power is dissipated as heat at the cathode and anode
surfaces, in
the two molecule interface layer, at least i*0.6 V worth on both
the cathode
and anode where these potential drops occur.
Not so Horace, I made the same flawed reasoning the other day (and
later acknowledged it was flawed), a total power of i*1.54V is indeed
consumed at the interfaces, but it is not dissipated there, it is
converted to chemical energy, which can later be retrieved by
recombining the deuterium and oxygen.
Yes, here I think I made a mistake referring to the interface
potential drop as the source of heating. I confused knowing that
most all the potential drop in typical aqueous electrolysers is at
the interface with also knowing most of the heat is lost at the
electrodes. I know most of the heat is generated in the electrode
areas in electrolysers. Electrolysis is typically only about 50%
efficient for cells like the SPAWAR cells. Also, I think the heat
generation for hydrogen adsorbed is different from that of hydrogen
evolved. Hydrogen which evolves as a gas comes from primarily the
electronation of H3O+ radicals by electrons tunneling across the
interface, i.e. from:
H3O+ + e- -> H3O
H3O + H3O -> 2 H2O + H2
However, protons that make it to adsorbtion in Pd can't do so by
direct tunneling, so cross the interface by tunneling between H2O
molecules followed by their rotation. This process heats the
interface, but also depletes the kinetic energy of the hydrogen, and
thus a substantial overpotential is required to obtain good Pd
loading, and that results in interface heat.
Same thing for the electrical
energy required to pull the Pd out of solution, it shouldn't be
dissipated as heat. Energy released by PdD formation is another story,
that I expect to be released as heat.
Michel
Best regards,
Horace Heffner
http://www.mtaonline.net/~hheffner/
