On Aug 29, 2007, at 9:06 AM, Michel Jullian wrote:
Excess electrons don't hang around on the surface like that. They
are in conduction bands. Their waveforms are distributed throughout
conduction bands, not localized in blobs.
I never said they were hanging around, on the contrary I expect
them to move very fast on very deformed Pd orbitals, jumping easily
from one metal atom to the next.
I think the wave forms don't "jump around" or "move fast" any more
than they "hang around". Conduction band electrons have wave
functions that are *itinerant*, i.e. distributed throughout the
metal. Their probability distribution I think is localized when
ionically bound to adsorbed hydrogen, but only partially.
BTW I know electrons are not really little yellow blobs, it's just
that there are limits to what Jmol can draw, spheres basically :-)
My understanding is conduction band electrons are waveforms located
throughout orbital bands. They are not like free electrons which have
a more particle like existence and a de Broglie wavelength. When
you've drawn the atoms you've drawn the electron boundaries. The
electrons are not point like or individual ball like.
It was also a convenient way to show approximately the number of
excess electrons I expect to exist simultaneously at any given time
on a PdD lattice cube face's surface: a few tens
Multiple excess electrons per cell? This can't be right. As I
showed earlier, one excess electron per 3 angstrom cell produces a
surface field of 2x10^11 V/m, which is about 4 orders of magnitude
larger than a typical interface.
(as many as solvated deuterons facing that surface,
That is not a realistic number. Even if you run at an ampere per cm^2
current density, you will only have a tunneling rate of about 6000
per second per 3 angstrom cell, which is not fast enough to stack up
multiple charges. Charges beyond the interface have almost no
direct effect. Most of the cell potential drop is in the immediate
vicinity of the two molecules thick interface-interphase.
i.e. 4 times the number of layers of solvated deuterons
constituting the double layer's positive plate, i.e. 4 times
several = a few tens ;-).
"Solvated deuterons" are encased in the orbital electrons of the H3O+
ion. There are also theories that suggest proton solvation can also
occur in the form of Zundel (H5O2+) cations and Eigen (H9O4+)
cations, especially in the presence of some proteins. There are
typically 6 water molecules solvating the hydronium.
I agree it would be much better to show their wavefunctions but I
am not at the quantum modelling phase yet, I will have to learn how
to do that first. Have you practiced that yourself?
No. In fact, I think various "many body" problems are
computationally infeasible at present. It is my understanding that
for this reason and others electrochemistry is done on more of a
macro level.
It takes energy for them
to leave that state - so there is an energy barrier they tunnel
through to get to an ion in solution. Cathode electrons tend to
tunnel to H3O+ hydronium ions and thereby neutralize a hydrogen, i.e:
e- + H3O+ --> H2O + H
and they are never "free" during this process because they tunnel
through a barrier to their final state. This process/reaction is
called "electronation". Due to particle mass being in the exponent
of the number e in the tunneling probability, electronation is far
more likely than adsorbtion, which takes time for H3O+ rotations,
during which electronation can occur and thus form neutral H gas away
from the metal where it is not adsorbed.
This may be true, but a significant number of electrolyte deuterons
do make the jump, otherwise there would be no loading would there?
Yes, I didn't say they don't. However, the question remains as to
exactly to where those hydrogen atoms which are adsorbed jump, and
where the ionic bonding electrons for them are located. I think they
tunnel into the lattice from a hydronium at the surface of the
cathode. The protons tunnel/rotate their way through the 2 atom
thick cathode interface.
Electrons can easily
tunnel across several H2O molecules. Protons take a while to tunnel
from an H3O+ to an immediately adjacent H2O molecule because only a
small amplitude exists away from the H3O+ electron orbitals.
Protons do not tend to exist in a free state in solution.
That's why I have shown them in a solvated state.
I'm saying they don't exist in that state, like for example sodium
does. They are typically part of the hydronium ion, which is solvated.
They are
bound to the H3O+ ion by shared electrons, and the H3O+ ions are
surrounded by polarized H2O molecules.
If this can be distinguished visually (in a 3D molecular view) from
a multi-hydrated H+ (H+ surrounded by polarized H2O molecules), a
pointer to a picture or better a 3D rotatable structure will be
welcome.
It's an H3O+ surrounded by polarized H2O. However, in a field, the
proton can hop (tunnel) into adjacent H2O to make a new H3O+, which
then has to rotate before the next hop can take place. For pictures
and more info see:
http://en.wikipedia.org/wiki/Hydronium
In any case I expect the idealized spherical cushion of solvent
molecules to be highly deformed in the tremendous electric field of
the double layer capacitor, surely there must have been some work
on this.
Most of the potential drop is across the first two layers of
molecules at the interface.
The proton merely works its way toward the cathode by its method of
conduction, ending its journey when it is either electronated, or
reaches the electrode surface and is adsorbed.
The pictures show my (open to improvement) model for the PdD/
electrolyte interface at the atomic scale. Tools used: Excel and
Jmol. Atoms drawn at ~25% of their vanDerWaals radii on left view
for clarity.
<028901c7e9ce$2ec13fe0$3800a8c0>
<028a01c7e9ce$2ec13fe0$3800a8c0>
Grey-blue: unit cell of 0.41nm side Face Centered Cubic Pd lattice
in PdD1.0. The loaded deuterons (white) fill all its octahedral
sites, forming an identical FCC lattice shifted inwards by 0.205 nm
(half the cube side). One of them, colored light yellow, is about
to deload via a pyramidal surface site through the excess electron
layer in order to ultimately bubble up.
I think in FCC lattices D tends to tunnel through the square face
holes, but I'll need to find a ref. on that to be sure.
Yes that's what I meant. Those holes have a pyramidal shape (half
an octahedron) haven't they?
I think we understand each other here. I think of the interior of
the pyramidal shape as not being a "hole" but rather a space, or
volume. I refer to the triangular face "hole" as being the irregular
3 sided hole in the pyramid surface (imagined in planar cross
section). These are the triangular holes (constriction bounded by 3
lattice ions) through which hydrogen must tunnel to diffuse. There
is also a square "hole" (a constriction bounded by 4 atoms) through
which tunneling is much more likely to occur, and there may in fact
be complex states maintained between two D located opposite each
other through the square hole, or bordering each other through a
triangular face hole.
Red and white D2O molecules, being polar, solvate positive
electrolyte deuterons on their O side and negative surface Pd ions
on their D side.
This is not right. The positive deuterons end up as H3O+ ions. The
spare deuteron moves through the solution in an E field by tunneling
to an adjacent H2O atom which then must rotate 180 degrees before the
next tunneling event. The H3O+ ions are surrounded by a polarized
blanket of H2O.
As I said I would expect this spherical geometry not to be
maintained near the cathode, where there is "real" charge around,
I think the proton migrates through the interface the same way it
conducts through the electrolyte in a field. It doesn't bring the
polarized water layer with it, it migrates through it, creating a new
one as it goes.
not just the slight charge assymmetry of neutral water molecules.
What we really need is a molecular view of the first hydrated ionic
layer of the double layer based on quantum modelling not on
imagined things. Anyone seen such a model?
The bulk size of D2O (occupies on average a cube of side 0.31 nm,
cubic root of molecular mass 20E-3Kg/6E23 over density 1106Kg/m3
when unstressed) should allow it to be shoehorned by the field into
the 0.29 nm pitch of the surface Pd atoms, i.e. the structures and
thus the ion channels on both sides of the excess electron layer
should be aligned.
The excess electrons are waveforms in the metal, not little blobs on
the surface.
Oh, and they are not yellow?? ;-) I expect the orbitals to bulge
out of the metal (beyond the vDW radius) under the pull of the high
field, don't you?
Bulge yes, but not by much. It is still orbital to orbital contact
everywhere throughout the cathode interface.
D2O acts as the dielectric of the double layer capacitor
This double layer of atoms at the cathode surface is commonly called
the "interface".
I did mean the double layer capacitor: the cathode surface on the
negative side, _several_ layers of hydrated cations on the positive
side. The same capacitor you have drawn in your equivalent circuit
for Ron's 3-electrode cell.
The molecular layer at the surface of the cathode is called the
interface. It is comprised of a layer of (almost entirely) polarized
H2O dipoles, mostly oriented with the hydrogen side toward the
cathode, called the hydration sheath, and a second layer consisting
of typically hydrated cations. The locus of the centers of the
hydrated cations is called the outer Hemholtz plane. The cation
solvation sheaths, i.e. water molecules, weakly bond to the hydration
layer to, in effect, form a second hydration layer. These two
hydration layers I typically refer to as the 2 molecule thick
interface. It appears you refer to the cathode plus hydration layers
plus Helmholtz layer as the double layer capacitor. The potential
drops very rapidly through the interface. Thermal vibrations
overwhelm any remaining field not far from the interface.
The proton differs from other heavier cations in that it can migrate
through the hydration layers, as a hydronium, via tunneling plus
hydronium rotation. The deuteron is not quite as adept at this but
obviously still gets the job done.
The first few layers of metal+adsorbents is called
the interphase.
by separating its negative plate (cathode surface with its excess
electrons) from the multilayer positive "plate" formed by the
electrolyte deuterons
Deuterons in the interface when the cathode becomes a cathode
Uh?
It is only momentarily that a high concentration of cations and
anions can exist up close to the cathode once a charge is applied to
the cathode. In the instant following the initial cell turn on an
equilibrium is established and the interface forms and stays there.
are
immediately either electronated or adsorbed.
Those are the ones we are concerned with for putative fusion
events, aren't they?
Here I was just talking about how the interface gets initialized.
More ions come along, but they have to make their way through the now
fully formed interface, which has a substantial potential drop across
it. There is no evidence I know of for electrolyte based fusion in
low voltage electrolysis, and plenty for cathode lattice based fusion.
For this reason, the
interface is generally not shown containing H3O+ ions because it
consists almost entirely of the two molecule thick water interface.
Why two water molecule thick and not just one water molecule thick
as I have shown it?
There is indeed a first layer of polarized water.
Or maybe you mean the next water layer, beyond the one I have
drawn, which indeed I expect to be two molecule thick (first D2O
layer has its Ds away from the surface
I think the Ds, being positive, face the cathode surface, which is
negative. The next hydration layer is the solvation layer for the
attracted cations (e.g. hydronium), which are positive. The
solvation molecules have their oxygen side facing the cations, and
their hydrogen side facing the first hydration layer, so this makes
for a moderate bond between the first and second hydration layers. I
know it is not all that simple, and things can happen due to random
walk, etc, but I think this is a fairly accurate picture for our
purposes.
, next has its Ds towards the surface, no deuteron can be hydrated
in between).
I think the Ds can actually migrate right through the hydration
layers by tunneling plus hydronium rotation, if they don't get
electronated first.
Anyway, a picture will be welcome.
Ions other than hydrogen are kept even further away by their own
polarized blankets of water molecules.
That's fine, we don't want them to muddy the waters (literally)
on the right (only the first of many one D2O molecule thick layers
is shown). One of those d's, colored light yellow, is about to jump
from the positive "plate" to the negative one, going for one of the
cathode's excess electrons in order to discharge and ultimately
bubble up.
This doesn't happen. If the D+ makes it to the plate it is adsorbed,
not evolved as gas. The D that ends up as gas is electronated by
tunneling electrons.
Possibly. An incoming d anyway, if it fuses it doesn't matter much
what it was supposed to do instead.
Well, I don't think the adsorbed D fuse right there, they just get a
big electron waveform associated with them. I believe there is a lot
of data that shows fusion occurs at least a few layers into the
electrode.
My (not yet debunked on CMNS) theory for CF is that a significant
number of such outgoing+incoming deuteron pairs reach the electron
layer at the same time and fuse instead of their usual bubbling up,
One problem with this theory is the old "where is the electron"
problem. When a deuteron enters a the first plane lattice site,
which is by tunneling because a D atom doesn't fit through the holes,
Agreed
it is joined by an electron which doesn't quite fit. If (and when -
that lattice vibrates) the orbital does not fit, the orbital becomes
a dual state thing, having an orbital amplitude, and a conduction
band conduction band amplitude. I personally think it also has a
deflated state amplitude with the adsorbed hydrogen. This means any
D that can potentially tunnel to this lattice site has some
probability of tunneling into a fused state. That is deflation
fusion.
So you think any fusion occurs in those octahedral sites, whereas I
suspect it occurs at the excess electron layer.
Yes. Very special sites at that. Ed Storms calls the special
conditions there the nuclear active state (NAS). See:
http://lenr-canr.org/Introduction.html
with the help of excess electron screening and channel alignment.
An experiment where the probability of such encounters would be
increased, e.g. by back-loading the cathode to maintain a steady
front deloading flux of deuterons while electrolyzing,
This is just the back side cell concept with the terms "front" and
"back" reversed.
Possibly, ref please.
Done.
would support the hypothesis if it yielded enhanced heat or
reaction products wrt previous P&F experiments (where simultaneous
deloading and electrolysis hasn't been particularly sought).
One problem with this is the helium ends up in the Pd, just below the
surface
No, as I said I expect the fusion to occur just above the surface.
There is a wealth of data to the contrary.
- typically in little crystalline domains - possibly due to
their ability to handle the hydrogen fugacity without cracking.
Another is that lots of experiments have been done using DC combined
with AC or pulsed waveforms having a wide variety - all without
special or repeatable results.
Ah, but has it been tried to deload _continuously_ (using back
loading) while DC-electrolyzing on the same surface that is
deloading (two reasons to bubble)? I expect this to be better than
any DC+AC scheme.
Actually, I think the AC or AC + DC back side de-loading scheme might
have some chance to work measurably, based on anode glow experiments
I've done. I think useful devices will run too hot for electrolysis
though.
Further, this has been done using a
huge range of voltages. In fact I referred you to some HV
experiments documented on my web page. I didn't use D2O for those,
but others have.
What kind of HV, do you mean hundreds of V i.e. Mizuno type
experiments?
Yes. I ran up to 1 kV, but not generating plasma arcs. I think
anodization is key to running at high voltages.
I have taken part in one of those, but it was DC, or rather as much
DC as it can be when you make sparks in water! Also it was tungsten
in my case, can palladium stand the temperature without melting
immediately???
I didn't use Pd. I used Al, Zr, and other things.
Horace Heffner
http://www.mtaonline.net/~hheffner/
