On Feb 9, 2010, at 2:09 AM, Michel Jullian wrote:
Hi Horace, sorry for the late response, my comments below.
2010/2/7 Horace Heffner <hheff...@mtaonline.net>:
On Feb 7, 2010, at 4:42 AM, Michel Jullian wrote:
2010/2/7 Horace Heffner <hheff...@mtaonline.net>:
Two things to consider: (1) reversing the current *does*
"dissolve" the
Pd
surface,
True, but extremely slowly I believe. A Pd anode is known to
dissolve
relatively fast in acidic electrolytes such as D2SO4, but I don't
think that's what they used. It is doubtful whether they reverted
the
current long enough to dissolve more than a few atomic layers.
I think the experimenters were competent. They knew what they were
doing.
Using a Faraday constant of 96,485 C/mol, and (conservatively) a
valence of
4, n for moles produced, I for current = .2 A, t for time = 1 s,
we get:
n = I * t / (96,485 C/mol * 4)
n = (0.2 A)*(1 sec) / (385940 C/mol) = 5.182x10^-7 mol
This means that at 200 mA/cm^2, 5.182x10^-7 mol/s is removed, or
3.12x10^17
atoms per second.
We also have for Pd: (12.38 g/cm^3)/(106.42 g/mol) = 0.1163 mol/
cm^3 =
7.006x10^22 atoms/cm^3. The atomic volume is 1.427x10^-23 cm^3,
and the
atomic dimension is 2.426x10^-8 cm. The amount of Pd removed per
second is
(3.12x10^17 atoms per second) * (1.427x10^-23 cm^3 per atom) =
4.45x10^-6
cm/s, or 445 angstroms per second. The number of layers of atoms
removed is
(4.45x10^-6 cm/s)/(2.426x10^-8 cm) = 183/s.
If this is correct (highly suspect! 8^), then at a current density
of 200
mA/cm^2 we have a thickness of 183 atoms removed per second, or 445
angstroms per second.
This would be correct if palladium, when driven as an anode, did
dissolve in an alkaline electrolyte (they classically used LiOD in
that M4 experiment, according to their original report at
http://newenergytimes.com/v2/archives/1998epri/TR-107843-V1.PDF ,
thanks to Steve Krivit for the link), which it doesn't, see the Pd/H2O
Pourbaix diagram at
http://www.platinummetalsreview.com/jmpgm/data/datasheet.do?
record=532&database=cesdatabase
which shows that such corrosion only occurs in an acidic
electrolyte (pH <3).
It has been pointed out to me privately that hydrogen charge
transport has to be accounted for as well, i.e. that hydrogen
evolution reduces the effective "corrosion" current. However, since
the reversed current cleaning process was carried out in part to
degass the Pd, I expect the hydrogen contribution to the positive
surface charge of the Pd anode would be extremely diminished in the
latter part of this cleaning process.
Well, this is indeed an interesting electrochemical problem. My
experience is that nothing, including platinum, totally avoids anodic
corrosion if there is a current present. Passification works in part
by eliminating the current at the potential at which the passivation
is occurring, or less, i.e. by building an insulating layer. I do
not think passification of *highly loaded* Pd is possible. The
evolving hydrogen would prevent accumulated oxidation of the Pd
surface. I have done various passivation experiments (not with Pd
though) and my experience has been that passification takes
considerable time, even for metals that are not loaded with hydrogen,
and once it does occur, the current is highly reduced. Further, if a
constant current source is used then the voltage rises to the point
where the passified surface is breached.
Beyond that, and this is a fairly irrelevant point I know, I think Pd
corrodes as an anode in the presence of current in neutral Ph salt
electrolytes.
The EPRI article states: "They accomplished loading with a
combination of initial low cathode current densities of ~20-50 mA/
cm2, followed by current ramps up to ~1.0 A/cm2. Current reversals to
deload or “strip” the cathodes of D and clean the surface by
temporarily making it an anode resulted in high loadings."
It seems to me the Pd would be dissolving during the deloading
process when the current is reversed. Also, apparently my estimate
of 200 mA/cm^2 was too low - it was probably 1 A/cm^2.
It would be interesting to actually do an experiment with Pd wire,
loading and then reversing the current repeatedly for a long period
and then weighing the wire.
It seems to me the experimenters would not have gone thorough this
procedure if the current reversal did not actually clean the
electrode surface, i.e. expose a pure Pd surface. A fully passified
surface would not be effective at loading hydrogen as a cathode
because it would not even be conductive. If pure Pd was exposed to
the electrolyte as an anode it seems to me certain that Pd was being
dissolved in the process.
One thing I take to be self evident to anyone who has practical
experience with electrochemistry experiments. If you have current
through an anode then *some* of that anode is going into solution,
and that includes platinum. I think that is far more true with metals
like Pd or Ni.
What I find interesting is the lack of comment about what this
implies. If the effective fusion zone was indeed fully dissolved by
the current reversal then the energy produced was *way over* that
which would be expected by D + D --> 4He + 24 MeV.
And this is exactly what must be expected if heavy element LENR is
occurring. The average energy produced per deuterium atom consumed
should well over 10 MeV. See (warning this is a 2 MB report) "Report
A - Energetically Feasible Aneutronic X + n D* --> Y + Z Reactions, n
= 1 to 12 - Creating Stable Isotopes Y and Z With No Weak Reactions",
pages 404-438, located at:
http://www.mtaonline.net/~hheffner/RptA.pdf
For a shorter report showing the energy released *just in forming
compound nuclei with Pd*, in the process of cluster fusion, see:
http://www.mtaonline.net/~hheffner/PdFusion.pdf
The above report was created based on and in a format to be
compatible with that used by Ed Storms and Brian Scanlan in: Storms,
E and Scanlan, B 2010, “What is real about cold fusion and what
explanations are plausible”, To be published by AIP:
http://www.lenr-canr.org
which has the important information regarding heavy element LENR in Pd.
Now, most important, note that any reactions that are not *nuclear
catalytic*, i.e which do not create helium from fission of the
compound nucleus, *do not create helium atoms*. Therefore all the
energy produced per D atom consumed can not be reflected in the
helium atom count. If the energy produced by D + D --> 4He is fixed,
then the energy created by heavy element transmutation must add to
the energy per helium atom. The observed energy must be larger than
24 MeV/4He, and depending on the overall experiment environment, i.e.
the tendency to create heavy element LENR vs fusion, it could be much
larger.
What is of further interest, and this to me is miraculous, is the
fact that radioactive nuclei do not typically result, despite the
fact that many of the initial compound nuclei are necessarily
radioactive, and many have radioactive daughter products. This, to
me, rules out explanations that merely involve neutron exchanges, or
resonant tunneling of deuterons. I think this necessitates electrons
in the compound nucleus that prolong the reactions, and produce an
initially highly de-energized nucleus, in particular as shown
(approximately) on page two of:
http://www.mtaonline.net/~hheffner/PdFusion.pdf
I should also note that the EPRI article demonstrates the sparseness
and low certainty of data regarding energy/4He ratios. Due to the
cost and complexity of helium measurement experiments, the number of
experiments is few, and due to repeatability problems as well, no
highly reliable energy/4He number is available in my opinion. The
"RESULTS OBTAINED" section merely states that "Taking these results
at face value, the excess heat and He formation are apparently
connected, via a multiplier of several tens of MeV/He." "While not
definitive due to high background levels, the data obtained in
experiment M4 are consistent with a quantitative, but temporally
displace correlation between excess heat and helium-4 production."
This is characteristic of the helium measurement problem. Quantities
in large excess of background are required, because confidence
intervals grow very large when counts are near background levels.
Speaking of that, it would be nice to know if the brackets shown in
the Violante graphics represent 3 sigma or one sigma confidence
intervals on the helium measurements.
Best regards,
Horace Heffner
http://www.mtaonline.net/~hheffner/
