On Nov 25, 2009, at 1:05 PM, Michel Jullian wrote:
Horace,
My comments below, some things are still wrong
2009/11/25 Horace Heffner <[email protected]>:
Gad. It still isn't right! Corrections below. I have vertigo at
the
moment and can't think straight. I've actually done half of this
experiment, though decades ago, and it is interesting how the
concentration
gradient wanders, it doesn't follow what you would expect for any
kind of E
field.
On Nov 23, 2009, at 2:48 AM, Michel Jullian wrote:
See: http://sci-toys.com/scitoys/scitoys/echem/fuel_cell/
fuel_cell.html
I had no idea an ultraclean rechargeable battery could be done so
simply!
Supplies:
<<- One foot of platinum coated nickel wire, or pure platinum wire.
Since this is not a common household item, we carry platinum coated
nickel wire in our catalog.
- A popsickle stick or similar small piece of wood or plastic.
- A 9 volt battery clip.
- A 9 volt battery.
- Some transparent sticky tape.
- A glass of water.
- A volt meter.>>
It seems to me a small amount of lye would help the reaction
along. No
matter, the intent is apparently not to create a working cell,
i.e. generate
power, it is merely to generate a voltage.
I see they sell the wire for $14.41 plus shipping. A bulk source
for wire
and mesh might be:
http://www.gerarddaniel.com/
H2 and O2 are produced by short electrolysis runs, after which the
bubbles clinging to the electrodes are catalytically recombined
by the
electrode surface material (platinum) to generate electricity :)
1/ The article features nice "explanations" of how it works, but how
does it _really_ work? In particular, in the generating (fuel cell)
phase, they don't say what makes the positive hydrogen ions climb
"uphill" from the negative electrode to the positive one, anyone can
explain this miracle? ;-)
2/ It seems to me a much higher capacity (and perhaps even
practical)
rechargeable battery could be made by using a hydrogen
absorbing/desorbing material e.g. Pd for the negative electrode, and
by making gaseous oxygen available at the anode. Storing the
latter is
not required of course, O2 from the air is fine... maybe a floating
support which would keep a grid or flat serpentine shaped positive
electrode at the surface of the water or just below?
Michel
The explanation looks bogus to me. I think the cell works by
reversible
reactions, not recombination.
Bockris states that conduction in an electrochemical cell in the
volume
between the interface layers is almost entirely due to concentration
gradients. That is because almost all the potential drop is in the
interface
layers themselves. The E field in the bulk of the cell is very
small.
I expect the cell actually operates by creating even *more*
bubbles, not
consuming the gas already there in the form of bubbles.
In the course of the brief electrolysis by battery, the volume of
water
around the *anode* is preferentially filled with H3O+ ions, as the
OH- ions
release their electrons and form O2 and H2O2, and the volume
around the
*cathode* is filled with OH- ions as the H3O+ ions present at the
cathode
surface are electrolyzed. This can actually be viewed by use of a
dilute
electrolyte, plus a pH indicator like phenolphthalein, which is
colorless in
acidic electrolytes, and pink in basic solutions. To do this
first add the
(liquid) phenolphthalein to distilled water. Connect the
battery. To view
the creation and migration of OH- ions: add a little bit of boric
acid to
the water, and stir. Repeat the process until you can see the
electrolyte
turns pink in the vicinity the *cathode* (- electrode) once the
electrolyte
settles down. Boric acid was chosen because it is commonly
available from
pharmacies. To view the creation and migration of H3O+ ions add a
little
bit of lye to the water and stir. Repeat the process until you can
see the
electrolyte is pink, but when the electrolyte settles down you can
see the
volume around the *anode* (+ electrode) gradually turing clear. It
can take
a little fooling around with concentrations to get the effect to work
quickly and dramatically. The diffusion occurs slowly but at a
clearly
visible pace.
I agree with the above paragraph now, but putting it right has broken
your explanation for the generating phase two paragraphs below.
This is the same principle I had in my original explanation.
Here is the original explanation, less the garbled indicator test
information:
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
"I think the cell works by reversible reactions, not recombination."
"In the course of the brief electrolysis by battery, the volume of
water around the anode is filled with H3O+ ions, and the volume
around the cathode is filled with OH- ions."
"You can demonstrate the reversibility of the reactions by reversing
the battery. Note, however, that the diffusion occurs in a somewhat
random manner."
"In any case I doubt it is actually recombination that causes the
potential at the electrodes. It is the presence of the high
concentration of ions in solution that makes the residual potential
when the battery is disconnected. The H3O+ ions take on electrons
through the wire originally releasing hydrogen at the site where the
hydrogen was generated, the anode, thus making *more* hydrogen
bubbles. Similarly, the OH- ions donate electrons to make H2O2 and
*more* O2 at the site where O2 was generated prior."
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
Still looks right to me, despite the fact I remain dizzy!
You can demonstrate the reversibility of the reactions by
reversing the
battery. Note, however, that the diffusion occurs in a somewhat
random
manner. It doesn't typically blossom out in a perfectly spherical or
cylindrical manner (depending on the electrode shape). Reversing the
reaction is thus not a perfect process either. I tried some of
this decades
ago in a feeble attempt to make a display technology. I got a nice
red
stream of ions coming from a copper anode in a basic solution.
In any case I doubt it is actually recombination that causes the
potential
at the electrodes. It is the presence of the high concentration of
ions in
solution that makes the residual potential when the battery is
disconnected.
The H3O+ ions take on electrons through the wire originally
releasing
hydrogen at the site where the hydrogen was generated,
No, the ions there are OH- ions
No, the ions that remain there in excess when the battery is
disconnected are H3O+ ions. The anode takes away the electrons from
the OH- radicals, thus consuming them. This leaves an excess of H3O+
ions there near the anode. Thus a + potential remains there, at
least for a while. It can even draw a small current against it.
the anode, thus
making *more* hydrogen bubbles. Similarly, the OH- ions donate
electrons to
make H2O2 and *more* O2 at the site where O2 was generated prior.
No, the ions there are H3O+ ions.
No, the ions remaining in excess around the cathode when the battery
is disconnected are OH- ions. Thus a - potential remains there at the
cathode, at least for a while. Likewise, it can even draw a small
current against it. Together the two environments make for a very
feeble short lived battery. Some of the ions make it back through
the solution by diffusion, but the shortest path for balancing
charges is through the wire. When acting as a battery, the vicinity
of the wire quickly becomes ion neutral and thus the largest
diffusion gradient are right near the wire, along with the highest
concentration of the right kinds of ions to drive the battery.
Want to give it another try? :)
No, that's my story and I'm sticking to it! 8^)
Michel
The meter is probably a 10 megohm meter, meaning registering the 2 V
potential requires generating 0.2 microamps of current, and thus 0.4
microwatts of power. Not much of a fuel cell!
It would be interesting to run the current for a while until a
significant
concentration gradient can be visualized, and then disconnect the
battery to
see what effect the current generated through the meter has on the
visible
gradients.
Note that the concentration gradients of H3O+ and OH- particles
does not
necessarily require an E field to maintain them, provided there
are other
radicals in the electrolyte. Salt buffers can be used to increase
conductivity without driving Ph to extremes. The presence of
additional
radicals can balance the charges to neutral, or to match the E
field in the
electrolyte. For example, if the electrolyte is NaOH, the Na+ can
redistribute to neutralize the charges. If boric acid is used, the
B(OH)4-
radical will balance the charges.
Best regards,
Horace Heffner
http://www.mtaonline.net/~hheffner/
Best regards,
Horace Heffner
http://www.mtaonline.net/~hheffner/
