On Nov 26, 2009, at 6:55 AM, Michel Jullian wrote:
Horace,
2009/11/26 Horace Heffner <hheff...@mtaonline.net>:
<snip>
Here is the original explanation, less the garbled indicator test
information:
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
- - -
...
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!
<snip>
Well no, the site where the hydrogen was generated (which was the
cathode BTW, not the anode,
Oh yes. That was a typo. I actually do know hydrogen is generated
at the cathode in an electrolytic cell! 8^)
let's call it the negative electrode
rather, as anode and cathode names switch sides when current direction
is reverted) was surrounded by OH- ions, and the site where O2 was
generated prior (which was the anode, let's call it the positive
electrode from now on) was surrounded by H3O+ ions. Therefore it can't
be a case of "more H2 where H2 was already bubbling and more O2 where
O2 was already bubbling", agreed?
Michel
Agreed! For my scenario to be a valid explanation the polarity shown
at the meter would have to change. It doesn't change.
Interesting! So it appears there there has to be a reversal of ion
flow in the electrolyte, the ions meeting in the middle and
recombining. The H3O+ leaving the interface frees up electrons
trapped on the other side of the electrolytic cell cathode
interface. It remains a source of electrons for the meter.
Similarly, the OH- leaving the vicinity of the electrolytic cell
anode essentially leaves a net positive charge there to accept
electrons. It would be interesting to see what an indicator like
phenolphthalein would show when the battery is disconnected. There
would be an immediate current in the correct direction due to a
roughly 0.2 F/m^2 capacitance of the double layer.
I don't know what size the wire is, but guessing at 0.5 mm diameter,
that is 1.57 mm circumference, by 130 mm height, that's 1.99x10^-4
m^2 per wire or about 4x10^-4 m^2 total area, and thus (4x10^-4 m^2)
(0.2 F/m^2) = 8x10^-5 F, which at 9V can only support a charge of
7.2x10^-4 coulombs. I estimated the need to drive 2 microamps
current to register 2 V on the meter, which is about (7.2x10^-4
coulombs)/(2x10^-6 coulombs/sec) = 360 seconds. It looks like
interface capacitance discharge may actually account for the
current. The cell could be merely acting as a capacitor.
It would be interesting to see what the "charging" time is - i.e. to
compare discharge time to charge time.
I'd like to see what happens to the bubbles when the battery is
disconnected. If it really is a fuel cell it should be possible to
bubble O2 and H2 (from another cell) around the separate wires and
get a sustained current.
It would also be interesting to connect two half cells together by an
electrolyte bridge and remove the bridge prior to disconnecting the
battery. No current should flow at all except for a brief rebalancing
of charges due to the 9 V potential difference.
This looks like an interesting high school science project.
Best regards,
Horace Heffner
http://www.mtaonline.net/~hheffner/
