by MarkinAustralia » Sat Jul 29, 2006 5:06 am
Hi Alsaka and others,
I have been quiet for a while but thought I would paste this document by a friend who is doing some interesting work in this area.
Electrostatically Assisted Hydrogen Electrolysis
(Greg Watson 15 May 2006)
There are two electrochemical reactions taking place in the conventional electrolysis cell:
1) Oxidation is occurring at the anode
2 H2O (l) ® O2 (g) + 4 H+
(aq) + 4 e-
2) Reduction is occurring at the cathode
2 H2O (l) + 2 e- ® H2 (g) + 2 OH-
(aq)
To keep the numbers of electrons balanced, the cathode reaction must take place, twice as
much as the anode reaction. If the cathode reaction is multiplied by 2 and the two reactions
are added together we get:
6 H2O (l) + 4 e- ® 2 H2 (g) + O2 (g) + 4 H+
(aq) + 4 OH-
(aq) + 4 e-
Note the 4 H+
(ag) and the 4 OH-
(ag) which remain in the water. If we allow enough time after
the removal of power (both electrodes) the H+ and OH- combine in the water to form H2O
and cancel species that appear on both sides of the arrow, we get the overall net reaction:
Net: 2 H2O (l) ® 2 H2 (g) + O2 (g)
So during the application of power we get the creation of both H2(g) and O2(g) bubbling up
as well as H+
(ag) and OH-
(ag) in the water. As the number of + and – ions in the water are
balanced and there is no change in the Redox of the water.
However if we just disconnect the positive lead from the power supply, we stop the OH-
(ag)
from donating an electron via the anode. BUT the electron rich negative cathode lead is still
attached, causing the H+
(ag) to be electrostatically attracted to the cathode, gladly accepting
an electron and thus forming H(ag) which then combines with another H(ag) to form H2 (g)
which then happily bubbles to the surface. Note this electron comes from the negative
connection to the power supply. No positive connection is needed for this to happen. This
electron rich source can be electrostatic in nature.
As this process is reducing the number of H+
(ag) charges in the water, the Redox of the water
shifts negative due to the unreacted OH-
(ag) in the water which has no electron poor (+) place
to donate the excess electrons to. This OH-
(ag) also supports improved electrical conductivity
during subsequent positive power pulses to the anode and increases the amount of seeding
H+
(ag) and OH-
(ag) in the water. The ideal power “On” pulse length and applied voltage
remain to be determined.
It would appear that to generate a Redox negative state (much more OH-
(ag) than H+
(ag))
requires the application of pulsed DC to the cell with much longer “Off” period than “On”
periods. If too long a delay occurs between the “On” pulses the H+
(ag) and OH-
(ag) in the
water may naturally recombine to form H2O. If too short a delay then insufficient time is
allowed to attract H+
(ag) ions to the cathode, accept an electron and form H2 (g).
So during the DC pulse “On” period we get the generation of H+
(ag) ) and OH-
(ag) and during
the “Off” period we get 2 H+
(ag) + 2 e- ® H2 (g) with the solution being forced to Redox
negative due to the unaffected OH-
(ag).
While this reaction is not OU, it does suggest the electron source applied to the cathode
during the power off cycle may be from a negative electron rich electrostatic charge and
thus it may be possible to generate Hydrogen from atmospheric electrostatic charges instead
of using conventional power.
Kind reagrds
mark
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