Mike Monett wrote:

> Re: CS>basic dumb question
> From: Marshall Dudley
> Date: Fri, 28 Jan 2005 10:08:31
> http://escribe.com/health/thesilverlist/m77263.html
>
>   > Here is  a reference that shows the solubility of AgOH to  be more
>   > than 100 ppm if the ph is less than 9!
>
>   > How the heck do you get known good information when  the "experts"
>   > don't seem to be able to agree?
>
>   > Marshall
>
>   Silver hydroxide is insoluble in dw.
>
>   Take two  12 ga electrodes spaced 0.050 inches apart.  Fill  a glass
>   with dw to a wetted length of 4 inches. Apply 670uA/sq.in. current.
>
>   Here are the observations.
>
>   Fri Mar 25, 2005, 05:30:21 am
>
>   Setup Parameters
>   ~~~~~~~~~~~~~~~~
>   volume               : 300ml
>   initial conductivity : 1.1uS
>   wetted length        : 4.63 inches
>   constant current src : 775.0 uA
>
>   Observations
>   ~~~~~~~~~~~~
>          Time/Date            Cell Voltage
>
>   Fri Mar 25, 2005, 05:41:11 am 4.0V
>   Fri Mar 25, 2005, 05:42:00 am 2.4V
>   Fri Mar 25, 2005, 05:44:58 am 2.5V <- cathode is black
>   Fri Mar 25, 2005, 05:48:00 am 2.5V <- anode starting to get black
>
>   Terminated the run.
>
>   Dw final conductivity is 1.0uS
>
>   The side  of  the  cathode facing the  anode  is  coated  black. The
>   material is very soft and easily wiped from electrode. A  shiny spot
>   appeared on  the  tissue after rubbing  the  electrodes. Examination
>   under a  stereo  zoom  microscope showed  a  patch  of  black powder
>   embedded in the tissue. The central portion was shiny. Several other
>   shiny areas were apparent under the microscope.

Yep, done that myself, the black powder is finely divided silver which will
be black. This can be easily confirmed by applying pressure to the powder,
and since silver is very mallable, the powder will be forced together forming
a larger particle which will regain it's silver appearance.

>
>
>   During the  short run, no Tyndall effect was visible in the dw  in a
>   dark room.  It  was not possible to check  for  Tyndall  between the
>   electrodes since they were spaced so closely that the laser beam hit
>   one or the other and the brilliant reflection made it  impossible to
>   see anything.
>
>   The tissue  was  placed in a 3oz glass and  covered  with  H2O2. The
>   black areas disappeared in a few minutes.

Yep, done that myself. As I have reported before H2O2 will oxidize silver
metal producing silver oxide which dissolves quite readily.

>
>
>   The test took 410 seconds. Silver hydroxide appeared on  the cathode
>   at 226  seconds.

No, what appears on the cathode is silver metal being precipitated out.

When current is applied to pure water the Ag+ goes toward the cathode and a
hydrogen is released at the cathode and the remaining OH- goes toward the
anode. At that point you have silver hydroxide in the water.  As electrolysis
continues then some of the OH- makes it to the anode, and some of the Ag+
makes it to the cathode.  At the cathode the Ag+ gains an electron and
precipitates out as very finely grained particles which will be black. At the
anode the OH- will combine with a second OH-, lose two electrons to the
anode, and produce H2O and O.  The O being very reactive will tend to react
with the silver making a tan colored silver oxide at the anode.

So after a while we will have metallic silver power on the cathode, and
silver oxide on the anode.  The colors are correct for these two and so is
the chemistry.  Now if we reverse the polarity, the H+ will go to the cathode
which was the anode before. The hydrogen will react with the silver oxide on
the cathode, and produce water and silver metal again. The fine silver metal
powder on the new anode will form Ag+ ions and go toward the cathode. So
reversing the polarity tends to minimize buildup but swapping the reducing
elecrode with the oxidizing electrode.

So, since the cathode is receiving H+ and Ag+ ions, it is absolutely
impossible for silver hydroxide to form at the cathode.  If there were any
silver hydroxide at the cathode, the H+ would immediately react with it
forming water and silver atoms.


> The amount of silver liberated from  the  anode in
>   this time  was  0.655ppm,  but not all of  it  was  involved  in the
>   production of silver hydoxide (See calculations at the end.)
>
>   Conclusions
>   ~~~~~~~~~~~
>   Silver ions  released  from  the anode  quickly  reached  the nearby
>   cathode.

That is why one should use stirring.

>
>
>   Silver hydroxide  formed in the Nernst Diffusion layer  next  to the
>   cathode:
>
>   Ag(+) + OH(-) --> AgOH

Uh, you have silver hydroxide to start with, the silver from the anode and
the OH- from the cathode form a solution of silver hydroxide.  This is
throughout the liquid, not just at the cathode. At the cathode the Ag+
reaches the cathode, gains and electron and becomes silver metal again.

>
>
>   Some of  the  particles stuck to the cathode  and  formed  a visible
>   black layer.

Yes, the finely powdered silver will form a black layer.

>
>
>   Pressure during rubbing decomposed the hydroxide to silver metal.
>

No, there is no silver hydroxide there.  It is silver metal to start with and
pressure is just turning it into a large ingot which will appear silver.

>
>   H2O2 dissolved the hydroxide and silver metal back to ions.

No H2O2 reacted with the silver metal producing silver oxide, which
dissolves.

>
>
>   The solubility  of  silver  hydroxide   is  less  than  0.655ppm and
>   probably can be taken as zero.

If the solubility of silver hydroxide were less than .655, then it would be
impossible to produce ionic silver any greater than that. Since 20 ppm of
silver hydroxide can be produced via the EIS method, the solubility must be
at least that much.

Marshall

>
>
>   --------------------------------------------------------------------
>   Calculations
>
>   Unit Conversions
>   ~~~~~~~~~~~~~~~~
>   Coulombs     = I * Seconds            ; total number of Coulombs
>   CoulombsGram = 107.868 / 96485        ; Coulombs per gram of silver
>   ElectronsSec = I / 1.60217733e-19     ; electrons per second
>   Grams        = CoulombsGram * I * Seconds ; Faraday's equation
>   Hours        = Seconds / 3600
>   IonsPerNano  = IonsSqInSec / 6.45e14  ; ions per square nanometer/sec
>   IonsSqInSec  = ElectronsSec / SquareIn    ; ions per sq. in. per sec
>   Litres       = 3.785 * Gallon         ; convert gallons to litres
>   Litres       = Millilitres / 1000     ; convert millilitres to litres
>   Milligrams   = Grams * 1000           ; convert grams to milligrams
>   Millilitres  = 29.57 * Ounces         ; convert ounce to milliliters
>   Minutes      = Seconds / 60           ; minutes
>   NumberIons   = ElectronsSec * Seconds ; number of silver ions
>   SquareIn     = 0.25 * Length          ; convert length to SquareIn
>   uASqIn       = 1e6 * I / SquareIn     ; current density in uA per sq in
>   uS           = Milligrams / Litres    ; 1 uS is 1 milligram per litre
>   uSPerHr      = uS / Hours             ; uS per hour
>
>   Input Parameters
>   ~~~~~~~~~~~~~~~~
>   Hours       = .063
>   I           = 775e-6
>   Length      = 4.63
>   Millilitres = 300
>
>   Solution
>   ~~~~~~~~
>   Coulombs    = 0.175
>   Grams       = 0.000
>   Hours       = 0.063
>   Litres      = 0.300
>   Milligrams  = 0.196
>   Millilitres = 300.0
>   Minutes     = 3.780
>   Ounces      = 10.14
>   Seconds     = 226.8
>   SquareIn    = 1.157
>   uASqIn      = 669.5
>   uS          = 0.655
>   uSPerHr     = 10.39
> ----------------------------------------------------------------------
>
> Mike Monett
>
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