Author Topic: First Colloidal Silver run today  (Read 11116 times)

Offline kephra

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Re: First run today
« Reply #45 on: August 16, 2015, 01:25:17 PM »
Well, its really not important what the byproduct is, as what is important is that it reduces silver.  It is a curiosity though.  The fact that gas is given off means that the reaction that is supposed to happen does not, or that a secondary reaction is occurring.  I do not have the means to investigate that further.  Unless and until I do, it remains a curiosity.
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Offline kephra

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Re: First run today
« Reply #46 on: August 16, 2015, 01:59:22 PM »
Quote
The color that is reflected by the beam is different with each.
Some reflect different shades of blue in the beam and some reflect white or grey in the beam
The reflected color come from large silver particles.  Blue is caused by particles smaller than the wavelength of green and red light.  Gray is from particles at least as large as the wavelength of red light.  Nano sized particles do not reflect any visible light.
The reflected color is not caused by any side reaction byproducts as they would be dissolved and atomic sized.
There is the unknown and the unknowable.  It's a wise man who knows the difference.

Offline RickinWI

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Re: First run today
« Reply #47 on: August 16, 2015, 05:11:47 PM »
Which do you think would be better? A smaller amount of grey light reflected or a larger amount of blue light reflected.

Since it is larger silver particles that are doing this reflecting then I assume those particles are probably orders of magnitude larger in size from 14 nm.

(The type of turbidity I am talking about can be clearly seen in the pics that Sancho posted in reply # 19 of this thread [pg. 2]. In pics though you really can't see the difference in color of the beam like you can in person.)
So many VARIABLES & so little TIME.

Offline PeterXXL

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Re: First run today
« Reply #48 on: August 16, 2015, 06:24:06 PM »
Thats all well and good, but cannot be the case.  You have to account for the gas production which your formulas do not.


Maybe it's so that when the silver oxide is reduced to silver nanoparticles with glucose, that glucose will also be oxidized to gluconic acid. But since we have an excess of glucose in the solution, that extra glucose will in fact be reduced with hydrogen to glucitol (sorbitol), so that we actually both have some glucose to be oxidized and some to be reduced, and as result we will both have silver nanoparticles, gluconic acid and glucitol (sorbitol) in the colloidal silver solution.

Offline PeterXXL

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Re: First run today
« Reply #49 on: August 16, 2015, 11:05:59 PM »
Thats all well and good, but cannot be the case.  You have to account for the gas production which your formulas do not.


Maybe like this...

First reaction (silver ions are reduced by Glucose, while Glucose is oxidized to Gluconic Acid):

Silver Oxide + Glucose => Silver + Gluconic Acid

Ag2O + C6H12O6 => Ag2 + C6H12O7

And in water Gluconic Acid hydrolyses in equilibrium with the cyclic ester Glucono Delta-Lactone (but higher pH and heat makes more Gluconic Acid):

C6H12O7 <=> C6H10O6 + H2O

Second reaction (Silver will be a catalyst for the reaction of Gluconic Acid AND Glucono Delta-Lactone to a sugar alcohol and another monosaccharide):

Silver (as catalyst) with Gluconic Acid will react to the sugar alcohol Xylitol plus Carbon Dioxide:

Ag2 + C6H12O7 => C5H12O5 + CO2

Silver (as catalyst) with Glucono Delta-Lactone will react to the monosaccharide Deoxyribose and Carbon Dioxide:

Ag2 + C6H10O6 => C5H10O4 + CO2

But due to the heat and higher pH, very little of the last reaction will occur.

Offline kephra

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Re: First run today
« Reply #50 on: August 16, 2015, 11:18:31 PM »
Cool!  That could do it.  Originally, I found the reaction in a book by Pigman from 1948.  I no longer have access to the book, so cant look it up again. 
There is the unknown and the unknowable.  It's a wise man who knows the difference.