Author Topic: Amount in mg of Maltodextrin for chemical reduction of Gold ions  (Read 1034 times)

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Offline PeterXXL

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Kephra. At http://www.cgcsforum.com/index.php?topic=757.0 last section. you wrote...

"1/2 teaspoon maltodextrin", which is an unspecified amount in weight.

The molar weight of Maltodextrin is between 900 and 3200, so...

Reducing 40 mg Gold requires...

Weight in mg for what should be reduced / Molar weight of what should be reduced x Molar weight of reducing agent...

40 / 196.96 x (900 to 3200) = 183 to 650 mg.

And since the gold ions need 3 electrons, and we assume that 1 molecule of maltodextrin only releases 1 electron, we need 3 times that weight of maltodextrin...

183 to 650 x 3 = 549 to 1950

And if we use the higher value and round it up, it will be 2000 mg (2 gram). In volume that is about 1 / 3 to 1 / 2 teaspoon. Is that how you ended up with "1/2 teaspoon maltodextrin" ?



Offline kephra

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Re: Amount in mg of Maltodextrin for chemical reduction of Gold ions
« Reply #1 on: July 25, 2015, 10:43:27 AM »
Yes,  I don't have a 1/3 tsp measure, and more carbo-gain does not hurt anything.
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Offline PeterXXL

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Re: Amount in mg of Maltodextrin for chemical reduction of Gold ions
« Reply #2 on: July 25, 2015, 10:59:04 AM »
Ok, thanks.


Also, isn't it possible that polysaccharides like Maltodextrin can release 1 electron for each glucose molecule in the chain, or is it always that only one electron is released?

Offline kephra

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Re: Amount in mg of Maltodextrin for chemical reduction of Gold ions
« Reply #3 on: July 25, 2015, 11:53:59 AM »
Maltodextrin and similar polysaccharides have only 1 reducing site in the chain no matter how long the chain. The free aldehyde group on the last glucose molecule (or first depending on perspective) is the reducing part. 

With silver:
Silver Oxide + Glucose --> Silver metal + Xylitol + Carbon Dioxide
 OR
Silver Oxide + Glucose --> Silver metal + gluconic acid  (this is the generally accepted reaction)
Which would react with the sodium hydroxide to form sodium gluconate and water.

This implies 2 electrons.

Ag2O + C6H12O6 --> Ag + C5H12O5 + CO2

However, the linear structure exists in equilibrium with the ring structure.  The ring structure does not have a free aldehyde, so it cannot reduce silver.  Therefore, only some of the sugar molecules are reducing agents.

Colloidal Silver is only a bargain if you make it yourself.

Offline PeterXXL

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Re: Amount in mg of Maltodextrin for chemical reduction of Gold ions
« Reply #4 on: July 26, 2015, 03:50:27 AM »
Thanks Kephra :)


But isn't Glucose ( C6H12O6 )oxidized to Gluconic Acid ( C6H12O7 ) ?

Ref: https://en.wikipedia.org/wiki/Gluconic_acid    Quote: "Gluconic acid, gluconate salts, and gluconate esters occur widely in nature because such species arise from the oxidation of glucose" End quote


Maltodextrin and similar polysaccharides have only 1 reducing site in the chain no matter how long the chain. The free aldehyde group on the last glucose molecule (or first depending on perspective) is the reducing part. 

With silver:
Silver Oxide + Glucose --> Silver metal + Xylitol + Carbon Dioxide
 OR
Silver Oxide + Glucose --> Silver metal + gluconic acid  (this is the generally accepted reaction)
Which would react with the sodium hydroxide to form sodium gluconate and water.

This implies 2 electrons.

Ag2O + C6H12O6 --> Ag + C5H12O5 + CO2

However, the linear structure exists in equilibrium with the ring structure.  The ring structure does not have a free aldehyde, so it cannot reduce silver.  Therefore, only some of the sugar molecules are reducing agents.

Offline kephra

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Re: Amount in mg of Maltodextrin for chemical reduction of Gold ions
« Reply #5 on: July 26, 2015, 06:21:48 AM »
Quote
But isn't Glucose ( C6H12O6 )oxidized to Gluconic Acid ( C6H12O7 ) ?
I used to think so, but it doesn't fit the experimental observation.
Every time I make colloidal silver with glucose based agents, it evolves a gas as evidenced by the bubbles which collect on the sides of my bottles.  Where does the gas come from?  It cannot come from making gluconic acid, so something different must be occurring.

There is no gas evolved if the glucose oxidizes to gluconic acid, so that reaction cannot be correct under the conditions we use to make colloidal silver.

To observe the gas production, make 20 ppm ionic silver cold, add corn syrup, heat to about 140F, and set the bottle aside.  30 minutes later, you will see tiny bubbles collected on the sides of the bottle.

But whether you make xylitol (or one of its cousins) or gluconic acid, the number of electrons transferred is the same.

Its also possible that both are true, and that the gluconic acid is an intermediate reaction.  I have no way to tell.  I have seen the reaction leading to C5H12O5 listed as a possible reaction in a journal though.

Edited to add:
I refound the reference to the oxidation of glucose resulting in CO2.  In the book "Chemistry of the Carbohydrates" by William Ward Pigman.
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