[PeterXXL]

Hi!

I have a question regarding the oxidation of monosaccharides when reducing silver oxide in ionic form (aqua solution).

A soluble form of Silver Oxide (Ag₂O) as 10 ppm (10 mg / liter) is created with electrolysis of Silver (Ag) in distiilled water with a small amount of Sodium Carbonate (Na₂CO₂) for the purpose as electrolyte and to raise the pH to 8.0 (to open the D-Glucosis ring).

Enough of D-Glucose (C₆H₁₂O₆) is also added prior to the electrolysis in order to reduce the Silver Oxide.

The result should be that the Silver Oxide is reduced to Silver NanoParticles and the D-Glucose is oxidized to D-Gluconic Acid (C₆H₁₂O₇).

To be sure that all the silver oxide is fully reduced, it's necessary to add somewhat more D-Glocose than necessary for the reduction of the Silver Oxide, so what will then happen will that extra D-Glucose? I assume that it will be reduced to D-Glutinol (C₆H₁₄O₆) by the released H.

I have it confirmed that all the Silver Oxide is fully reduced to Silver NP's, so the question is about the D-Glucosis.

D-Glucitol  <--- Reduction --- D-Glucose --- Oxidization ---> D-Gluconic acid 
C₆H₁₄O₆             +  2H              C₈H₁₂O₆          +  O                  C₈H₁₂O₇

Questions:

Will the reduced Silver Oxide oxidize the equivalent D-Glucosis to D-Gluconic Acid?

Will the extra D-Glucosis be reduced to D-Glucitol?

[Dan]

You need to watch your typos here - glutinol is a terpene (unlike glucitol) and glucosis is not a compound - did you mean glucose?

To be sure that all the silver oxide is fully reduced, it's necessary to add somewhat more D-Glocose than necessary for the reduction of the Silver Oxide

Yes, in preparative chemistry a slight excess of one reagent is almost always (deliberately) added. In this reaction Ag₂O is the limiting reagent.

so what will then happen will that extra D-Glucose? I assume that it will be reduced to D-Glutinol (C₆H₁₄O₆) by the released H.

The "H" released from what? What reducing agent do you propose will react with the glucose?

Will the reduced Silver Oxide oxidize the equivalent D-Glucosis to D-Gluconic Acid?

What do you mean by "the equivalent"? Are you asking if 1 mol Ag₂O reduces 1 mol of glucose? Start by writing a balanced chemical equation for the process.

Will the extra D-Glucosis be reduced to D-Glucitol?

By which reducing agent?

[PeterXXL]

Thanks for replying Dan!

Sorry for the typos (I'm unable to edit/modify my post to correct this)...

I'm referring to the monosaccharide Glucose, or more exactly D-Glucose. And I'm referring to the sugar alcohol Glucitol, or more exactly, D-Glucitol (which commonly is known as Sorbitol).

If D-Glucose is reduced, the result should be D-Glucitol (Sorbitol), and the reduction is with H, as two free hydrogen atoms will be added to each molecule of Glucose for this reduction to take place.

Glucose is a common reducing sugar, and for so called "green synthesis" of nanoparticles, glucose is the best choice, both because so little is needed, and because when glucose is oxidized in the process (of reducing silver ions), the result is gluconic acid, which is a biological natural ingredient.

My question is however, will the excess of Glucose in the pH 8.0 water solution (which is full of OH^-) be reduced or not. And if it's reduced, will it be reduced to Glucitol (Sorbitol), or to what?

[Dan]

My question is however, will the excess of Glucose in the pH 8.0 water solution (which is full of OH^-) be reduced or not. And if it's reduced, will it be reduced to Glucitol (Sorbitol), or to what?

My question to you is: if the excess glucose were to be reduced to to glucitol, what would the reducing agent be that participates in that process? Where do you envisage the "H" coming from?

[PeterXXL]

My question is however, will the excess of Glucose in the pH 8.0 water solution (which is full of OH^-) be reduced or not. And if it's reduced, will it be reduced to Glucitol (Sorbitol), or to what?

My question to you is: if the excess glucose were to be reduced to to glucitol, what would the reducing agent be that participates in that process? Where do you envisage the "H" coming from?

During the electrolysis under heating and constant stirring, and with added sodium carbonate, there will be enough of both H⁺ and OH^- in the water, which theoretically could act as reducing agents, especially since Ag will act as a cathalyst for chemical reactions. Just my humble thoughts.

Or, maybe the reaction will be as follows...

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.

I just want your professional reflection about the above, as I want to be sure exactly what the result will be beside the Ag-NP that for sure are formed with an estimated size of 5 - 20 nm (based on color and clearance).

[Babcock_Hall]

H⁺ is not a reducing agent; there are no electrons.

[PeterXXL]

H⁺ is not a reducing agent; there are no electrons.

Yes, I know as all + ions lacks electrons and all - ions have an excess of electrons. So here only OH^- could be considered.

But forget about that now, and re-read my post #4.

Q1) Are the reactions there correct, or?

Q2) What happens with the excess of Glucose (or any reducing sugar) in a solution where we both create Ag₂O from electrolysis of Ag in a pH 8.0 (added 1-mol Na₂CO₃ til we get pH 8.0) solution at a temp. around 90°C (Ag  2AgOH  Ag₂O + H₂O) where the Ag₂O reduction then is reduced with a reducing sugar, like Glucose (Ag₂O + C₆H₁₂O₆    Ag₂ + C₆H₁₂O₇)?