Conversion of AgCl via NaOH and Karo Question

[Yves_Michel]

Butcher's equation is: "2AgCl (s) + 3NaOH (aq) + C6H11O6 --> 2Ag(s) + 2NaCl (aq)+ C6H11O7 Na (aq)"
My question is how many electrons the C6H11O6 (glucose or fructose) the aldehyde group gives? Correct me if I am wrong but is not suppose to be only one. Therefore we have 2 Ag+ that need one electron each the equation should be written "2AgCl (s) + 3NaOH (aq) + 2C6H11O6 --> 2Ag(s) + 2NaCl (aq)+ 2C6H11O7 Na (aq)" ? That is doubling the amount of the glucose.

Yves

 

[butcher]

2AgCl (s) + 3NaOH (aq) + C6H11O6 --> 2Ag(s) + 2NaCl (aq)+ C6H11O7 Na (aq)

Those are moles of units of silver and moles of glucose...
one mole of silver is 107.8 grams of silver, so in the equation we are talking about converting two moles of silver or 215.6 grams of silver metal from the silver chloride, using one mole of C6H11O6 karo syrup.

so we need more than one electron to react with the silver chloride to convert almost 7 troy ounces of silver metal, and one mole of glucose should contain enough electrons to handle the job.

the molecular weight of one mole of C6H11O6
C =12g/mol
H = 1g/mol
O = 16g/mol
we have
C6 (6x12g/mol = 72g for 6mol carbon)
H11 (11x1g/mol = 11g for 11mol hydrogen)
O6 (6x16g/mol = 96g for 5mol oxygen)
or
72+11+96=179 grams

NaOH
Na = 22.9g/mol
O = 16g/mol
H = 1g/mol
NaOH =40g/mol
3NaOH = 120grams

so 179 grams of glucose needed to react with 2moles of silver in the silver chloride.

silver chloride
AgCl
Ag = 107.86g/mol
Cl = 35.4g/mol
AgCl =143.3g/mol
We have two moles of silver chloride so 2 x 143.3g =286 grams of silver chloride in the formula.

So from the equation we see that two moles of silver chloride 286 grams, and 179 grams of glucose and 120 grams of sodium hydroxide will give us about 7 troy ounces of silver metal (with billions of atoms in their shell all with a full shells of electrons.

 

[g_axelsson]

Butcher, that was a really detailed and exhaustive answer... sadly it's wrong and not what was asked. The question was in other words "To convert 2Ag⁺ to 2Ag you need two electrons. One comes from C₆H₁₁O₆ that turns into C₆H₁₁O₇^-+Na⁺, where does the second electron come from?"

Well, maybe your answer wasn't totally wrong, but at least it contains an error.

2AgCl (s) + 3NaOH (aq) + C6H11O6 --> 2Ag(s) + 2NaCl (aq)+ C6H11O7 Na (aq)

Both sides contains only neutral compounds so obviously the electrons should balance out, but does the elements balance?

2AgCl + 3NaOH + C₆H₁₁O₆ --> 2Ag + 2NaCl + C₆H₁₁O₇Na

2 Ag on both sides, 2 Cl on both sides, 3 Na on both sides, 9 O on left but 7 on right, 14 H on left but 11 H on right, 6 C on both sides.
You have forgotten 3 H and 2 O on the right hand side, sounds like an H₂O and OH (Without charge! :shock: ).

I suspect that the equation is more like (Where the aldehyde group -H-(C=O) > turns into a carboxylic acid (-COOH)) and the Na is actually Na⁺ associated with OH^-
2AgCl + 3NaOH + C₆H₁₁O₆ --> 2Ag + 2NaCl + C₆H₁₁O₇ + H₂O + NaOH

A second step could turn the carboxylic acid group (-COOH ) and the NaOH into (-COONa) and H₂O so the total equation is (changes in red)

2AgCl + 3NaOH + C₆H₁₁O₆ --> 2Ag + 2NaCl + C₆H₁₀O₇Na + 2H₂O

The answer to the question is : The change in oxidation state of the 2Ag⁺ reduced into 2Ag is powered by the aldehyde group is oxidized into carboxylic acid by an oxygen atom.

Those are moles of units of silver and moles of glucose...

It doesn't matter if we are talking about electrons and atoms or moles of electrons and atoms. The equations must balance in the end.

Disclaimer : I'm a mathematician and physicist so my chemical knowledge is nothing but a couple of google searches and wikipedia, but I can count and spot an error. So I'm standing by my statement that there is an error but my solution could be just as wrong or even worse. Time to let the chemists check my reasoning.

Göran

 

[butcher]

g_axelsson,
Thanks, I was only trying to show there were a lot of atoms and electrons in the equation, and it was a formula of moles, so there are definitely more than one electron.

I believe this is more of a chemical reaction,used to reduce silver back to metal not just an electron exchange.

 

[butcher]

Actually I am still unsure how these long carbon chains would react, and what the actual byproduct from the reaction would be.
http://en.wikipedia.org/wiki/Glucuronic_acid
http://en.wikipedia.org/wiki/Category:Sugar_acids
http://en.wikipedia.org/wiki/C6H12O6
http://www.chemspider.com/Chemical-Structure.23253896.html
https://www.google.com/#q=C6H11O7

 

[g_axelsson]

g_axelsson,
Thanks, I was only trying to show there were a lot of atoms and electrons in the equation, and it was a formula of moles, so there are definitely more than one electron.

I believe this is more of a chemical reaction,used to reduce silver back to metal not just an electron exchange.

We never use formulas of moles, it is just chemical formulas. If we are using single atoms, single moles or thousands of atoms as an unit is of no consequence, it's just a scale factor.

If we create water from oxygen and hydrogen we use the formula 2H₂ + O₂ -> 2 H₂O. This is the most basic formula of this equation, but nothing says anything about the scale of the reaction, if it involves atoms or moles. A mole is just a large number, 6.02214129×10²³, but instead of mole you can use any number you like... or if you really insist on counting with moles...

1.204428258x10²⁴ H₂ + 6.02214129×10²³O₂ -> 1.204428258x10²⁴ H₂O

If you are missing an electron in the simple formula, then you are missing a mole of electrons if you scale it up to moles. Just because there are a lot of electrons doesn't mean it is enough of electrons.

By the way, 99% of chemistry is electron transfers, the rest is folding and chirality.

Göran

 

[moose7802]

I need to start understanding how these equations work. Could someone point me in the direction of a basic chemistry book that I could start learning this from? Sorry for the outside post!

Tyler

 

[butcher]

I often use formulas of moles all the time, the formulas and equations of moles, so that I have an idea of how much of one chemical I need to react with another chemical to get the reaction I am trying to achieve.

I tried to find the formula for glucose sugar conversion of silver chloride in a caustic solution, from this paper the byproduct is thought to be, or presumed to be Gluconic acid (C6H11O7), but they did not really know for sure either.
I do not know but the gluconic acid may also be converted to sodium gluconate (NaC6H11O7).

2AgCl + 3NaOH + C6H12O6 --> 2Ag + C6H11O7.Na +2NaCl + 2H2O

http://library.deerfield.edu/pdfs/ChemAgAckerman.pdf

http://en.wikipedia.org/wiki/Glucose

http://en.wikipedia.org/wiki/Gluconic_acid

http://en.wikipedia.org/wiki/Sodium_gluconate


Basically to me, just by looking at this the sodium is taking the chlorides from the silver, this may be where most of the electron transfer occurs although the glucose carbon compound is reduced also in the reaction,this looks like somewhat of a replacement reaction assisted by the reduction of the sugar, I cannot explain what is actually happening here, I have enough trouble trying to learn and understand just a little bit of the inorganic chemistry, and have not studied the organic chemistry, of these long chain carbon compounds like sugar, I will leave that to the biochemist,or chemist who learn it in school, I have enough trouble just trying to get the silver back from the silver chloride.

edit word from inorganic to organic

 

[samuel-a]

I think the equations above are over simplified, thus making it hrad for us to wrap our mind around the reaction dynamics.

I believe there are several reactions taking place, some in sequential order and some simultaneously:
- Silver Chloride is converted to Silver Oxide by adding Lye.
- Glucose react: 1. with lye solution to form Sodium Gluconate 2. with Silver oxide to produce Gluconic acid and Silver metal
- Sodium Gluconate reacts with Silver Oxide to form Silver metal and H2/CO2

The process seems to produce vast amounts of H2 and/or CO2 gases, and i see no mentioning of that in Ackerman's report... Though there's might be other side reactions i can't think of.

To be honest, i couldn't find any reliable source who can give us a direct answer as to what exactly going on in this reaction.

 

[butcher]

I think this mystery will probably continue at least unless we learn more about the complicated organic chemistry.

I believe samuel-a is correct on the reaction of silver chloride forming an silver oxide in solution as an intermediate reaction, and here in our formula with karo syrup we see we also have excess sodium hydroxide (1 mole excess) (after Ag2O forms) for the sugar to react with the excess sodium hydroxide in solution to react with the silver I oxide that was formed.

2AgCl + 2NaOH --> Ag2O + 2NaCl +H2O


In our formula for silver chloride reduction using sodium hydroxide and glucose (karo syrup):
2AgCl +3NaOH + C6H12O6 --> 2Ag + NaC6H11O7 +2NaCl + H2O

In looking at the glucose reduction we see the glucose gives up an H in its chain and gains an O in its place.
I assume with the reaction of that extra mole of NaOH and the Ag2O.

I could not draw the chain here, but if you look at the pictures here (wiki link below),for glucose, then for sodium gluconate we see this.
We Know the silver gives up it chloride, and sodium ends up with it as salt water, we also assume the glucose sugar forms sodium gluconate in solution.

http://en.wikipedia.org/wiki/Glucose
http://en.wikipedia.org/wiki/Sodium_gluconate

 

[butcher]

Again with no knowledge of organic chemistry, some of these things have me wondering and wanting to learn more. While confusing me more, the more I try to figure it out.

Apparently glucose (sugars) contains an aldehyde group.

I wonder if this does not play a major part in converting silver chloride to metal when using glucose (Karo syrup and NaOH method of reduction of silver chloride to silver) similar?
Similar to the formaldehyde reduction methods?
Or
Something similar to oxidation of formyl oxxidation in making mirrors with formaldehydes?

The formyl group readily oxidized in an oxidation reaction is the basis of the silver mirror test. In this test, an aldehyde is treated with Tollens reagent, which is prepared by adding a drop of sodium hydroxidesolution into silver nitrate solution to give a precipitate of silver(I) oxide, and then adding just enough dilute ammonia solution to redissolve the precipitate in aqueous ammonia to produce [Ag(NH3)2]+ complex. This reagent will convert aldehydes to carboxylic acids without attacking carbon-carbon double-bonds. The name silver mirror test arises because this reaction will produce a precipitate of silver whose presence can be used to test for the presence of an aldehyde.

A further oxidation reaction involves Fehlings reagentas a test. The Cu2+ complex ions are reduced to a red brick coloured Cu2O precipitate.
https://www.google.com/#q=aldehyde+group+reduction+silver+salts
Aldehydes are strong reducing agents and reduce the metal ion and are oxidised in the process
i.e. RCHO + [O] ==> RCOOH
Reduction of silver(I) ion to silver metal
RCHO + 2Ag+ + H2O ==> RCOOH + 2Ag + 2H+
Or
Reduction of copper(II) to copper(I) i.e. the blue solution of the Cu2+ complex changes to the brown/brick red color of insoluble copper(I) oxide Cu2O.
RCHO + 2Cu2+ + 2H2O ==> RCOOH + Cu2O + 4H+
Oxidation sugars:
http://butane.chem.uiuc.edu/pshapley/GenChem2/B7/book.pdf
http://en.wikipedia.org/wiki/Reducing_sugar


I am sure there is a simple explanation as to what happens in our reaction of reducing silver chloride to metal with glucose (karo syrup) and sodium hydroxide.
But I do not see it yet.
This mystery still alludes me.

 

[Lou]

Butcher,

To answer your question:

What does the reduction is the aldehyde (http://en.wikipedia.org/wiki/Aldehyde) functionality on a sugar. Simple sugars with an aldehyde group are called aldoses ( http://en.wikipedia.org/wiki/Aldose ). More complex sugars that have an aldehyde or can form one via hydrolysis (basically OH- from caustic attacking the glycoside bond at the anomeric carbon, breaking the hemiacetal by ring opening) are called reducing sugars ( http://en.wikipedia.org/wiki/Reducing_sugar ).

This is the chemical basis for the Tollens' test, where Ag₂O oxidizes that aldehyde into a carboxylic acid.

This means that the better sugars to use for this process are simple, small sugars that are easily oxidized and do not have to undergo hydrolysis to make the reducing part of it accessible.

Generally, it is preferred to use simple sugars rather than plain aldehydes like formaldehyde because of the ease in handling, and their inherent low toxicity.

A simple effective sugar for silver reduction is:
glucose/dextrose

These sugars polymerize as well to yield active reducing sugars.

 

[samuel-a]

Lou,

Can you breake it down into simple redox equations?
Considering side reactions (if there are any), what would the stoichiometric ratio be?
What would be the optimal water volume per mole AgCl to accommodate the reaction?


On a side note, what is your take on reducing Agcl with Sodium borohydride (other than the fact it is expensive)?

 

[butcher]

Lou, thank you very much for taking the time out of your busy day to add helpful information to this topic.


The organic chemistry is a real mystery to me, I begin to get real confused with so many H’s and O’s tied up in those long chains attaching themselves to those carbons, and the funny drawings of those complicated circles, and all of those big long funny names, and trying to follow what those scientist who write about them are trying to say.

I see where glucose C6H12O6 can be expressed as C6 (H2O)6, showing the hydrolyzed carbon.

I was just wondering if sodium gluconate NaC6H11O7could not be expressed something like this?
My presumption NaC6(H2O)5 (HOO)
(If some knows the correct way to express sodium gluconate showing it’s hydrolyzed carbon formula (something like this) I would like to know how it is written properly).

If so this kind of shows where in reduction, one of the (H2O) in glucose carbon chain, becomes HOO replacing a hydrogen in its chain with an oxygen in the glucose reduction to gluconate, I also understand the (H2O) is not water, and is just used expressed in this manner to show hydrolyzed carbon (sugar).

Another assumption on my part would be writing the formula something like this:

AgCl + 3NaOH + [C6(H2O)6] --> 2Ag + [Na C6(H2O)5 HOO] + 2NaCl +2H2O

I wonder if this would be correct to express the formula for reduction of silver using karo syrup (glucose) and sodium hydroxide this way.

 

[g_axelsson]

I wouldn't write a formula with (H₂O) as a grouping, that implies that water is an integral part of the molecule. By writing H₂O it implies that the hydrogen atoms is connected to the oxygen atom. In the case of glucose it is apparent that no H₂O molecules are involved in the molecule, any oxygen is either part of an OH group or double bounded to a carbon atom.

Disclaimer: I'm writing this under the influence of C₂H₅OH

Göran

 

[butcher]

hydrolysis of carbon does implies that water is an integral part of the molecule.
hydrolysis the breaking of water H+and OH- bonds.
Although the bonds with carbon are no longer water, when bonded to the carbon chain.
Cn(H2O)n is just another way to show the hydrolysis of carbon in a formula.

Glucose can be written as C6H12O6 or as C6(H2O)6 either is correct, neither of these show how the water is broken or how the H or OH are in bond with the carbons in the chain.


I was just unsure of the proper way to express sodium gluconate as NaCn(H2O)n in the formula.

I thought also by expressing the equation in this fashion:
AgCl + 3NaOH + [C6(H2O)6] --> 2Ag + [Na C6(H2O)5 HOO] + 2NaCl +2H2O
We may be able to see easier where the H in the chain is replace by an O, as written in a chemical formula expressing the reaction.

 

[Lou]

Butcher:

The stoichiometry of the reaction is dependent on if one is using a sugar that doesn't have a free aldehyde; if it's a disaccharide, you will need an equivalent of base to open it up, above and beyond what is needed for the conversion of AgCl--> Ag2O.* Based off of Butcher's mechanism he kindly provided, you'll need just two equivalents of base alone for the aldehyde.

For the reduction, it's one electron (Ag+ --> Ag(0)) so it will take one equivalent of silver for every equivalent of sugar. It will take 1 equivalent of caustic soda for every two equivalents of silver. Then one should add an excess of caustic to help bring the Ag2O into solution more as its soluble hydroxide complex, which helps shuttle silver into the aqueous phase where it is reduced.

@Samuel_a:

Borohydride is a great, nonselective reducing agent. It's what I call a kitchen sink reductant, because it goes after pretty much every metal above -700 mV. Issue with it is, it's transient and hydrolyses quite rapidly in acidic milieu. Basic solutions like Dow's Venmet, are 12% solutions in 50% v/v NaOH, and store for years. Unlike N2H4 which only works well on PGMs (with perhaps the exception of the easy to reduce Pd) when there is plenty of base in solution, borohydride works best from pH 1-4, optimally at 2-2.5. The reduction is instant, and quantitative when excess is used.

As it is capable of a 4 electron reduction and has a low molecular weight, it's very efficient on a gram for gram basis. Nickel and copper (which are reduced by borohydride, to their respective borides, which then hydrolyze and reduce further to the metal). I don't know how expensive the powdered material is, but 5 gallons of 12% material is about $300 and will reduce hundreds of thousands of dollars in PGMs. I think it would be a waste to do with silver.

I don't even know why people are so caught up on the sugar reduction. Makes much more waste. I think it's preferable to just make the silver oxide (which won't quantitatively give you your silver, it is slightly soluble, particularly in excess base) and filter that, then melt with borax/carbonate.

 

[samuel-a]

Thanks Lou,

I too share your thought about the actual need to use the sugar reduction. But i do use it in cases when time doesn't permit the long long evaporation -> drying -> decomp. of Ag2O. This is just my point of view according to my specific setup capabilities.
I really like how the silver powder comes down heavy and chunky after dextrose reduction and is quick to dry too. I usually never even wait for it to dry completly and put it damp in a hot crucible and let it steam out (important to say, i never add damp powder to an already molen metal).


As to the NaBH4, i can get dry powder (reagent grade) for 35$/Kg.
I'm not quite sure what the reaction is exactly,b but i deduct it will cost me appx. 14$ per Kg (assuming an 15% excess) of silver reduced.
Unless i did the calculation wrong I agree it's quite borderline for Silver refining usage.

 

[butcher]

Although from a different reaction (Tollens reagent) reduction of a silver complex, this shows the aldehyde group of glucose (portion of the carbon chain)being converted to gluconic acid as silver is reduced from the complex.
http://www.walkingitaly.com/tuserg/tuserg_metallo/D-Tollens-e.htm

 

[Lou]

You're getting borohydride for that cheap? I'll be glad to buy some of it--Dow has been ripping me off!


I don't think it'll cost that much.


37.83 g/mol gives four electron-mol reducing power.

So 0.03783 kg will (theoretically: there are side reactions) reduce about 0.431 kg of Ag.

More like $5/kg Ag at the top end.

Like I said, we don't use for silver (between pyro and cell, no need to use any hydrometallurgy).