# What is exactly chemical process and why gold drops?



## Alentia

There are 3 main precipitants most people on this forum use.

After analyzing some of the actions, there are following "reasons" for gold to drop:

1. SMB - does not drop gold itself, but rather creates a chemical reaction to release SO2, which in fact as described in Hoke's book, and Na replaces the Au with Cl in the solution

Na2S2O5+AuCl+?=>Au+NaCl+Na(?)+2SO2 (something wrong here, not sure what make up here)

2. Oxalic - what exactly does it do in the solution?

H2C2O4 + AuCl +H2O =>

3. Copperas, it is actually seems to use the same principle as SMB 
FeSO4+AuCl => SO2 + Au + FeCl + O2 (not sure if I did correctly) with a bit of a difference Fe replaces Au with Cl

What does oxalic do during chemical reaction as it seems to be carbonated water?
Can anyone correct formulas if possible?


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## lazersteve

Check out my post with the various gold reactions:

Gold Equations

The entire thread is a great read also.

The oxalic acid gives up it's hydrogen to make HCl (with the chlorine from the gold chloride) and carbon dioxide (CO2) which makes the fizz. The gold chloride gives up the chlorine and becomes the solid metallic gold.

Steve


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## butcher

Dissolving gold:

Au + HNO3 + 4 HCl --> HAuCl4 + NO + 2 H2O

Or 

Gold in nitric
Au + 3 NO3 + 6 H --> Au3(+) + 3 NO2 +3 H2O
And then in HCl
Au3(+) + 4 Cl(-) --> AuCl4

Or

Au + 3 HNO3 + 4 HCL --> HAuCl4 + 3 NO2 + 3 H2O

Or 

Just aqua regia
HNO3 + 3HCl --> NOCl + Cl2 +H2O

Or
Another just aqua regia:
4 H(+) + 3CL(-) + NO3 --> NOCl + Cl2 + 2H2O 


Some chemical formula’s using reagents to precipitate gold:


Sodium bisulfite and gold in aqua regia:
2 HAuCl4 + 2 NaHSO3 --> 2 Au + 8 HCl + Na2SO4 + SO2

Sodium metabisulfite in gold solution:
Na2S2O5 + H2O --> 2 NaHSO3 
Notice we actually make sodium bisulfite
Then 
2 AuCl3 + 3 NaHSO3 + 3 H2O --> 2 AU + 6HCL + 3 NaHSO4

Ferrous sulfate (iron sulfate) (copperas) (FeSO4) and gold in chloride solution:
2 AuCl3 +6 FeSO4 --> 2 AU +2 Fe2(SO4)3 + Fe2Cl4


Oxalic acid and gold solution:
2 HAuCl4 + 3 H2C2O2 --> 2 Au + 8 HCl + 6CO2

Correction in above formula:
Oxalic acid and gold solution:
2 HAuCl4 + 3 H2C2O2 --> 2 Au + 8 HCl + 6CO2
Or
2AuCL3 + 3 H2C2O2 --> 2 Au + 6 HCl + 6 CO2 

So as you can see from above there can be several way’s to express a chemical reaction with a written formula, and as many times the real reaction is much more complicated than the formula we write to express the reaction there can also be a lot more going on inside that chemical flask, and many products also can depend on conditions of the reaction involved, and in the real world we may have other things involved like base metals in our solutions, so it is actually more complicated than what we show with these simple-xified formulas.


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## MysticColby

butcher said:


> Sodium metabisulfite in gold solution:
> Na2S2O5 + H2O --> 2 NaHSO3
> Notice we actually make sodium bisulfite
> Then
> 2 AuCl3 + 3 NaHSO3 + 3 H2O --> 2 AU + 6HCL + 3 NaHSO4





Alentia said:


> 1. SMB - does not drop gold itself, but rather creates a chemical reaction to release SO2, which in fact as described in Hoke's book, and Na replaces the Au with Cl in the solution
> 
> Na2S2O5+AuCl+?=>Au+NaCl+Na(?)+2SO2 (something wrong here, not sure what make up here)



these can't both be true, can they?
Alentia's reaction wouldn't account for the HCl in butcher's reaction...
I found this on the internet (complete version of Alentia's reaction):
Na2S2O5 + H2O -> 2Na + 2 HSO3
HSO3 + H -> SO2 + H2O

(Then: AuCl3 + 3Na -> Au + 3NaCl)

My question: if the gold is precipitated because sodium takes it's place with the chloride, wouldn't SMB also precipitate other metal salts that happen to be in solution? (like copper chloride) Of course, copper should be removed before digesting the gold, but I'm just curious of the mechanism and possibilities (can SMB be used to precipitate silver nitrate?).


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## lazersteve

Butcher,

Two small corrections:

1) your equation for oxalic should read 6HCl, not 8 HCl.
2) in the same equation your formula for Auric Chloride should be AuCl3

The link I provide has some other gold dissolving and precipitation equations as well.

Steve


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## butcher

OOPs

Thanks good eye Steve,
Forgot the hydrogen in the acidic solution,
Could blame it on all timers but I have had it all my life.

Correction this should be right:

Oxalic acid and gold solution:
2 HAuCl4 + 3 H2C2O2 --> 2 Au + 8 HCl + 6CO2

Look better?

Or Steve suggested formula could also be proper in more dilute solution.

2AuCL3 + 3 H2C2O2 --> 2 Au + 6 HCl + 6 CO2 

Sorry about that will correct above post in color.


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## Alentia

Butcher, Steve,

Thank you very much. It does make much more sense looking at the formulas and understanding what is going on in the solution at specific moment.

Till this moment I did not realize after gold precipitation we actually have HCl as end product which is saturated by other metals. 

Question on SMB if

2 AuCl3 + 3 NaHSO3 + 3 H2O --> 2 AU + 6HCL + 3 NaHSO4

where that SO2 smell is coming from?


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## butcher

Alentia, and MysticColby,

Where in the heck is the sulfur dioxide SO2 gas, well I thought I knew until I tried balancing these formulas, and All I can get is sulfur trioxide SO3 gas.

now you have me asking the same question. 

MysticColby, 

"My question: if the gold is precipitated because sodium takes it's place with the chloride, wouldn't SMB also precipitate other metal salts that happen to be in solution? (Like copper chloride) Of course, copper should be removed before digesting the gold, but I'm just curious of the mechanism and possibilities (can SMB be used to precipitate silver nitrate?)."


The way I thought I understood this is the gold is reduced by SO2 sulfur dioxide gas generated in the reaction, 
But further looking I believe it to to be sulfur trioxide gas is formed. Now I am getting confused. I thought Ihad learned it formed SO2 sulfur dioxide gas in our solution.
MysticColby, Problem is you have me thinking and trying to balance my equations.and from this I am seeing SO3 sulfur trioxide is formed, ( I am getting confused now).

Sodium metabisulfite in gold solution:
Na2S2O5 + H2O --> 2 NaHSO3 

Maybe if I showed the equation, balanced, like this and accounted for the gas that escapes the reaction.
Na2S2O5 + H2O --> 2 NaHSO3 
Notice we actually make sodium bisulfite
Then 
2 AuCl3 + 3 NaHSO3 + 3 H2O --> 2 AU + 6HCL + 3 NaHSO4 
And the gas evolved:
2NaHSO4 --> Na2SO4 + SO3 + H2O


Sodium bisulfite in gold solution:
is also shown :
2HAuCl4 + 2 NaHSO4 --> 2 Au + Na2SO4 + 8 HCl +SO2 (gas) ???


My problem is after looking at this it actually looks like SO3 is formed????

I may have to spend more time scratching my Pea size Brain, but I believe the SO3 gas is actually formed from this not the SO2 as I thought.

Now I wonder if these also come into play here?
sulfur in the +4 oxidation state, sulfur dioxide is a reducing agent It is oxidized by halogens to give the sulfuryl halides, such as sulfuryl chloride.
SO2 + Cl2 → SO2Cl2 
Sulfur trioxide also reacts with sulfur dichloride to yield the useful reagent, thionyl chloride. 
:SO3 + SCl2 → SOCl2 + SO2 
I am starting to get lost in this and either need to study this more later getting too tired now and getting nowhere on it..


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## samuel-a

butcher said:


> OOPs
> 
> Thanks good eye Steve,
> Forgot the hydrogen in the acidic solution,
> Could blame it on all timers but I have had it all my life.
> 
> Correction this should be right:
> 
> Oxalic acid and gold solution:
> 2 HAuCl4 + 3 H2C2O2 --> 2 Au + 8 HCl + 6CO2
> 
> Look better?
> 
> Or Steve suggested formula could also be proper in more dilute solution.
> 
> 2AuCL3 + 3 H2C2O2 --> 2 Au + 6 HCl + 6 CO2
> 
> Sorry about that will correct above post in color.



Butcher

It is now balanced and indeed a theoretically possible redox reaction.
But it is worth mentioning that the reduction action called for in most litrature is the reduction with the use oxalate (or Hydrogenoxalate HC2O4-) anions of Na/K/NH4.
With good reason for it too.... if you add just Oxalic acid to gold solution and boil, nothing will happen... a complete reduction could take months or even years.
Hoke was wise enough to choose Ammonium Oxalate, the reason for it is simple as it's more solubale then the corresponding Na / K salts.

you can look at it as:

the Hydrogenoxalate of Na -
3NaHC2O4 +﻿ 2HAuCl4 = 2Au + 3NaCl + 6CO2 + 5HCl
or as the fully neutralized salt -
3Na2C2O4 +﻿ 2HAuCl4 = 2Au + 6NaCl + 6CO2 + 2HCl

edit: spelling


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## butcher

Thanks Sam, I am gonna rest my head all this is making it throb like a drum.


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## samuel-a

butcher said:


> Thanks Sam, I am gonna rest my head all this is making it throb like a drum.




I hear ya !
Writing, balancing and understanding equations makes my head swell...


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## MysticColby

I think the first thing to make clear: When you add AR to gold, does it form AuCl3 or HAuCl4?
it sounds like steve thinks it's AuCl3:


lazersteve said:


> 2) in the same equation your formula for Auric Chloride should be AuCl3


What is HAuCl4? when is it made?



butcher said:


> Maybe if I showed the equation, balanced, like this and accounted for the gas that escapes the reaction.
> Na2S2O5 + H2O --> 2 NaHSO3
> Notice we actually make sodium bisulfite
> Then
> 2 AuCl3 + 3 NaHSO3 + 3 H2O --> 2 AU + 6HCL + 3 NaHSO4
> And the gas evolved:
> 2NaHSO4 --> Na2SO4 + SO3 + H2O


I think this might be close to right. the last line is the only part I think might be wrong
From wikipedia: Na2SO4 + H2SO4 ⇌ 2NaHSO4
So, going the other way, if sodium bisulfate is in solution, it will form sodium sulfate and sulfuric acid (will reach an equilibrium)
And what about Alentia's reaction, where SMB + water = sodium + sulfur dioxide + water, and sodium + gold chloride = gold + sodium chloride? does that ever happen?
Are you sure there is actually a sulfur smell (SO2)? Maybe you're smelling HCl and sulfuric acid? acids are VERY piercing, painful, smells while sulfur would smell like rot.
I came across this thread because I tried refining gold for the first time, and was hit with an unexpected whiff of unpleasantness soon after adding SMB. I would consider that smell more acid than sulfur...

I WANT TO ADD: please do not go smell your solution. inhaling acids is bad for you. it was a mistake that I did, but we learn from mistakes. I just brought it up because it sounded like you had alread smelled a sulfur smell, so I was asking to clarify that memory of the smell.


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## lazersteve

From wiki:

Auric Chloride

and

Chloroauric acid

The SO2 is formed in situ and is can only be smelt after the solution saturates with SO2.

The equation for the SO2 gas with Gold III Chloride is :

3SO2 + 2AuCl3 + 3H2O = 3SO3 + 6HCl + 2Au

This reaction only occurs as a step in the overall process or if you combined sulfites with water or acid:

Sodium Sulfite with water to Sodium Bisulfite:

Na2SO3 + H2O = 2NaHSO3

Sodium Bisulfite with AuCl3 to Sodium Bisulfate:

3NaHSO3 + 2AuCl3 + 3H2O = 3NaHSO4 + 6 HCl + 2Au

SMB with Acid: 

Na2S2O5 + 2 HCl → 2 NaCl + H2O + 2 SO2 

Steve


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## Alentia

> SMB with Acid:
> 
> Na2S2O5 + 2 HCl → 2 NaCl + H2O + 2 SO2
> 
> Steve



This looks like where SO2 is coming from. When there are free HCl molecules released during gold precipitations SO2 is formed and should theoretically enhance precipitation or acting as a catalyst?


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## butcher

Gold is at the bottom of the reactivity list of metals, it does not oxidize easily (loss of electrons), it will not corrode or dissolve in any one acid like HNO3 nitric acid, or HCl acid hydrochloric acid, but it will in combination of these two acids, the HNO3 acts as oxidizer (to assist the gold atom in giving up its electron) it gives this electron (reluctantly in these acid combination) to the HCL hydrochloric acid, this forms a soluble salt of gold in solution (a gold chloride soluble salt).

We need to give back the gold its missing electrons to get back our gold as metal from this solution (precipitate the metal gold), we can do this with several methods.

By adding a metal higher in the reactivity series which will gladly give up its electrons to reduce (gain of electrons) the gold, in doing so this added metal lost its electrons to the gold so it replaces the gold in solution, as a soluble metal salt, this metal lower in series can be a metal like copper or zinc or many other metals, or it can sometimes come from a dissolved salt of a metal in solution (like iron dissolved in H2SO4, called copperas or ferrous sulfate FeSO4), this would be considered a replacement reaction.

We can also reduce gold from solution by giving the gold missing electrons in a chemical redox reaction, using a chemical like SO2 gas, sodium bisulfite or sodium metabisulfite, and so on, here the chemical reactions give gold back the missing electrons, through a chemical reaction.

And as we know once we reduce (gain of electrons) the gold to metal (powders), the gold metal is not soluble in acids like HCl, or H2SO4.

But as we also know this reduced gold metal can have some chloride salts with it, this gold powder was reduce from being a chloride in solution, so if we added nitric acid we would again dissolve some of our gold back into solution, as the acid give some of its hydrogen to the chloride salt and again makes HCL in this HNO3 solution, so again our gold gives up the electrons, this is why we incinerated the gold if we came from a chloride solution and go back to a nitrate solution and do not wish to dissolve the gold , which sometimes we may do if we have base metals we missed in first treatments of our gold.


We could also lower the pH with sodium hydroxide and reduce the gold, to form a precipitant of gold hydroxide:

AuCl3 + 3NaOH --> Au(OH)3 + 3NaCl

Here we did not reduce gold to elemental metal but precipitated it as a hydroxide, this gold hydroxide can be dissolved if we used too much caustic soda (NaOH), to form NaAuO2 in solution.

Also if we heated gently the Au(OH)3 we could form gold II oxide Au2O3 and water vapors, gently heating further we would again form gold metal.

Strong Hydrogen peroxide can also reduce gold from solutions, as well as some carbon or carbon-based organics, these I cannot comment on how, as I have not research them much.


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## lazersteve

Great description of the reactions of gold in solution Butcher.

I would like to add a side note:

Strong bases like sodium hydroxide will not only precipitate gold hydroxide from solutions, but it can also precipitate nearly all other metals from the same solution as hydroxides. This leads to a mixture of metal hydroxides in the solids and is not selective for gold. To put it another way, your gold will need refining again.

Steve


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## MysticColby

lazersteve said:


> 3SO2 + 2AuCl3 + 3H2O = 3SO3 + 6HCl + 2Au
> Na2SO3 + H2O = 2NaHSO3
> 3NaHSO3 + 2AuCl3 + 3H2O = 3NaHSO4 + 6 HCl + 2Au
> Na2S2O5 + 2 HCl → 2 NaCl + H2O + 2 SO2



Thanks for clearing it up, steve! From what you said, I think these are the two chains that lead from gold chloride to gold (amounts adjusted so molar amounts are the same from one to the next):
3 sodium sulfite (Na2SO3) + 3 water (H2O) = 6 sodium bisulfite (NaHSO3)
3 sodium metabisulfite (Na2S2O5) + 3 water (H2O) = 6 sodium bisulfite (NaHSO3)
6 sodium bisulfite (NaHSO3) + 4 auric chloride (AgCl3) + 6 water (H2O) = 6 sodium bisulfate (NaHSO4) + 12 Hydrochloric (HCl) + 4 gold
6 sodium bisulfate (NaHSO4) ⇌ 6 sodium sulfate (Na2SO4) + 6 sulfuric acid (H2SO4)

3 sodium metabisulfite (Na2S2O5) + 6 Hydrochloric (HCl) = 6 sodium chloride (NaCl) + water (H2O) + 6 sulfur dioxide (SO2)
6 sulfur dioxide (SO2) + 4 auric chloride (AgCl3) + 6 water (H2O) = 6 sulfur trioxide (SO3) + 12 Hydrochloric (HCl) + 4 gold

It depends on the speed of each reaction for which is more prevalent, but I would guess the first is more, as there is probably more water than HCl. Final products look to be gold, HCl, sodium bisulfate, sodium sulfate, sulfuric acid, sulfur trioxide, sodium chloride, and any unreacted sodium bisulfite and sulfur dioxide.
A shame this isn't a simpler reaction like cementing :| 
From all of this, I'm expecting SMB to NOT precipitate silver from silver nitrate (sodium is never formed). I do wonder if this would precipitate copper from copper chloride (and other chlorides), though. Does it?

AuCl3 and HAuCl4... even more complicated. But it looks like the specific one doesn't matter much:
Au + AR = AuCl3
AuCl3 + HCl + heat ⇌ HAuCl4


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## butcher

Steve,
Thank you for mentioning that the other metals in solution would also be converted to Hydroxides, if they were in solution, I was just trying to show an example where gold was not reduced to a metal, and was not suggesting the use of hydroxide in refining.

But I can see where someone reading it may get an idea, of an easy way to get there gold from a solution.

Also thank you for helping with the formula’s and equations.


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## lazersteve

butcher said:


> Steve,
> Thank you for mentioning that the other metals in solution would also be converted to Hydroxides, if they were in solution, I was just trying to show an example where gold was not reduced to a metal, and was not suggesting the use of hydroxide in refining.
> 
> But I can see where someone reading it may get an idea, of an easy way to get there gold from a solution.
> 
> Also thank you for helping with the formula’s and equations.



No problem Butcher,

We are all on the same team here and I'm a firm believer in two heads are better than one.

P.S.: You have really been doing a great job wth your posting.

I really like the attention to detail you put in your responses.

Steve


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## maynman1751

lazersteve said:


> *We are all on the same team here *and I'm a firm believer in two heads are better than one.
> 
> P.S.: *You have really been doing a great job wth your posting.**I really like the attention to detail you put in your responses.*Steve



Well said Steve! Butcher deserves recognition for all the help and knowledge that he provides here, as well as you Steve, Harold, and several others that seem to go out of their way to help people understand what they're doing. I can't name you all, but I think that I can *THANK YOU *from all of us here at GRF! :mrgreen: :mrgreen: :mrgreen:


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## Lobby

A chemical engineer doing chemistry is always an interesting thing, but sometimes we bring something to the table with our simple ways of looking at chemistry. 

For me it's been fascinating rationalizing the positions that the proponents of the various precipitating agents take on this site. We have proponents of SO2, of sodium sulfite, and of sodium metabisulfite (not to mention copperas, but that's a different issue).

Here's my 2 cents on the issue:

1. There will be excess HCl in our AR solutions after we've dissolved the gold. Both sodium sulfite and SMB are going to react with that HCl to make SO2:

Na2SO3 + 2HCl = 2 NaCl + H2O + SO2

Na2S2O5 + 2HCl = 2NaCl + H2O + 2SO2

This is how they tie into the "it's the same as using SO2" discussion.


2. * SO2* will, obviously, precipitate gold.

3SO2 + 2AuCl3 + 3H2O = 2Au + 3SO3 + 6HCl

two interesting issues here:

- we've now made more HCl (to react with the sulfites)

- the SO3 will immediately react with water to make sulfuric acid: SO3 + H2O = H2SO4

so the reaction can probably be better written like this:

3 SO2 + 2 AuCl3 + 6 H2O = 2 Au + 3 H2SO4 + 6 HCl


3. We can write similar equations for *sodium sulfite* going through SO2

3 Na2SO3 + 6 HCl = 6 NaCl + 3 H2O + 3 SO2

and

3 SO2 + 2 AuCl3 + 6 H2O = 2 Au + 3 H2SO4 + 6HCl

summing those reactions:

3 Na2SO3 + 2 AuCl3 + 3 H2O = 2 Au + 3 H2SO4 + 6 NaCl




4. And for *SMB*, through an SO2 intermediate

3 Na2S2O5 + 6 HCl = 6 NaCl + 3 H2O + 6 SO2

6 SO2 + 4 AuCl3 + 4 H2O = 4 Au + 6 SO3 + 12 HCl

summing the reactions (and including the H2SO4 step)

3 Na2S2O3 + 4 AuCl3 + 9 H2O = 4 Au + 6 NaCl + 6 HCl + 6 H2SO4



If one calculates the stoichiometry of the reactions, to obtain 1 g gold requires = 0.49 g SO2, or 0.96 g sodium sulfite, or 0.72 g SMB



Anyway, my 2 cents... (assuming I didn't mistype from my notes) :mrgreen:


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## Alentia

Sounds like Lobby has "cut the Gordian Knot"... 

The reaction with SMB seems to be very complex and all elements constantly participating in the reaction, if I understood it correctly.


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## Lino1406

It should be mentionned that free SO2 will 
also emerge out of solution, doing
nothing except smell.
Because I can't PM (only get) 
who is interested in e-book
please give an e-mail


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