Sciencemadness Discussion Board

Advanced Re-Visit of Silver Cleaning

AJKOER - 21-2-2020 at 06:41

I start with the usual simple depiction of the electrochemistry based removal of Ag2S from Sterling Silver. To quote from Wikipedia (https://en.wikipedia.org/wiki/Galvanic_corrosion) on the electrochemical cleaning of Silver from dark spots of silver sulfide:

"Silver darkens and corrodes in the presence of airborne sulfur molecules, and the copper in sterling silver corrodes under a variety of conditions. These layers of corrosion can be largely removed through the electrochemical reduction of silver sulfide molecules: the presence of aluminum (which is less noble than either silver or copper) in the bath of sodium bicarbonate strips the sulfur atoms off the silver sulfide and transfers them onto and thereby corrodes the piece of aluminum foil (a much more reactive metal), leaving elemental silver behind. No silver is lost in the process.[16]"

Next, I note that aqueous solutions of NaHCO3 tend to be alkaline due to the formation of carbonic acid and hydroxide ion:

HCO3(-) + H2O <--> H2CO3 + OH-

It is probably speculative if cold aqueous alkaline NaHCO3 can assist in the removal of the protective Al2O3 coating from the Al foil (but on heating, aqueous NaHCO3 attacks Al, including even your aluminum pots in your dishwasher, otherwise scrubbing with steel or copper wool may be required) based on the proposed reaction:

2 NaHCO3 (aq) + 5 H2O (l) + Al2O3 (s) -> 2 NaAl(OH)4 (aq) + 2 H2O (l) + 2 CO2 (g)

And if proceeds, a further actually cited reaction (https://chemiday.com/en/reaction/3-1-0-1818):

Na[Al(OH)4] + CO2 → Al(OH)3 + NaHCO3

which would imply a catalytic hydration of Al2O3 to Al(OH)3 and potential removal of the oxide coating from the Aluminum surface, namely:

Al2O3 + 3 H2O ?--Cold NaHCO3(aq)-->? 2 Al(OH)3

Next, in alkaline conditions, a reaction with the active exposed Aluminum (which can even very slowly be attacked by water liberating hydrogen):

Al + OH- -> Al(OH)3 + 3 e-

In the presence of solvated electrons, the creation of the hydrogen atom radical along with hydrogen gas:

H+ + e- ⇌ •H

•H + •H = H2 (g)

Further, water and the bicarbonate ion are potential sources of H+ ion:

HCO3- ⇌ H+ + CO3(2-)

Also, per a galvanic cell between the Aluminum and the Silver metal, the half-cell reaction at the anode:

Al -> A(3+) + 3 e-

At the cathode, the proposed half-cell reaction explaining the removal of Silver sulfide:

Ag2S(s) + 2e- + 2H2O -> 2Ag(s) + H2S(g) + 2 OH-

Source: Equation (6) at http://www.electrochemsci.org/papers/vol8/80507223.pdf

And, also based on commercial applications of the hydrogen atom radical (previously detailed in my comments and provided sources with illustrative commercial applications on PbS, as provided previously on SM at http://www.sciencemadness.org/talk/viewthread.php?tid=116637...). Effectively, •H can operate as the pair (e-, H+) on a salt. For example with Silver sulfide (https://en.wikipedia.org/wiki/Silver_sulfide),

Ag2S + 2 •H -> 2 Ag + H2S

With an illustrative reaction scheme (as Ag2S does not dissociate to any extent and relatedly nor does PbS, which is very poorly soluble (https://en.wikipedia.org/wiki/Lead(II)_sulfide) to explain the action of •H:

2 Ag+ + S(2-) + 2 (e-, H+) -> 2 Ag (s) + H2S (g)

Also the reaction:

H2S + 2 OH- -> 2 H2O + S(s)

Which is apparently confirmed by the cited source to quote:

"However, after the cleaning of heavily tarnished samples (immersed for 30 and 60 min) some small quantities of sulfur (~0.3wt.%) are detected on the surface by EDS analysis, and the colour is changed to light yellowish."

In total, this is one albeit complex (and advanced) explanation for the cleaning of silver from sulfide stains with Baking Soda and Aluminum. Note, this confirming source (http://www.electrochemsci.org/papers/vol8/80507223.pdf) states to quote:

"The nascent hydrogen may also act as a reducing agent as it evolves from the metal surface. "

Note also, the same source states the formation reaction of Ag2S can be reversed, to quote:

"In that case silver suffers attack by dissolved oxygen and present sulphide species according to reaction (5):

4Ag (s) + 2HS- (aq) + O2(g) <--> 2Ag2S (s) + 2 OH- "

So, without Aluminum-based electrochemistry, the reverse reaction implies the use of a strong base along with heating (to drive off oxygen moving the reaction to the left) may also be able to effect some cleaning of the spotted silver.

[Edited on 21-2-2020 by AJKOER]

[Edited on 21-2-2020 by AJKOER]

unionised - 21-2-2020 at 10:12

Quote: Originally posted by AJKOER  
I
Next, I note that aqueous solutions of NaHCO3 tend to be alkaline due to the formation of carbonic acid and hydroxide ion:

HCO3(-) + H2O → H2CO3 + OH-


[Edited on 21-2-2020 by AJKOER]


You might want to look at the equilibrium constant for that reaction.

AJKOER - 21-2-2020 at 10:35

Quote: Originally posted by unionised  
Quote: Originally posted by AJKOER  

Next, I note that aqueous solutions of NaHCO3 tend to be alkaline due to the formation of carbonic acid and hydroxide ion:

HCO3(-) + H2O → H2CO3 + OH-


[Edited on 21-2-2020 by AJKOER]


You might want to look at the equilibrium constant for that reaction.


Correct, I should have the reaction expressed as:

HCO3(-) + H2O <--> H2CO3 + OH-

Clearly, on heating or other conditions that contribute to the decomposition of carbonic acid, the above equilibrium moves to the right.

H2CO3 <--> CO2 + H2O

There is also the equilibrium (see, for example, http://www.skyscrubber.com/Carbon%20Dioxide%20from%20Seawate...):

H2CO3 (aq) <--> H+ + HCO3-

which, I cite, as in the presence of solvated electrons (from say the galvanic cell), H+ is potentially removed:

H+ + e- <-> •H

•H + •H -> H2

which could result in the removal of H+ (and carbonic acid per above equilibrium), resulting in a rise in pH (at least locally at the electrode).
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I will adjust my opening thread as some may extract parts of the discussion as presented.

[Edited on 21-2-2020 by AJKOER]