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I know two distinct recipes for cleaning silver:

  • Put them into a pot with some club soda and a piece of aluminium foil and pour over a hot water.
  • Gently clean silverware with a fine chalk powder (obtained from a pharmacy).

I'd be happy to know:

  1. What happens chemically during these procedure - why does it work, and what are the byproducts of these reactions?
  2. How toxic (if at all) are the byproducts?
  3. How much silver is lost in such a procedure?
  4. Are these procedures "idempotent"? I mean, if I apply one of these procedures to an already cleaned item, will it consume some more silver, or will it remain intact? In other words, do the reactions consume just tarnish or do they consume pure silver too?
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Put them into a pot with some club soda and a piece of aluminium foil and pour over a hot water.

I am familiar with this process. I am not familiar with the other. Perhaps someone else can provide an answer for it.

What happens chemically during these procedure - why does it work, and what are the byproducts of these reactions?

This process cleans silver by a redox process. Aluminum is a more active metal than silver. This means that aluminum metal, Al(s), will reduce silver salts (salts of Ag+). The most common type of tarnish on silver is silver oxide, Ag2O. The reaction is:

$$\ce{3Ag2O(s) + 2Al(s) -> 3Ag(s) + Al2O3(s)}$$

The club soda - containing sodium bicarbonate - plays two roles. First it acts as a supporting electrolyte so that electrons can be conducted through the solution. Second, the carbonic acid in it dissolves the passivation layer of aluminum oxide that protects the aluminum foil. This produces aluminum carbonate, which is unstable in water and decomposes into amphoteric aluminum hydroxide, which is soluble through the formation of a number of complex ions.

$$\ce{Al2O3(s) + 3H2CO3(aq) -> Al2(CO3)2(s) + 3H2O(l)-> 2Al(OH)3 (aq) + 3CO2(g)}$$

The net reaction is then:

$$\ce{3Ag2O(s) + 2Al(s) +3H2CO3(aq) -> 3Ag(s) + 2Al(OH)3(aq) + 3CO2(g)}$$

The byproducts are carbon dioxide and aluminum hydroxide.

  • How toxic (if all) are the byproducts?

Not very. Aluminum salts have very low health impact on you or the environment. Additionally, you produce only very small amounts of aluminum salts this way.

  • How much silver is lost in such a procedure?

None (Almost). The process coverts solid silver oxide into solid silver metal, both of which are negligibly soluble in club soda (or any other aqueous solution). The $K_{sp}$ of AgOH (which is what the surface of Ag2 looks like when exposed to water) is $1.52\times10^{-8}$. All of the action happens at the surface of both metals. The only common silver salts of appreciable solubility in aqueous solutions are silver nitrate and silver sulfate.

  • Is this procedure "idempotent"?

Yes. This procedure converts silver oxide into metallic silver. It has no effect on metallic silver: metallic silver does not react with most acids; it cannot be reduced farther by aluminum metal; and it is not soluble in water or most aqueous solutions.

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  • $\begingroup$ Thanks for a thorough answer. I observed that the aluminium foil gets black during the process, so I thought they are some silver compounds. But if there are no silver compounds as the result and everything reduces atomic silver, what is it then? (Just a minor thing, formatting got wrong at * How much silver is lost in such a.) $\endgroup$
    – Petr
    Oct 30, 2012 at 20:31
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    $\begingroup$ The black stuff is aluminum salts: aluminum carbonate, aluminum hydroxide, or some other salts depending on what else is in the club soda. Maybe aluminum phosphate. $\endgroup$
    – Ben Norris
    Oct 30, 2012 at 22:34
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    $\begingroup$ Or aluminum sulfide if the tarnish on the silver is silver sulfide. $\endgroup$
    – Paul J. Gans
    Oct 31, 2012 at 1:32
  • $\begingroup$ @PaulJ.Gans - Of course, if the tarnish is silver sulfide, the chemistry changes a little, but the most important parts of the answer - no silver is consumed and the procedure does nothing to untarnished silver - remain the same. $\endgroup$
    – Ben Norris
    Nov 1, 2012 at 21:15

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