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Isolation of Silver

Isolation of Silver

Silver sometimes occurs in large nuggets (see the figure below) but more frequently in veins and related deposits. At one time, panning was an effective method of isolating both silver and gold nuggets. Due to their low reactivity, these metals, and a few others, occur in deposits as nuggets. The discovery of platinum was due to Spanish explorers in Central America mistaking platinum nuggets for silver.

When the metal is not in the form of nuggets, it often useful to employ a process called hydrometallurgy to separate silver from its ores. Hydrology involves the separation of a metal from a mixture by first converting it into soluble ions and then extracting and reducing them to precipitate the pure metal. In the presence of air, alkali metal cyanides readily form the soluble dicyanoargentate(I) ion, \({[\text{Ag}{(\text{CN})}_{2}]}^{-},\) from silver metal or silver-containing compounds such as Ag2S and AgCl. Representative equations are:

\(\text{4Ag}(s)+{\text{8CN}}^{\text{−}}(aq)+{\text{O}}_{2}(g)+{\text{2H}}_{2}\text{O}(l)\;⟶\;4{[\text{Ag}{(\text{CN})}_{2}]}^{\text{−}}(aq)+{\text{4OH}}^{\text{−}}(aq)\)

\({\text{2Ag}}_{2}\text{S}(s)+{\text{8CN}}^{\text{−}}(aq)+{\text{O}}_{2}(g)+{\text{2H}}_{2}\text{O}(l)\;⟶\;4{[{\text{Ag(CN)}}_{2}]}^{\text{−}}(aq)+\text{2S}(s)+{\text{4OH}}^{\text{−}}(aq)\)

\(\text{AgCl}(s)+{\text{2CN}}^{\text{−}}(aq)\;⟶\;{[\text{Ag}{(\text{CN})}_{2}]}^{\text{−}}(aq)+{\text{Cl}}^{\text{−}}(aq)\)

This figure contains two images. The first is of a small clump of bronze-colored metal with a very rough, irregular surface. The second shows a layer-like region of silver metal embedded in rock.

Naturally occurring free silver may be found as nuggets (a) or in veins (b). (credit a: modification of work by “Teravolt”/Wikimedia Commons; credit b: modification of work by James St. John)

The silver is precipitated from the cyanide solution by the addition of either zinc or iron(II) ions, which serves as the reducing agent:

\(2{[\text{Ag}{(\text{CN})}_{2}]}^{\text{−}}(aq)+\text{Zn}(s)\;⟶\;\text{2Ag}(s)+{[\text{Zn}{(\text{CN})}_{4}]}^{2-}(aq)\)

Example

Refining Redox

One of the steps for refining silver involves converting silver into dicyanoargenate(I) ions:

\(\text{4Ag}(s)+{\text{8CN}}^{\text{−}}(aq)+{\text{O}}_{2}(g)+2{\text{H}}_{2}\text{O}(l)\;⟶\;4{[\text{Ag}{(\text{CN})}_{2}]}^{\text{−}}(aq)+{\text{4OH}}^{\text{−}}(aq)\)

Explain why oxygen must be present to carry out the reaction. Why does the reaction not occur as:

\(\text{4Ag}(s)+{\text{8CN}}^{\text{−}}(aq)\;⟶\;4{[\text{Ag}{(\text{CN})}_{2}]}^{\text{−}}(aq)\text{?}\)

Solution

The charges, as well as the atoms, must balance in reactions. The silver atom is being oxidized from the 0 oxidation state to the 1+ state. Whenever something loses electrons, something must also gain electrons (be reduced) to balance the equation. Oxygen is a good oxidizing agent for these reactions because it can gain electrons to go from the 0 oxidation state to the 2− state.

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