# Tin identification

Tin is identified as one of group 2 cations. It is part of the arsenic subgroup.

If you need a small resume, you can quicly check the summary of arsenic subgroup analysis with this scheme.

Now it is time to understand how is it possible to identify tin in an acqueous solution... don't worry, it is going to be really easy and interesting!

Before the test: antimony removal

As tin is annoying during antimony identification so antimony is annoying when you perform assays to identify tin.

Do not forget that we are working on a solution of group 2 analysis;  it has to be "cleaned" from sulfide ions S2-, because their presence do not allow the set up of specific identification assays (it is easily oxidized to S°).

Then, we heat the test tube containing arsenic subgroup solution to ease the removal of  H2S (gas). At the same time, we keep a piece of paper soaked with lead acetate just above the test tube. As long as H2S comes out from the test tube, the paper will turn black because of formation of PbS formation.

• Test with elementary iron and HgCl2

Iron plays actually a double role. At the same time it reduces  antimony (+3 or +5) to elementary antimony and tin (IV) to tin (II).

Tin (II) is the real reactant that we use to carry out the specific identification test.

• E°Fe2+/Fe°  = -0.44V
• E°Sn+4/Sn2+ = 0.15 V
• E°Sb3+/Sb = 0.16 V
• E°Sb5+/Sb3+  = 0.6 V

From a practical viewpoint, we add a spatula of iron filings (elementary iron) to the test tube;

Sn4+ + Fe $\dpi{120}&space;\rightleftharpoons$ Sn2+ + Fe2+

The iron reduces all of the Sn4+ to Sn2+

2Sb3+ + 3Fe $\dpi{120}&space;\rightleftharpoons$ 2Sb°↓ + 3Fe2+

while antimony if present gives a black precipitate (elementary antimony) which can be separated by centrifugation. We add then mercuric chloride, HgCl2 . Since tin (II) is a very sensitive species easily oxidized to tin (IV) even by atmospheric oxygen, HgCl2 should be added as soon as possible, to avoid side reactions.

Sn2+ + 2HgCl2 $\dpi{120}&space;\rightleftharpoons$ Sn4+ + Hg2Cl2  + 2Cl-

Sn2+ + Hg2Cl2 $\dpi{120}&space;\rightleftharpoons$ Sn4+ + 2Hg↓ + 2Cl-

The reaction is, as you can see, based on the reducing properties of Sn2+. The mercuric chloride has to be the limiting reagent so that the reduction wil continue up to elementary mercury, black.

Anyways, if we got just a white precipitate it has to be verified whether it is mercurous chloride or not.

Thus we isolate the precipitate and we treat it with few drops of NH3, which induces (it is a very well known reaction, check group 1 analysis) the dismutation of the chloride in mercuric amidochloride and elemental mercury, black.