Description of experiment
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Needed compounds:
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stannous chloride : SnCl2
sodium sulfide : Na2S . 3H2O
hydrochloric acid : HCl
potassium tetrachloroplatinate (II) : K2PtCl4
Class:
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elem=Pt,Sn,S
redox
Summary:
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Platinum(II) is not reduced by sulfide in acidic environments. When it is
treated with stannous ions at the same time, then a dark brown compound
is formed, but probably this is not metallic platinum but stannous sulfide.
What speaks against this is that stannous chloride does not precipitate
with H2S in acidic environments (see sequence 2), while with the Pt-compound
a dark brown compound is formed. More research will be needed to resolve
this.
Description:
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Sequence 1:
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Add a very small amount of solid sodium sulfide (technical grade, yellow
flakes) to dilute HCl (appr. 10% by weight): The solid quickly dissolves,
while fizzling a little bit. Liquid remains clear and colorless.
Add a drop of a solution of K2PtCl4 (appr. 35 g/l): Liquid hardly changes.
Color becomes very light red/brown, almost colorless.
Add some SnCl2: The solid dissolves, the liquid becomes dark brown.
Sequence 2:
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Add a very small amount of solid sodium sulfide (technical grade, yellow
flakes) to dilute HCl (appr. 10% by weight): The solid quickly dissolves,
while fizzling a little bit. Liquid remains clear and colorless.
Add some SnCl2: The solid dissolves, the liquid remains almost clear and
colorless, it just becomes a little white opalescent.
Sequence 3:
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Add a solution of SnCl2 to a solution of sodium sulfide (now without the
HCl): Formation of a dark brown precipitate.