Description of experiment
Below follows a plain text transcript of the selected experiment.
Needed compounds:
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sodium hydroxide : NaOH
sulphuric acid : H2SO4
stannous chloride : SnCl2
nitric acid : HNO3
molybdenum trioxide : MoO3
Class:
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elem=Mo
redox
Summary:
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When molybdate (VI) is reduced by a small amount of reductor, then a blue
compound is formed. When a larger amount of reductor is available, then a
black compound is formed.
Description:
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Prepare a solution of MoO3 and NaOH: This is a colorless solution, containing
molybdate and excess alkali. This liquid is used as the starting point for
the two sequences, described here. This liquid is called MOLYBDATE.
Sequence 1:
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Add an excess amount of HNO3 (52% by weight) to MOLYBDATE: Liquid remains
clear and colorless.
Add some solid SnCl2: The solid dissolves, the liquid becomes very dark (almost
black). When the liquid is black, a gas is evolved and the liquid then becomes
yellow/brown and a little turbid.
Add more SnCl2: Liquid again becomes black. Again, however, the liquid becomes
yellow/brown and a gas is evolved.
Remark on sequence 1:
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Apparently, SnCl2 quickly reduces the molybdate (VI) and the nitric acid in
turn oxidizes the reduction product (black solid) to a light brown compound,
itself being reduced to a gaseous product.
Sequence 2:
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Add an excess amount of H2SO4 (2 mol/l) to MOLYBDATE: Liquid remains clear
and becomes very light blue (almost colorless, but nonetheless clearly
visible blue).
Add some SnCl2: The solid dissolves and the liquid becomes intense dark blue.
Add much more SnCl2: Now the liquid becomes turbid and black.
Decant the liquid: A black solid is sticking to the glass. The glass cannot be
cleaned with running water.
Add some solid NaOH and a small amount of water: The NaOH dissolves and the
black solid dissolves. The liquid becomes light brown and somewhat turbid.
Now the glass can be cleaned by rinsing with water.