Description of experiment
Below follows a plain text transcript of the selected experiment.
Needed compounds:
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sodium sulfite : Na2SO3
starch : (C6H10O5)1
potassium iodide : KI
sodium hydroxide : NaOH
sulphuric acid : H2SO4
molybdenum trioxide : MoO3
Class:
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elem=Mo
redox
Summary:
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Molybdenum trioxide dissolves well in strongly alkaline liquids. It does
not dissolve (or just a little bit) in plain water. Once dissolved, it
can be kept in solution, even when pH is lowered.
Molybdates are capable of oxidizing iodide and sulfite.
Description:
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Add some solid MoO3 to water: The solid does not dissolve.
Add some solid NaOH: All NaOH and MoO3 quickly dissolve. The resulting
liquid is clear and colorless.
Add an excess amount of sulfuric acid (appr. 15% by weight): Liquid remains
colorless and clear.
Add some solid KI: The solid dissolves, the liquid turns yellow and clear.
Boil liquid for appr. 1 minute: The liquid turns blue/green. Probably this
color is due to a mixture of brown/yellow iodine and blue poly-molybdate
Mo(VI)/Mo(V) compounds. When some starch is added, the liquid turns intense
blue, this shows the presence of free iodine.
Add some solid sodium sulfite to the blue/green liquid: The liquid becomes
more blue, but a greenish shade remains. Smell of SO2.
Add an excess amount of NaOH (relative to the sulfuric acid): The liquid
becomes hot, the solid dissolves and the final color of the liquid is bright
blue. The liquids looks like a concentrated solution of aqueous copper (II).
Add an excess amount of NaOH to the blue/green liquid, which is formed after
boiling the liquid with KI: The liquid becomes (almost) colorless again. This
is a strong difference with the situation that Na2SO3 is added, before the
NaOH is added. Probably the iodine, present in acidic solution, which is
converted to iodide, hypoiodite and iodate, is capable of back-oxidizing the
blue Mo(VI)/Mo(V) compounds to Mo(VI)-only compounds.