Description of experiment
Below follows a plain text transcript of the selected experiment.
Needed compounds:
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zinc : Zn
sodium sulfite : Na2SO3
sulphuric acid : H2SO4
vanadium pentoxide : V2O5
Class:
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elem=V,Zn
redox
Summary:
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Vanadium pentoxide can be easily dissolved in acid, when a reductor is
present. Then it is dissolved and at the same time reduced to vanadyl
(vanadium (V) to vanadium (IV)). With very strong reducing compounds,
further reduction to vanadium (III) or even vanadium (II) is possible.
Description:
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Add some solid V2O5 to dilute sulfuric acid (appr. 15% by weight) and boil
gently for a few minutes: Part of the solid dissolves, the liquid becomes
clear and yellow.
Add some solid sodium sulfite: The solid quickly dissolves and a strong smell
of SO2 is observed. When heating and shaking continuously, the rest of the
solid V2O5 also dissolves and the liquid becomes sky-blue, looking like a
fairly concentrated aqueous copper (II) solution.
Keep on boiling, even when all V2O5 is dissolved and the liquid is blue: The
color does not change anymore. All SO2 is driven off, without further
reacting with the vanadyl ions in the liquid.
Add a piece of zinc and shake vigorously: A lot of gas bubbles are formed at
the piece of zinc. The liquid slowly turns green (fairly dark green). When
the liquid is left alone, with the piece of zinc still bubbling vigorously,
then the color slowly shifts to lavender (purple/blue). When all zinc is
dissolved, then the color shifts back to green, when the liquid is shaken
again.
Explanation:
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Solid V2O5 is reduced by SO2 in acidic environment and blue vanadyl [VO]2+
enters the solution. SO2 is not capable of further reducing this compound
(oxidation state IV). When zinc is added, then the green V3+ ion is formed.
The green V3+ is even further reduced to lavender V2+.
The lavender V2+ is oxidized by oxygen from the air to green V3+ very
quickly, when there is no more zinc. V2+ is a strong reductor.