Description of experiment
Below follows a plain text transcript of the selected experiment.
Needed compounds:
-----------------
potassium metabisulfite : K2S2O5
sulphuric acid : H2SO4
vanadium pentoxide : V2O5
Class:
------
elem=S,V
redox
Summary:
--------
Vanadium pentoxide dissolves in concentrated sulphuric acid with some heating.
A deep red liquid is formed, which, remarkably, only is reduced by sulfite in
the presence of some water.
Description:
------------
Add some vanadium pentoxide to concentrated sulphuric acid (98% H2SO4): No
visible reaction. The solid settles and the liquid turns turbid and yellow.
Heat, until a white fume from the hot sulphuric acid can be seen (be careful,
hot H2SO4 is REALLY bad for your skin!): The vanadium pentoxide dissolves, the
liquid becomes less viscous (the acid becomes hot), and it becomes totally
clear and dark red. All solid vanadium pentoxide disappears, all of it
dissolves.
Add some solid potassium metabisulfite (K2S2O5): The liquid starts foaming. The
reaction is not violent, but due to the viscosity of the liquid, the foam is
quite stable and it slowly fills up the entire test tube (this is a risk, keep
a bucket with water nearby, just in case when the foam really tends to go over
the rim of the test tube). The foam remains dark red. A strong smell of sulphur
dioxide can be observed.
Carefully add a few drops of water: A violent reaction occurs between the
concentrated still warm acid and the water. There is a hissing noise and the
water probably boils. The liquid now becomes dark green and remains clear.
Apparently part of the vanadium in its +5 oxidation state is reduced to
vanadium in its +4 oxidation state.
Add more water and shake well: The liquid becomes dark blue and remains clear.
This dark blue color is due to formation of vanadyl ions [VO]2+(aq), which
contain vanadium in its +4 oxidation state.
Dilute with more water: The liquid turns bright blue, this is the well known
color of vanadyl ion.