Table 2
- Questions
1. Write a balanced chemical equation to represent each of the following reactions.
a) Ammonium metavanadate and dilute sulphuric acid
b) Ammonium metavanadate and dilute sulphuric acid with iron(II) sulphate
c) Ammonium metavanadate and dilute sulphuric acid with potassium iodide
d) Ammonium metavanadate and dilute sulphuric acid with aqueous sulphur dioxide
e) Ammonium metavanadate and dilute sulphuric acid with copper powder
f) Ammonium metavanadate and dilute sulphuric acid with zinc powder
g) Ammonium metavanadate and dilute sulphuric acid with tin powder
2. Which of the reactions mentioned above is the best one to prepare each of the following
solutions? Explain briefly.
a) Vanadium(II)
b) Vanadium(III)
c) Vanadium(IV)
3. Explain why it is necessary to add sodium thiosulphate to the reaction mixture in which
potassium iodide has been added.
4. Account for any differences between the theoretical results and the observed results.
-
a) VO3- + 2H+ → VO2+ + H2O
b) Fe2+ + VO2+ + 2H+ → VO2+ + H2O + Fe3+
c) 2VO2+ + 4H+ + 2I- → 2VO2+ + 2H2O + I2
d) 2VO2+ + SO2 → 2VO2+ +SO42-
e) 2VO2+ + 4H+ + Cu → 2VO2+ + 2H2O + Cu2+
f) 2VO2+ + 8H+ + 3Zn → 2V2+ + 4H2O + 3Zn2+
g) VO2+ + 4H+ + Sn → V3+ + 2H2O + Sn2+
- a) The reaction of ammonium metavanadate and dilute sulphuric acid with zinc powder.
It's the only reaction of ammonium metavanadate reduces to vanadium(II) ions. After the
reaction, zinc sulphate is one of the products. It is soluble in water and is colourless
solution. So the purple colour of the reaction mixture indicates the presence of
vanadium(II) ions and vanadium(II) solution is in more pure form.
b) The reaction of ammonium metavanadate and dilute sulphuric acid with tin powder.
It's the only reaction of ammonium metavanadate reduces to vanadium(III) ions. Sn2+ is
colourless. The reaction mixture only gives out the color of vanadium(III) ions. The
physical properties of vanadium (III) ions will not be affected.
c) The reaction of ammonium metavanadate and dilute sulphuric acid with aqueous sulphur
dioxide. It's because SO42- is colourless while in other reactions which produce VO2+ will
give out colored ions. As SO42- is colourless, vanadium(IV) ions can exists in its blue
color.
(3) Potassium iodide reacts with ammonium metavanadate to produce iodine. The color of
iodine is brown/reddish-brown. Then the reaction mixture turns to brown color at first. In
order to test the presence of vanadium(IV) ions, it needs to add sodium thiosulphate
solution. The sodium thiosulphate solution can reduce iodine to iodide ions which are
colourless. The following chemical equation shows the reaction between them.
I2 + 2S2O3 2- → 2I- + S4O6 2-
As iodine becomes colorless iodide, the only color exists is blue and it proves there are
VO2+ ions.
(4) In reaction of ammonium metavanadate and dilute sulphuric acid with iron(II) sulphate,
iron(III) ions and VO2+ are produced. For theoretical result, the color of the mixture is blue
and it shows VO2+ . But the observed result is the color of the mixture is bluish-green. Since
Fe3+ is yellow in color, yellow mix with blue to give out green.
In reaction of ammonium metavanadate and dilute sulphuric acid with copper powder,
copper(II) ions and VO2+ are produced. For theoretical result, the color of the mixture is blue
and it shows VO2+ . But the observed result is the color of the mixture is bluish-green.
However, Cu2+ is blue in color. There may be reaction that reduction of VO2+ to V3+ which is
green in color.
VO2++ 2H+ + e- → V3+ + H2O +0.340V
Cu2+ + 2e- → Cu +0.337V
Although the reduction potentials of above two reactions have very small difference, some
VO2+ may reduce to V3+. So the mixture turns to bluish-green color.
In reaction of ammonium metavanadate and dilute sulphuric acid with potassium iodide,
iodine and VO2+ are produced. For theoretical result, the color of the mixture is blue
and it shows VO2+ . But the observed result is the color of the mixture is brown. Iodine
gives brown color. So the blue color is covered by brown color.
- Discussion and sources of error
In this experiment, I can identify the color of vanadium compounds with different oxidation
states. From the experiment, I can find out which reaction is the best to prepare
vanadium(II), vanadium(III) and vanadium(IV) ions. It's helpful because we realize how to
prepare those ions from ammonium metavanadate in school laboratory or in industry.
Besides, different oxidation states of vanadium compounds shows its property in reduction.
They are strong oxidizing agent.
From the experiment, we have to consider the standard reduction potential of the reaction.
It's because if the differences of reduction potentials between oxidation and reduction
reaction is too small, the reaction will not occur. Hence all the experiment carried out will
not turns VO2+ to V2+ except the zinc case. It's because the reduction potential of the
reduction of Zn2+ is more negative than the the reduction of VO2+ to V2+.
There are some sources of error in the experiment. I added more amount of some chemicals,
like Cu and Sn. It make some Cu and Sn are not soluble in the mixture and they sink in the
bottom of test tubes. It make the color of the reaction mixtures in dark blue because of the
initial color of the metal powder. I should add little amount of metal powder each time and
observe the color change of mixture in each time. It's accurate to obtain the color change of
the mixture.
The oxidation state of vanadium is easily oxidized back to vanadium(III) or higher. Air
rapidly oxidize the vanadium(II) ions because there are hydrogen ions in the solution.
V3+ + e- → V2+ -0.260V
2H+ + 2e- → H2 0V
The difference of reduction potentials between two reaction is large. The V2+will be
oxidized to V3+and hydrogen ions are reduced to hydrogen. Therefore, some reaction
produce V2+ may have color change from purple to green. So the reaction must be covered
with cotton wool to prevent air contact with mixtures.
The same case occurs when air contact with V3+. Oxygen in air will oxidize the solution to
vanadium(IV) state VO2+ ions. Then the solution turns from green to blue. The reaction of
ammonium metavanadate and dilute sulphuric acid with tin powder shows blue reaction
mixture at last although the reaction produce V3+ions. So the test tube should be covered
with cotton wool.
- Conclusion
VO3-, with oxidation state +5, is yellow in color.
VO2+, with oxidation state +5, is yellow in color.
VO2+, with oxidation state +4, is blue in color.
V3+, with oxidation state +3, is green in color.
V2+, with oxidation state +2, is purple in color.
