Colorful oxidation states of vanadium
Vanadium can be in oxidation states +2 to +5 in aqueous solution. Each of the oxidation states has its own bright color. In this experiment, all oxidation states will be obtained and can be displayed next to each other.
Procedure for performing the experiment
Take two spatulas full of sodium hydroxide and put one spatula full of vanadium pentoxide in a test tube. Add a a few ml of water and dissolve all sodium hydroxide. Now, the test tube contains the solution of sodium hydroxide with the solid vanadium pentoxide suspended in it.
Gently heat the contents of the test tube, while swirling it slowly and carefully. Now the vanadium pentoxide dissolves in the hot solution. The solution becomes colorless. It might be that a small amount of a grey flocculent solid remains and that the liquid cannot be made perfectly clear. This is of no concern. If yellow/brown vanadium pentoxide remains, then a little more sodium hydroxide should be added carefully.
Let liquid cool down and then add dilute hydrochloric acid, such that one gets a test tube, approximately halfway full of liquid. The liquid will be light yellow. This is the vanadium in the +5 oxidation state.
Divide the liquid over four test tubes and set one test tube aside.
To another one of the test tubes add a spatula full of sodium sulfite and dissolve this in the liquid. The liquid slowly turns bright blue through different shades of green and remains bright blue. This is the vanadium in the +4 oxidation state. Sulfite does not reduce vanadium further. If the reaction proceeds very slowly, then gently heat for a while.
To the other two test tubes add a small spatula of zinc filings or powder. The liquids start to effervesce and the color of the solutions slowly changes. The color goes from yellow to green and then to bright blue. If this point is reached, then the +4 oxidation state is reached. The color continues changing and at a certain point a grayish green color is obtained. At this point the +3 oxidation state is reached. Decant the contents from one of the test tubes to another test tube, without zinc. Finally, the last test tube is swirled, with the zinc in it, until the liquid has a lavender color. The zinc can be left inside the test tube, the vanadium will not be reduced further. When the zinc is removed, then the lavender liquid quickly turns grayish green again. Vanadium in the +2 oxidation state is very prone to oxidation by oxygen from the air.
The following picture shows the final result of this experiment.
Discussion of results
Vanadium pentoxide, V2O5, only very sparingly dissolves in water. In a strongly alkaline solution this dissolves much better, forming metavanadate, VO3-, or orthovanadate, VO43-, depending on the precise concentration of the hydroxide. These anions are colorless. Some heating may be needed to get all V2O5 dissolved. The following reactions occur:
V2O5 + 2OH- → 2VO3- + H2O
V2O5 + 6OH- → 2VO43- + 3H2O
When acid is added, then the vanadates are converted to pervanadyl ions, having formula VO2+. When the acid is added slowly, then one can observe a change of color from colorless to yellow, through a deep shade of orange. Vanadium in the +5 oxidation state has a very strong tendency to form large condensed ions, with many vanadium atoms in them. When pH is gradually decreased from 14 to 0, then first polyvanadate anions are formed, which are yellow. The more vanadium atoms in such an anion, the deeper the color. At a pH near 3, the anions become very large and their color is deep orange. At sufficient concentration of vanadium, even a flocculent precipitate may be formed of a compound with net formula xV2O5·yH2O, with x and y being very large and not of constant value throughout the mixture. When pH is decreased even further, then large cationic species are formed, which break down in smaller pieces when pH is decreased more and more. At the other end of the pH-scale, the simple cation VO2+ is formed. This has a yellow color.
Vanadium in its +5 oxidation state already has interesting properties, but on reduction, especially in acidic media, new colorful species can be produced. Vanadium in its +5 oxidation state is a moderately strong oxidizer in acidic media. Sulphur dioxide is oxidized to sulfate ions and the yellow pervanadyl is reduced to blue vanadyl, VO2+. In this experiment, the sulphur dioxide is from the sulfite. On addition of sodium sulfite to the acidic vanadium (+5) solution the following reactions occur.
Dissolving of sodium sulfite:
Na2SO3(s) → 2Na+(aq) + SO32-(aq)
Reaction of sulfite with acid:
SO32- + 2H+ → SO2 + H2O
Reduction of pervanadyl to vanadyl by sulphur dioxide:
SO2 + 2VO2+ → SO42- + 2VO2+
Vanadyl is the most stable vanadium cation in acidic environments. It hardly has any reducing power, nor any oxidizing power. In order to reduce vanadium further, beyond its +4 oxidation states, a strong reductor is needed.
Further reduction of vanadium can be reached by addition of zinc. When the zinc is added, then a gas is produced and another reaction is reduction of vanadium.
The gas produced is hydrogen, due to reaction between zinc and acid.
The target reactions in this experiment are the following:
Zn + 2VO2+ + 4H+ → Zn2+ + 2V3+ + 2H2O
Zn + 2V3+ → Zn2+ + 2V2+
The cationic species V3+ is grayish green and the species V2+ is lavender. The latter is oxidized back very easily to V3+. The species V3+ is much more resistant to air-oxidation, but when an acidic grayish green solution of a vanadium (III) salt is left in contact with air, then in the course of several days its color slowly turns blue, due to aerial oxidation to VO2+. In order to get vanadium (III), without an appreciable amount of vanadium (II) or vanadium (IV), one can reduce a solution with vanadium, such that it becomes lavender and then shake the solution for a few minutes, while in contact with air.