oxidation states of vanadium experiment

Vanadium Oxidation States Experiment

This chemistry experiment shows you how to produce red, orange, green, yellow, blue and purple all in one solution, all the colors of the rainbow – without dyes or indicators. How is that possible? It’s possible using the amazing vanadium oxidation states .

We show you how to produce the amazing oxidation states of the element vanadium in one of the most exciting experiments we have ever done. Most versions of this experiment use expensive and hard to find reagents like ammonium vanadate or zinc-mercury amalgam, but not us! We show you an easier route. We also take you both forward and back though the various oxidation states , and show exciting intermediary colors.

Vanadium compounds are colorful because of the electrons in its d-orbital. The electrons can absorb light and move to a higher energy state, absorbing some colors but reflecting others. All vanadium compounds can absorb some spectra of light, which is why none of them are white.

Vanadium Oxidation States – Materials:

Vanadium Pentoxide 8.6 grams (OR Sodium vanadate OR Ammonium metavanadate) Sodium Carbonate 5 grams Distilled water 500ml Sulfuric Acid 98% 50 grams Zinc granules or lumps, or zinc mossy (a small handful) Potassium permanganate (4 grams in 100ml of distilled water) Beakers Flask Hotplate / Stirrer Thermometer

Where to get the reagents?

You have two choices. If you have a sizeable budget, you can purchase everything you need from your favorite chemical distributor or retailer. If you don’t, you can purchase vanadium pentoxide very cheaply from most pottery supply stores. In the USA, you can get 93% sulfuric acid from hardware stores as drain cleaner. We’ve seen zinc granules and potassium permanganate for sale online for as little as $6 USD each.

Chemistry Experiment Directions:

Making sodium vanadate.

If you don’t have sodium or ammonium vanadate, you’ll need to make it. Don’t worry, it’s easy to make sodium vanadate. Combine equimolar amounts of vanadium pentoxide and sodium carbonate, mix with water. Use a ratio of 1.72 to 1, for example 17.2 grams of vanadium pentoxide and 10 grams of sodium carbonate. You can even make your own sodium carbonate by heating sodium bicarbonate in an over at 450F for an hour (learn how in this video ). TIP – vanadium pentoxide costs about 20 times less than ammonium metavanadate.

For our procedure, we used 8.6 grams vanadium pentoxide and 5 grams of sodium carbonate. We used pottery grade vanadium pentoxide and homemade sodium carbonate, which worked just fine!

Heat at 90 degrees Celsius using a hotplate stirrer until the solution turns from yellow to green. It should take 5 or 10 minutes. When it turns green, you have sodium vanadate!

MAKING VANADYL SULFATE

Add a small amount of sulfuric acid to your sodium vanadate solution. We added 50 grams of concentrated sulfuric acid to 500ml of sodium vanadate solution. When the solution turns yellow, which should happen immediately, you now have vanadyl sulfate, which has vanadium in the +5 oxidation state. The vanadyl ion formula is VO2+.

MAKING THE COLORS OF VANADIUM

To produce the different colors of vanadium compounds, pour the vanadyl sulfate solution into a large flask, and put it on low heat on a hotplate / stirrer. And a generous amount of zinc granules. As the zinc reduces the vanadium, the colors will change to green, then blue, then dark-green, and finally purple. You have to shake the zinc around in the solution to get the last 2 color changes. Using a magnetic stirrer can help this process.

As the colors change, your can pour some of the solution out of the flask into a beaker. Have several 100ml beakers lined up, ready to go! TIP – Using zinc granules instead of powder means you don’t have to filter anything, just pour and the pieces of zinc remain in the flask. TIP – getting from V+3 to V+2 can take 15 – 40 minutes, using more zinc, keeping it warm, and swirling it around a little can keep the time down to a minimum. No need for expensive and dangerous zinc-mercury amalgam!

TIP – the beautiful purple color of the V+2 ion won’t last long in air, as the oxygen will oxidize it back to V+3. Enjoy it while you can!

Reversing the vanadium colors, making the rainbow

By adding a potassium permanganate solution, a strong oxidizing agent , you can reverse the colors. The colors will change from purple, to dark-green, to electric blue, to yellow. Adding more permanganate at this point will produce orange, and then red.

Safety Notes for this Experiment:

Vanadium compounds are toxic, do not breathe them in, and wear gloves if there is a danger of coming into contact with the solution. Sulfuric acid is highly corrosive – avoid all contact. Potassium permanganate is a very strong oxidizer – never mix with fuels or organic compounds. Wear goggles during this experiment. Read our lab safety rules .

If you purchase vanadium pentoxide from a pottery store, you may need to transfer it into a proper plastic bottle. We recommend doing this outdoors while wearing gloves and an N95 mask to avoid breathing in the dust.

Vanadium compounds can be harmful to marine organisms, do not wash large quantities down the drain (no more than 1 gram of compound per day), larger quantities should be treated as heavy metal waste. You can precipitate vanadium compounds as a hydroxide or sulfide, and then filter and dry the solid compound.

In this experiment, you will start with the +5 oxidation state of vanadium, generate the other oxidation states by redox reactions, and separate them using ion exchange chromatography. Ammonium metavanadate, NH4VO3 will be the source of the +5 oxidation state. Treatment of the ammonium metavanadate with hot hydrochloric acid partly reduces the vanadium to the +4 oxidation state in the form of the VO 2 + ion. The two species are separated using an ion exchange resin. The VO 2 + ion is then treated with mossy zinc generating the V 3+ and V 2+ which also are separated by ion exchange chromatography.

Ion Exchange Chromatography (IEC)

Many natural water sources in the Willamette Valley are hard water sources meaning that they have significant amounts of magnesium and calcium ions along with dissolved iron. These ions react are responsible for deposits that build-up in pipes and that stain sinks, showers and tubs. If this hard water is passed through a column containing and ion exchange resin having sodium ions in its active sites, the undesired ions replace the sodium ions and are immobilized on the resin lowering the concentration of these ions in the water. Learn more about ion exchange and water systems here .

Lab Procedure

Part I. Preparation of the Chromatography Column

• Push a small plug of glass wool down to the stopcock of a 25-mL buret.
• Close the stopcock and fill the buret with an aqueous slurry of AF50W-X2 cation exchange resin (100-200 mesh, H+ form). Allow the resin to settle. Open the stopcock and continue adding slurry to the column until the column of settled resin is 5-6 inches high. Be sure to keep the level of liquid above the top of the resin during the column packing process. Make sure the column is tightly packed. Wash the column of resin with distilled water until the eluate is clear.

Part II. Separation #1

• In the fume hood, add 2.0 mLconcentrated hydrochloric acid to 0.2 g of NH 4 VO 3 in a large test tube. Heat the mixture in a boiling water bath for 5 minutes. Add 10 mL distilled water and mix well. The solution should change from its original yellow color to a bright green.
• Carefully pour the green solution onto the top of the cation exchange column. Drain the column via the stopcock until the level of the liquid falls just to the top of the resin.
• Fill the buret with 0.4 M HCl. Open the stopcock to begin chromatography. Add 0.4 M HCl as needed to keep the buret full. Two bands will appear on the column. The upper band, although blue, will appear to be green as seen through the orange color of the resin. The lower band is yellow which is hard to see on the column because of the resin color. When the eluate from the column becomes yellow, begin collecing fractions of 5-6 mL in test tubes. When all of the yellow species has been collected from the column (the eluate becomes green or blue), change the eluting solvent to 1.0 M HCl. Continue collecting fractions until all of the second band has been eluted. Combine all the blue fractions together (should be approximately 25-30 mL). Combine all the darkest yellow fractions together in another container. Dispose of any fractions that are yellow-green or blue-green as they are mixtures of the two ions. Be sure to leave a solvent head on the column.

Part III. Reduction and Separation #2

• Transfer 15-20 mL of the blue solution to a 25-mL Erlenmeyer flask containing about 3 g of mossy zinc. Put a cork into the flask and swirl for 5-10 minutes until the color becomes lavender in color and no longer changes. The color should become green before turning lavender.
• Decant the solution and pour it onto the chromatography column. Drain the column via the stopcock until the level of the liquid falls just to the top of the resin.
• Elute with 1.0 M HCl. Distinct green and lavender bands should appear. Collect in ~5 mL fractions in testtubes. After all of the lavender species has been collected, change the eluting solvent to 3.0 M HCl and elute the green species. Combine solutions of the same species.

Part IV. Characterization of Products.

• Place a 1 mL sample of each solution into a different test tube and add 0.001 M KMnO 4 dropwise. Note which solutions decolorize the permanganate and offer an explanation for your results.
• Place a 1 mL sample of each solution into a different test tube and add iodine water dropwise. Note which solutions decolorize the iodine and offer an explanation for your results.
• Acquire a visible spectrum of each of the four ions, noting the absorbance maxima for each oxidation state. You need to find the literature values for the absorbance maxima of the ions and compare your results to the literature.
• Reduce a portion of the green solution from Separation #1 directly with mossy zinc to V(II). Exposure of the resulting solution to air will oxidize V(II) to V(III).
• Wash the resin by passing 20-30 mL of water through the column. Remove the resin from the buret and collect the resin by gravity filtration. Return the resin to the original container.

You will write a formal, typed laboratory report for this experiment. Guidelines for this report can be found here .

Additional References

General informationon the chemistry of vanadium can be found in the following sources:

• Comprehensive Inorganic Chemistry, Vol. 3, Chapter 34. (in WOU library)
• Cotton & Wilkinson, Advanced Inorganic Chemistry, Chapter 25.

Electrochemistry:

• Pourbaix, Atlas of Electrochemical Equilibria in Aqueous Solutions (in WOU library)
• Any general chemistry text will have a section on balancing redox equations.
• Pourbaix diagrams are useful in predicting the stabilities of various species. A discussion of these diagrams can be found in J.Chem. Educ. 1969 , 46 , 90.
• An explanation of how to get information from a Pourbaix diagram can be accessed here .

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Oxidation States of Vanadium

(d) demonstrating the five oxidation states of manganese (ah).

Transition metals are well known for having more than one stable oxidation state.

Vanadium, like many transition metals, has several different oxidation state and, like several, these oxidation states are of different colours

Also like other transition metals, these oxidation states are coloured.

In this demonstration / experiment vanadium 5+ is reduced to give the 2+, 3+ and 3+ states and then sequentially oxidised back to the 5+.

Oxidation states of vanadium – Instructions

Oxidation states of vanadium – Risk Assessment

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Note:   If you have learnt to use E° values using some more complicated method (and there are lots of unnecessarily complicated methods around!), it would pay you to follow the link above to find out how to do it all much more easily.

So . . . if you mix together zinc and VO 2 + ions in the presence of acid to provide the H + ions:

That converts the two equilibria into two one-way reactions. You can write these down and combine them to give the ionic equation for the reaction if you want to.

Note:   If you aren't happy about building up ionic equations like this, you could usefully follow this link.

Use the BACK button on your browser to return quickly to this page.

The other stages of the reaction

Here are the E° values for all the steps of the reduction from vanadium(V) to vanadium(II):

. . . and here is the zinc value again:

Remember that for the vanadium reactions to move to the right (which is what we want), their E° values must be more positive than whatever you are reacting them with.

In other words, for the reactions to work, zinc must always have the more negative value - and that's the case.

Zinc can reduce the vanadium through each of these steps to give the vanadium(II) ion.

Using other reducing agents

Suppose you replaced zinc as the reducing agent by tin. How far would the set of reductions go this time?

Here are the E° values again:

. . . and here is the tin value:

In order for each reduction to happen, the vanadium reaction has to have the more positive E° value because we want it to go to the right. That means that the tin must have the more negative value.

In the first vanadium equation (from +5 to +4), the tin value is more negative. That works OK.

In the second vanadium equation (from +4 to +3), the tin value is again the more negative. That works as well.

But in the final vanadium reaction (from +3 to +2), tin no longer has the more negative E° value. Tin won't reduce vanadium(III) to vanadium(II).

Important!   If this doesn't seem pretty obvious to you, then you don't really understand about redox potentials .

This link will take you to the redox potential menu. Spend some time by starting at the beginning of the sequence of pages you will find there.

If you follow this link, use the BACK button on your browser (or the History file or the Go menu) if you want to return to this page later.

Re-oxidation of the vanadium(II)

The vanadium(II) oxidation state is easily oxidised back to vanadium(III) - or even higher.

Oxidation by hydrogen ions

You will remember that the original reduction we talked about was carried out using zinc and an acid in a flask stoppered with a piece of cotton wool to keep the air out. Air will rapidly oxidise the vanadium(II) ions - but so also will the hydrogen ions present in the solution!

The vanadium(II) solution is only stable as long as you keep the air out, and in the presence of the zinc. The zinc is necessary to keep the vanadium reduced.

What happens if the zinc isn't there? Look at these E° values:

The reaction with the more negative E° value goes to the left; the reaction with the more positive (or less negative) one to the right.

That means that the vanadium(II) ions will be oxidised to vanadium(III) ions, and the hydrogen ions reduced to hydrogen.

Will the oxidation go any further - for example, to the vanadium(IV) state?

Have a look at the E° values and decide:

In order for the vanadium equilibrium to move to the left, it would have to have the more negative E° value. It hasn't got the more negative E° value and so the reaction doesn't happen.

Oxidation by nitric acid

In a similar sort of way, you can work out how far nitric acid will oxidise the vanadium(II).

Here's the first step:

The vanadium reaction has the more negative E° value and so will move to the left; the nitric acid reaction moves to the right.

Nitric acid will oxidise vanadium(II) to vanadium(III).

The second stage involves these E° values:

The nitric acid again has the more positive E° value and so moves to the right. The more negative (less positive) vanadium reaction moves to the left.

Nitric acid will certainly oxidise vanadium(III) to vanadium(IV).

Will it go all the way to vanadium(V)?

No, it won't! For the vanadium reaction to move to the left to form the dioxovanadium(V) ion, it would have to have the more negative (less positive) E° value. It hasn't got a less positive value, and so the reaction doesn't happen.

Note:   There are several possible half-reactions involving the nitric acid with a variety of E° values. Two of these actually have E° values more positive than +1.00 and so, in principle, nitric acid could seem to be able to oxidise vanadium(IV) to vanadium(V) - but involving different products from the nitric acid.

In practice, if you do this reaction in the lab, the solution turns blue - producing the vanadium(IV) state. Just because the E° values tell you that a reaction is possible, you can't assume that it will actually happen. There may be very large activation energy barriers involved, causing the reaction to be infinitely slow!

You can work out the effect of any other oxidising agent on the lower oxidation states of vanadium in exactly the same way. But don't assume that because the E° values show that a reaction is possible , it will necessarily happen.

Where would you like to go now?

To the transition metal menu . . .

To the Inorganic Chemistry menu . . .

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© Jim Clark 2003 (modified June 2015)