Sciencemadness Discussion Board - Preparation of Vanadium from its Pentoxide

Preparation of Vanadium from its Pentoxide

Lab Rat - 8-1-2019 at 20:05

(I'd be really surprised if there weren't already a thread like this elsewhere, but I promise I searched for a while and couldn't find one. Please feel free to just link me if this has already been discussed!)

I'm just another recreational chemist/element collector, and I've been looking to add vanadium to my collection for a while now. V2O5 is readily available, and I feel like one could definitely use it as a source of vanadium to isolate the element itself. It's certainly possible to reduce the oxide with magnesium or aluminum in a Goldschmidt reaction, but those are terribly messy and produce terribly messy products, especially on the small scale that I'd be working at. I'd much prefer an aqueous route. Unfortunately, vanadium's aqueous chemistry is much more unusual than, say, copper's, for instance. Just for analogy's sake, let's say we start with CuO; all you'd have to do is dissolve it in acid and submerge a strip of zinc or electrolyze. But with V2O5 as the starting material, uhh... I'm clueless. I feel like a general route would be to dissolve the pentoxide in acid and electrolyze the resulting salt solution, but from what I understand, V2O5 doesn't form simple vanadium salts upon acid reaction, so that likely won't work. Can anyone please give me some ideas or procedures? I'd even love to hear just some random information about vanadium chemistry~

j_sum1 - 8-1-2019 at 20:21

V2+ + 2 e− ⇌ V(s) −1.13
https://en.wikipedia.org/wiki/Standard_electrode_potential_(data_page)

Getting from vanadium (V) to vanadium metal is going to be a multiple step reduction process: vanadium has numerous oxidation states. The last step from V2+ to the metal has a reduction potential of -1.13 volts. This is outside the range where wet chemistry can be done easily. (+1.229 to -0.8277: see the bolded entries in the link above.)

So, your intended route is going to be infeasible.
Vanadium pentoxide can be reduced via a thermite. This is very energetic. https://www.youtube.com/watch?v=-j-7LxavEJ4

It can be done however with a bit of care. This is a better attempt: https://youtu.be/LsdesMWC37g?t=50

elementcollector1 - 8-1-2019 at 20:30

From my own experience with vanadium thermite, it also has the problem of being miscible with aluminum metal, forcing you to either use a different reducing agent or (in my case) starving the reaction of available aluminum in the hopes that the molten aluminum-vanadium alloy would react with additional vanadium pentoxide and purify itself before it solidified. I did get very brittle, crystalline chunks of vanadium metal from that, but I've no idea as to the purity - I'd suspect greater than 90%, but that's not much of a bar to set.

Aqueous vanadium chemistry is fascinating in its own right, but... good luck getting vanadium[0] out of it. I did manage to find one vague reference to the subject:
https://core.ac.uk/download/pdf/4814615.pdf

From this, the (again, extremely vague) procedure would go something like this:
-Dissolve vanadic anhydride (vanadium pentoxide) in a stoichiometric excess of sodium hydroxide.
-Neutralize excess sodium hydroxide with hydrochloric acid.
-Electroplate vanadium.

What little I can add would be that you would probably initially be looking at a solution of sodium orthovanadate (Na₃VO₄, colorless), which upon acidification might take on a orange or even red tint due to the formation of sodium decavanadate (Na₆[V₁₀O₂₈], orange) and Na₄[V₄O₁₂], and probably a few others. Even simple neutralization of pH can give rise to multiple oxidation species coexisting in solution, so it's best to leave the chemistry at this if isolation is your goal.

In regards to voltage, things also get complicated. Some relevant potentials are:

V³⁺ + e^- -> V²⁺ (-0.26 V)
V²⁺ + 2 e^- -> V (-1.13 V)
VO²⁺ + 2 H⁺ + 2 e^- -> V³⁺ + H₂O (+0.34 V)

Because you don't have pretty much any of these ions in your beaker, though, you're going to have to take a guess as to what the voltage might actually be. As always, going too high results in the splitting of water (and thus high loss of efficiency, not to mention bubbling activity at your cathode that'll disrupt the structure of any grown metal crystals). You might have to play around with it a bit and see at what voltage you can get vanadium to actually plate - and please report back to us what that is!

[Edited on 1/9/2019 by elementcollector1]

woelen - 9-1-2019 at 04:52

Aqueous chemistry of vanadium is amazingly complicated, especially for the metal in oxidation state +5.

V2O5 is not a basic oxide, but an acidic oxide. In water, it will dissolve (very sparingly), giving a somewhat acidic solution. Vanadium(V) oxo-species form all kinds of condensed species, with the total size of condensed anions (or cations) depending on pH and total vanadium concentration.

The extreme at the acidic end is the ion VO2(+), which is pale yellow. At increasing pH, these ions combine to form cationic species (xVO2(+), yOH(-), zH2O) (x > y) until at a certain pH (somewhere around pH = 3 to 4) we get very big and very complicated neutral structures with empirical formula (HVO3)x(H2O)z, with x and z very large (can be up to millions), and x/z in the order of magnitude of 1. This stuff, sometimes written as "HVO3" appears as orange or even red slimy solid and on drying forms an orange/yellow amorphous solid. On further increase of the pH, these big neutral structures break up again, now forming anionic species (xVO2(+), yOH(-), zH2O) (x < y) which become smaller and smaller at increasing pH. The color shifts from deep orange to colorless with diminishing size of the anionic species. At the basic far end we end up with VO2(+), 4(OH)-, usually written as VO4(3-)(aq). This can form salts, which on careful heating in dry air can lose their water and form the free ion VO4(3-).

At oxidation state +4 vanadium also forms a peculiar ion, the blue VO(2+) ion (not to be comfused with the pale yellow VO2(+) ion). In oxidation state +4 also condensed species can be formed, which are nearly black when neutral VO(2+)/OH(-) combinations form a solid. On further increase of the pH, this black solid breaks up again in anionic species, which are brown. At very high pH, we end up with brown V4O9(2-)(aq). The oxide VO2 is a neutral (amphoteric) oxide, which forms blue VO(2+) when dissolved in acid and which forms brown V4O9(2-) when dissolved in alkalies.

Only the oxides V2O3 and VO are true basic oxides, which form salts with simple acids. VO, however, is a very unstable species and I doubt whether it exists in the pure state, I think that real samples will be something impure between VO and V2O3. Solutions of V(2+) ions also are unstable. They reduce water and within minutes the vanadium is oxidized to oxidation state +3 and hydrogen is formed. This makes it very hard to obtain metallic vanadium from aqueous solutions. Vanadium has a fascinating redox chemistry and also a beautifully colorful one, but making the pure metal from its most common oxides V2O5 and VO2 is very hard. It is not without reason that samples of pure vanadium metal are so expensive on eBay, while V2O5 is quite affordable.

MrHomeScientist - 9-1-2019 at 07:02

I know you want to avoid the Goldschmidt reaction, but that's all I have experience with and I wanted to share some tidbits. I wrote up a blog post on it at: http://thehomescientist.blogspot.com/2010/04/experiment-vana...

I did this reaction with fine mesh aluminum powder (~400 mesh I think) and V₂O₅ powder right out of the bottle. The reaction goes to completion almost instantly, and blows much of the product out of the reaction vessel. So you definitely lose a lot via this route. However I did manage to recover a fairly large lump from one reaction. You could slow this down and maybe improve recovery by using coarse aluminum powder, or diluting the mix with some fluxing agent like fluorite.
I think thermite products are usually 80%-90% pure, being mixed with unreacted aluminum (as EC1 mentioned). I haven't tested mine for purity. Is sure looks different than Al, though, so for now that's good enough for me. I did find a really nice sample of crystalline V on eBay a while back, so that's what I use for display.

XeonTheMGPony - 9-1-2019 at 07:32

Quote: Originally posted by MrHomeScientist

I know you want to avoid the Goldschmidt reaction, but that's all I have experience with and I wanted to share some tidbits. I wrote up a blog post on it at: http://thehomescientist.blogspot.com/2010/04/experiment-vana...

I did this reaction with fine mesh aluminum powder (~400 mesh I think) and V₂O₅ powder right out of the bottle. The reaction goes to completion almost instantly, and blows much of the product out of the reaction vessel. So you definitely lose a lot via this route. However I did manage to recover a fairly large lump from one reaction. You could slow this down and maybe improve recovery by using coarse aluminum powder, or diluting the mix with some fluxing agent like fluorite.
I think thermite products are usually 80%-90% pure, being mixed with unreacted aluminum (as EC1 mentioned). I haven't tested mine for purity. Is sure looks different than Al, though, so for now that's good enough for me. I did find a really nice sample of crystalline V on eBay a while back, so that's what I use for display.

Has any one tried pressing it into a pellet then Igniting? perhaps slow the reaction interface down

fusso - 9-1-2019 at 07:40

Quote: Originally posted by XeonTheMGPony

Quote: Originally posted by MrHomeScientist

Has any one tried pressing it into a pellet then Igniting? perhaps slow the reaction interface down

What about covering the container with fireproof bricks so the molten stuff won't splash out?

woelen - 9-1-2019 at 08:06

That may be quite dangerous. You may get an explosion, due to confinement of hot pressurized material. In many reactions it is not wise to confine your reactants to a small space.

Lab Rat - 9-1-2019 at 08:47

You're all far too accommodating, thanks for all the replies so quickly! As much as I had wished it weren't so, my suspicions about the unfriendliness of aqueous vanadium chemistry were correct, and that route will indeed be completely unfeasible for me. I really do appreciate all this information, though, especially from woelen's detailed post.

MrHomeScientist, it was indeed your blog post that initially got me thinking about V2O5 in the first place, so it's super cool to hear from you about it! Looks like a thermite-type reaction is going to be my best bet for vanadium metal someday (short of buying it, of course, but I do try to avoid that. Buying pure elements feels like cheating to me, especially if there's any potential for me to isolate it myself).

I'm also a big fan of growing vibrantly colored crystals, so as long as I'm here I suppose I might as well ask about vanadyl sulfate. It looks like it might be synthesized by acidic reduction of V2O5 with ethanol, but does anybody know if it likes to crystallize nicely?

DraconicAcid - 9-1-2019 at 08:52

I don't know how well it crystallizes, but the easiest way to reduce vanadium(V) is with dilute sulphite ion in acidic solution. Pretty, pretty.

MrHomeScientist - 9-1-2019 at 12:17

That's great! It's super cool to hear from people that have been inspired by my work

If you go the thermite route, just be very careful around V₂O₅. It's very toxic, especially via inhalation of dust. When I worked with it, I wore a lab coat, gloves, and a dust mask, and worked very carefully and slowly. It's likely the reaction will throw some unreacted dust around too, although I never noticed any orange residue. I suppose I should have tried to clean up the area afterward. That may not have been very environmentally responsible of me, come to think of it...