Why can't I find a reference for it? The closest thing is vanadyl sulfate, VOSO4. So what happens when vanadium metal is exposed to H2SO4?
Nothing?Xenoid - 26-8-2013 at 19:33
Google has plenty of references to vanadium (II) sulfate heptahydrate!
From Mellor Vol.9 page 818;
A. Piccini prepared hypovanadous sulphate, VS04 .7H2 0, by reducing a sulphuric acid soln. of vanadium pentoxide by means of an electric current in
the
absence of air. The violet-blue soln., when evaporated in vacuo over sulphuric
acid furnished violet, monoclinic crystals of the salt. The crystals are isomorphous
with those of ferrous sulphate with which it forms solid soln.
Could be a bit tricky to make
[Edited on 27-8-2013 by Xenoid]elementcollector1 - 26-8-2013 at 21:24
The reason I ask is that I was wondering precisely what forms when vanadium metal (as a very impure alloy, think 95% iron with the rest being V and
Cr) is dissolved in sulfuric acid. VSO4 may be the answer, but if it had to be prepared in such a manner as described above, I'm not sure
it's very stable.woelen - 26-8-2013 at 22:41
If you put vanadium in aqueous acidic solution, then the metal dissolves and you get blue vanadyl ions:
V + H2O + 2H(+) --> VO(2+) + 2 H2
Vanadium exists in 4 non-zero oxidation states, the purple V(2+), the green V(3+), the blue VO(2+) and the yellow VO2(+):
The most stable oxidation states are +4 and +5. You need strong reductors to go to +3 or +2. Preparation of V2(SO4)3 or VSO4 can be done by means of
electrolysis of a solution of VOSO4, where first V(3+) is formed at the cathode and on longer electrolysis, V(2+) is prepared. Isolation and storage
of the vanadium(III) or vanadium(II) salts requires vigorous exclusion of air. Oxygen from the air very easily oxidizes these salts to vanadyl salts. AndersHoveland - 26-8-2013 at 22:57
It is very difficult to get vanadium to dissolve in hydrochloric acid or dilute sulfuric acid, but it is attacked by concentrated sulfuric acid,
nitric acid, and hydrofluoric acid.woelen - 27-8-2013 at 00:08
In my experience it is not that difficult. I have thin sheets of very pure vanadium (99.9%) and these dissolve in hydrochloric acid fairly easily,
giving a blue solution of vanadyl chloride. It might be that certain alloys of vanadium (containing a few percent of other metals) may be more
resistant to dissolving in acids.AndersHoveland - 27-8-2013 at 09:24
A question then, why does vanadium metal react with hydrochloric acid to form vanadium in the +5 oxidation state, yet vanadium +3 is stable in aqueous
solution?Wizzard - 27-8-2013 at 10:32
I have some electrolytically refined crystalline pure vanadium. If anybody would like a small amount for testing (or collection) let me know. I don't
have much, but I do have enough for my own collection.chornedsnorkack - 27-8-2013 at 22:02
I have more issues with lack of references to the salts of lower transition metals!
Yes, their prevalent, most stable oxidation state is the highest one - V for Nb and Ta, VI for Mo and W, VII for Tc and Re. All these oxidation states
are fairly acidic - group VII has strong acids, groups V and VI have fairly strongly acidic amphoteric oxides which form oxide cations in strong
acids. Reducing any of these metals to lower oxidation states takes strong reductants.
Yet it is possible!
Jones reductor - Zn amalgam - is quoted to reduce V and Cr to II, Ti, Nb, Mo, W and U to III.
So, what do Nb(III), Mo(III) and W(III) salts look like? Surely you could pass Nb, Mo and W through Jones reactor and then evaporate the solutions in
vacuum?
Also, V is famed for having a full series of oxidation numbers all stable against dismutation, between II and V - V(III) and V(IV) are both stable.
How about the lower transition metals? Are Nb(IV), Mo and W (IV) and (V) stable against dismutation in acidic aqueous solution, or are they not so
that gradual oxidation of Nb(III), Mo(III) and W(III) solutions out of Jones reductor would go straight back to the highest oxidation state?
Finally, is Ta(V) stable against Jones reductor?woelen - 27-8-2013 at 23:22
A question then, why does vanadium metal react with hydrochloric acid to form vanadium in the +5 oxidation state, yet vanadium +3 is stable in aqueous
solution?
Vanadium does not react with hydrochloric acid to form the +5 oxidation state, it forms the +4
oxidation state. In absence of air (oxygen), the +3 oxidation state is stable, also in aqueous solution. Oxidation states +4 and +5 both are stable,
also in air. In practice, it is hard to keep a vanadium(III) salt around, unless it is stored under a protective atmosphere, but for amateur chemists
this is not feasible. Salts of the +4 or +5 oxidation state of course can be kept around without problem.
I made all oxidiation states from V2O5. First dissolve V2O5 in dilute NaOH to make a solution of metavanadate. Add a lot of acid to convert this to
the yellow VO2(+) ion, which has vanadium in +5 oxidation state.
Next, add SO2 (or if you are lazy, add Na2S2O5) and carefully heat the liquid with the yellow VO2(+) ions in it. Slowly, this is converted to blue
VO(2+). Next, boil vigorously to get rid of all excess SO2.
Then add zinc-shavings and allow this to bubble away for a while. Slowly, the blue VO(2+) is converted to green V(3+). If you allow the zinc to act
upon the solution for a longer time, then the color shifts from green to purple/violet and you get V(2+) in solution. The latter, however, slowly
converts back to the green V(3+), even when no air is allowed to reach the solution. Apparently, V(2+) slowly reduces water or H(+) ions.ElectroWin - 29-8-2013 at 18:58
I have some electrolytically refined crystalline pure vanadium. If anybody would like a small amount for testing (or collection) let me know. I don't
have much, but I do have enough for my own collection.
I'm interested to find, identify, and collect samples from natural sources of vanadium in ontario, canada
