Originally posted by woelen
...
Conclusion: Vanadium does not catalyse the reaction. Vanadium in the +5 oxidation state is reduced by the S to vanadium in the +4 oxidation state
(bright blue color is due to vanadyl, VO(2+)). The further step, the formation of the ochre/yellow precipitate, is not clear at all to me. Is this
some sulfide of vanadium? I could not find info on such a compound. So, we have a new, interesting riddle, but absolutely no catalyst.
The sulphur is indeed the reductor, because without sulphur, vanadium is not reduced, the boiling hot solution remains deep red.
|