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Vanadium(III) sulfate

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(Redirected from V2(SO4)3)
Vanadium(III) sulfate[1]
Names
IUPAC name
Vanadium(III) sulfate
Other names
Vanadium trisulfate
Divanadium trisulfate
Identifiers
ChemSpider
ECHA InfoCard 100.033.827 Edit this at Wikidata
EC Number
  • 237-226-6
UNII
Properties
V2(SO4)3
Molar mass 390.074 g/mol
Appearance Yellow powder
Melting point 400 °C (752 °F; 673 K) Decomposes
Slightly soluble
Except where otherwise noted, data are given for materials in their standard state (at 25 °C [77 °F], 100 kPa).

Vanadium(III) sulfate is the inorganic compound with the formula V2(SO4)3. It is a pale yellow solid that is stable to air, in contrast to most vanadium(III) compounds. It slowly dissolves in water to give the green aquo complex [V(H2O)6]3+.

The compound is prepared by treating V2O5 in sulfuric acid with elemental sulfur:[2]

V2O5 + S + 3 H2SO4 → V2(SO4)3 + SO2 + 3 H2O

This transformation is a rare example of a reduction by elemental sulfur.

When heated in vacuum at or slightly below 410 °C, it decomposes into vanadyl sulfate (VOSO4) and SO2. Vanadium(III) sulfate is stable in dry air but upon exposure to moist air for several weeks forms a green hydrate form.

Vanadium(III) sulfate is a reducing agent.

References

[edit]
  1. ^ Lide, David R. (1998), Handbook of Chemistry and Physics (87 ed.), Boca Raton, FL: CRC Press, pp. 4–93, ISBN 0-8493-0594-2
  2. ^ Claunch, Robert T.; Jones, Mark M.; Wolsey, Wayne C. (1963). "Vanadium(III) Sulfate". Inorganic Syntheses. Vol. 7. pp. 92–94. doi:10.1002/9780470132388.ch28. ISBN 978-0-470-13166-4.