10.5. Chemistry of Vanadium V, Z=23, 1s²2s²2p⁶3s²3p⁶3d³4s² 

data comparison of vanadium with the other members of the 3d–block and transition metals

Extended data table for VANADIUM

• Uses of VANADIUM

  • Vanadium is one of many transition metals alloyed with iron to make specialist steels.

• Vanadium(V) oxide, V₂O₅, is used as a catalyst in the 'Contact Process' in the production of sulphur trioxide for the manufacture of sulphuric acid.

  • The catalysing of the conversion of sulphur dioxide into sulphur trioxide is explained via change in oxidation state changes i.e. some classic transition metal chemistry.

  • 2SO_2(g) + O_2(g) ==> 2SO_3(g) 

  • The mechanism, somewhat simplified, goes via the catalytic cycle ...

    • (i) SO₂ + V₂O₅ ==> SO₃ + V₂O₄, then (ii) V₂O₄ + ¹/₂O₂ ==> V₂O₅ 

    • The vanadium changes oxidation state from +5 to +4 and back to +5 in the catalytic cycle, a classic combination of two characteristics of transition metals – variable oxidation state and catalytic properties.

    • This is an example of heterogeneous catalysis – reactants (g) and catalyst (s) in different phases.

The Chemistry of VANADIUM

• The electrode potential chart highlights the values for various oxidation states of vanadium.

• Vanadium shows a 'classic' display of variable oxidation states of varying colours when a solution of e.g. ammonium vanadate(V), is reduced by a zinc/dilute sulphuric acid mixture. do picture

  • You go from the vanadium(V) vanadate(V) ion ==> vanadium(IV) oxovanadate(IV) ion ==> vanadium(III) ion ==> vanadium(II) ion

  • Acidification changes the vanadate(V) ion into the pale yellow oxo–cation VO₂⁺ (oxovanadium(V) ion)

  • VO₄^3–_(aq) + 4H⁺_(aq) VO₂⁺_(aq) + 2H₂O_(l) [an acid–base reaction, NOT a redox change]

    • Note: Highly charged cations >3+ rarely exist as the simple 'hydrated' tetra or hexa–aqua ion.

    • The theoretical polarising power of the 'central metal ion' is so strong that they form oxocations (see above) or oxyanions e.g.

    • orange dichromate(VI) Cr₂O₇^2–, yellow chromate(VI) CrO₄^2–, purple manganate(VII) MnO₄^– etc.

    • For transition metals they may be coloured even if electronically the theoretical 'central metal ion'  has a noble gas structure e.g. [Ar] in its maximum oxidation state like V(V), Cr(VI) and Mn(VII).

    • These oxyanions are called charge transfer complexes and the theory is beyond pre–university chemistry.

  • Three successive reduction steps then follow to eventually give V²⁺ ions, shown as half–cell equations:

  • (i) V(V, +5) ==> V(IV, +4): VO₂⁺_(aq) + 2H⁺_(aq) + e^– VO²⁺_(aq) + H₂O_(l)

    • E^Ø_{half–cell potential} = +1.00V, pale yellow to the blue oxovanadium(IV) ion

  • (ii) V(IV, +4) ==> V(III, +3): VO²⁺_(aq) + 2H⁺_(aq) + e^– V³⁺_(aq) + H₂O_(l)

    • E^Ø_{half–cell potential} = +0.34V, blue to the green vanadium(III) ion

    • Here the vanadium(III) ion, V³⁺, is actually the green hexaaquavanadium(III) ion,

      • [V(H₂O)₆]³⁺

    • Both V(IV) and V(III) species are slowly oxidised by dissolved oxygen back to the V(V) compound in acid solution.

      • (see electrode potential comments later).

  • (iii) V(III, +3) ==> V(II, +2): V³⁺_(aq) + e^– V²⁺_(aq)

    • E^Ø_{half–cell potential} = –0.26V, green to the purple–violet vanadium(II) ion.

    • V²⁺_(aq) is powerful reducing agent and is unstable in the presence of air.

    • Any dissolved oxygen will oxidise V²⁺_(aq) back to the vanadium(III) cation.

    • V²⁺ is actually the purple–violet hexaaquavanadium(II) ion, [V(H₂O)₆]²⁺

  • Note

    1. The standard electrode potential E^Ø_{Zn(s)/Zn2+(aq)} is –0.76V, so the reducing power of zinc is sufficient to effect any of the three vanadium oxidation state reduction changes described above.

       • See electrode potential chart VO₃⁺/VO²⁺ (+1.00V), VO²⁺/V³⁺ (+0.34V) and V³⁺/V²⁺ (–0.26V) are less positive than O₂/H⁺/H₂O +1.23V in acid solution so the vanadium(II/III/IV) cations can theoretically be oxidised back to vanadium(V) i.e. VO₂⁺.

       • Electrode potentials (half–cell potentials) and how to work out E^Ø_reaction, which must be a +ve voltage for the reaction to be feasible, are all explained in Equilibria Part 7 Redox Equilibria and Electrode Potentials. How to use half–cell reactions to work out full redox equations is explained on Redox Reactions Part II section 6 Constructing redox equations.

    2. The reduction occurs on the surface of the zinc metal i.e. the site of electron transfer and you can write the above reductions as a fully balanced complete equations ...

       • (i) 2VO₂⁺_(aq) + 4H⁺_(aq) + Zn_(s) ==> 2VO²⁺_(aq) + 2H₂O_(l) + Zn²⁺_(aq)

         • E^Ø_reaction = E^Ø_reduction – E^Ø_oxidation = +1.00 – (–0.76) = +1.76V

         • The half–cell reaction of the reduction will have the most +ve E^Ø_potential.

       • (ii) 2VO²⁺_(aq) + 4H⁺_(aq) + Zn_(s) ==> 2V³⁺_(aq) + 2H₂O_(l) + Zn²⁺_(aq)

         • E^Ø_reaction = +0.34 –(–0.76) = +1.10V

       • (iii) 2V³⁺_(aq) + Zn_(s ==> 2V²⁺_(aq) + Zn²⁺_(aq)

         • E^Ø_reaction = –0.26 – (–0.76) = +0.50V

         • BUT the vanadium(II) cation is unstable in the presence of dissolve oxygen in air.

         • ¹/₂O_2(g) + 2H⁺_(aq) + 2e^–   H₂O_(l) has a standard electrode potential of +1.23V,

         • so, for the vanadium(II) oxidation reaction ...

         • ¹/₂O_2(g) + 2H⁺_(aq) + 2V²⁺(aq) ==> 2V³⁺_(aq) + H₂O_(l)

         • E^Ø_reaction = E^Ø_reduction – E^Ø_oxidation = +1.23 – (–0.26) = +1.49V

         • hence the if left standing open to air, the violet V²⁺_(aq) solution will gradually change to a green V³⁺_(aq) solution and in turn V³⁺_(aq) will revert back to VO²⁺_(aq) in the presence of air because of oxidation by dissolve oxygen unless protected by an inert atmosphere. (see Redox Electrode Potential Chart, V²⁺/V³⁺ and V³⁺/VO²⁺ potentials are less positive (below) that for O₂/H₂O/H⁺ potentials).

    3. You will see hydrogen formed simultaneously from the unavoidable metal–acid reaction.

       • Zn_(s) + 2H⁺_(aq) ==> Zn²⁺_(aq) + H_2(g)

• Does vanadium chemistry show an example of disproportionation?

  • This is just a little academic exercise using standard electrode potential data.

  • A disproportionation reaction is where a species in one oxidation state spontaneously and simultaneously changes into two species of different oxidation states – one higher and one lower in oxidation number.

  • Examples: disproportionation in manganese(VI) chemistry and disproportionation in copper(I) chemistry

  • Question: In terms of aqueous ions, is the disproportionation of vanadium(III) into vanadium(II) and vanadium (IV) feasible?

    • (i) VO²⁺_(aq) + 2H⁺_(aq) + 2e^– V³⁺_(aq) + H₂O_(l)   (E^Ø_VO2+/V3+ = +0.34V)

    • (ii) V³⁺_(aq) + e^– V²⁺_(aq)   (E^Ø_V3+/V2+ = –0.26V)

    • The disproportionation equation would be (iii) 2V³⁺_(aq) + H₂O_(l) V²⁺_(aq) + VO²⁺_(aq) + 2H⁺_(aq)

    • For equation (iii), (ii) will be the reduction half–cell equation and (i) reversed will be the oxidation half–cell reaction.

    • E^Ø_reaction = E^Ø_reduction – E^Ø_oxidation = = E^Ø_V3+/V2+ – E^Ø_VO2+/V3+ = (–0.26) – (+0.34) = –0.60V

    • showing the disproportionation is thermodynamically NOT feasible i.e. E^Ø_reaction is less than zero.

    • In fact what can actually happen is if you mix salt solutions of vanadium(IV) and vanadium(II) on an equimolar basis, you end up with a solution of vanadium(III) salts, a sort of 'anti–disproportionation' reaction!

• Summary of some complexes–compounds & oxidation states of vanadium compared to other 3d–block elements

• –

Scandium * Titanium * Vanadium * Chromium * Manganese * Iron * Cobalt * Nickel * Copper * Zinc * Silver & Platinum

keywords redox reactions ligand substitution displacement balanced equations formula complex ions complexes ligand exchange reactions redox reactions ligands colours oxidation states: vanadium ions V2+ V(+2) V(II) V3+ V(+3) V(III) V4+ V(+4) V(IV) V5+ V(+5) (V) SO2 + V2O5 ==> SO3 + V2O4 V2O4 + 1/2 O2 ==> V2O5 VO43– + 4H+ VO2+ + 2H2O V(V, +5) ==> V(IV, +4): VO2+ + 2H+ + e– VO2+ + H2O V(IV, +4) ==> V(III, +3): VO2+ + 2H+ + e– V3+ + H2O [V(H2O)6]3+ V(III, +3) ==> V(II, +2): V3+ + e– V2+ VO3+/VO2+ (+1.00V), VO2+/V3+ (+0.34V) and V3+/V2+ (–0.26V) 2 VO2+ + 4H+ + Zn ==> 2 VO2+ + 2H2O + Zn2+ 2VO2+ + 4H+ + Zn ==> 2V3+ + 2H2O + Zn2+ 2V3+ + Zn(s ==> 2V2+ + Zn2+ 1/2O2 + 2H+ + 2V2+ ==> 2V3+ + H2O V2+/V3+ and V3+/VO2+ potentials VO2+ + 2H+ + 2e– V3+ + H2OEØVO2+/V3+ = +0.34V) (ii) V3+ + e– V2+ (EØ V3+/V2+ = –0.26V) EØ V3+/V2+ – EØ VO2+/V3+ oxidation states of vanadium, redox reactions of vanadium, ligand substitution displacement reactions of vanadium, balanced equations of vanadium chemistry, formula of vanadium complex ions, shapes colours of vanadium complexes

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Alphabetical Index for Science Pages Content A B C D E F G H I J K L M N O P Q R S T U V W X Y Z

Scandium * Titanium * Vanadium * Chromium * Manganese * Iron * Cobalt * Nickel * Copper * Zinc * Silver & Platinum

Introduction 3d–block Transition Metals * Appendix 1. Hydrated salts, acidity of hexa–aqua ions * Appendix 2. Complexes & ligands * Appendix 3. Complexes and isomerism * Appendix 4. Electron configuration & colour theory * Appendix 5. Redox equations, feasibility, E^ø * Appendix 6. Catalysis * Appendix 7. Redox equations * Appendix 8. Stability Constants and entropy changes * Appendix 9. Colorimetric analysis and complex ion formula * Appendix 10 3d block – extended data * Appendix 11 Some 3d–block compounds, complexes, oxidation states & electrode potentials * Appendix 12 Hydroxide complex precipitate 'pictures', formulae and equations