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

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

Elect. pot. = standard electrode potential data for vanadium (E^Ø at 298K/25^oC, 101kPa/1 atm.)

na = data not applicable to vanadium

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 heterogeneous catalyst in the 'Contact Process' in the production of sulfur trioxide for the manufacture of sulfuric acid.

  • The catalysing of the conversion of sulfur dioxide into sulfur 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 

    • Adding (i) + (ii) gives (iii) SO_2(g) +  ¹/₂O_2(g) ===> 2SO_3(g)  (the summing up of the catalytic cycle)

    • This is an example of heterogeneous catalysis of a vanadium compound to reduce the activation energy by providing an alternative pathway via the active sites on the surface of the catalyst.

    • 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 oxidation states and electron configuration of vanadium in the context of the 3d block of elements

• Electrode potential chart for vanadium the context of the 3d block transition metals

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

The electrode potentials involving chromium ions correspond to hydrated complex ions where the ligands are water, oxide or hydroxide.

As you can see from the chart, changing either the ligand or the oxidation state, will also change the electrode potential for that half-reaction involving a vanadium ion.

Vanadium(II) compounds are readily oxidised to vanadium(III) and vanadium(IV) compounds.

The hexaaquavanadium(II) ion is a strong reducing agent.

The variety of vanadium's oxidation states

• 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 sulfuric acid mixture.

  • 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:

    • See the electrode potential chart for the data used in these calculations.

  • (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,

      • Electron configuration of V³⁺ is [Ar]3d²

      • e.g. in the 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(IV) or vanadium(V) ions i.e. [VO]²⁺ or  [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 fully balanced complete redox 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

keywords redox reactions ligand substitution displacement balanced redox equations formula of vanadium hexaaqua complex ions complexes ligand exchange reactions redox reactions ligands colours oxidation states: all the colour changes when a vanadium(V) salt is reduced in aqueous solution, 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 + 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 how to work out redox reactions of vanadium using electrode potentials and half-reactions to test for feasibility of a vanadium redox reaction