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.

PLEASE note: The electrode potentials (E^Ø) for vanadium chemistry in the text sometimes vary from the chart above - apologies, but data sources for vanadium chemistry can vary! Any discrepancy shouldn't significantly affect any electrode potential calculation outcomes e.g feasibility (email if concerned?).

5. Aspects of 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

6. The vertical connection of vanadium with the other d-block elements of Group 5 (IUPAC designation)