Reactions of vanadium and other transition metals

In this lesson, we will learn:

• The variety of oxidation states of vanadium and their associated colours.
• The redox reactions involved in converting vanadium between its oxidation states.
• The redox reactions involved in converting chromium between its oxidation states.
• The feasibility of vanadium and chromium redox reactions as shown by their reduction potentials.

• The variety of oxidation states (and associated colours) is a defining feature of the transition metals. This lesson looks at:
• Vanadium, as an example of this variety,
• The redox reactions used to prepare the oxidation states of vanadium,
• The feasibility of the redox reactions, due to their reduction potentials when we use zinc as the reducing agent,
• The redox reactions of some other transition metals in less detail.

IMPORTANT: When does ‘dark red’ become burgundy?
Remember that different sources can use different terms for the colour of a specific oxidation state of a specific transition metal. You should ALWAYS use the colour description in your course’s syllabus to avoid losing marks in an exam.


• Vanadium has four common oxidation states:
• +5 which is yellow in solution, for example in the dioxovanadium (V) ion, VO₂⁺. Vanadium oxide (V₂O₅) is another example of vanadium (V).
• +4 which is blue in solution, for example in vanadyl ion, VO²⁺.
• +3 which is dark green in solution.
• +2 which is violet in solution.

Vanadium ions can be oxidised by O₂ in the air, starting with vanadium in a higher oxidation state is more reasonable, such as its +5 oxidation state. A common source of this is from ammonium metavanadate, formula NH₄VO₃. When dissolved, VO₂⁺ is produced in solution with vanadium still in its +5 state.

The VO₂⁺ ions can be reduced by adding zinc metal and hydrochloric or sulfuric acid.
• This would cause a colour change from yellow (due to VO₂⁺ ) to blue, which is VO²⁺ .
• This reaction can proceed further to produce vanadium (III). This is possibly a mixture of [V(H₂O)₆]³⁺ and complexes with Cl^- or SO₄^2- ligand exchanges between H₂O ligands, depending on which acid was used earlier. Either way, the solution becomes a green colour.
• Going even further, a violet colour appears when vanadium (II) is formed, which is due to [V(H₂O)₆]²⁺.


• There are some features of the practical experiment that help these reactions take place:
• The conical flask is usually stoppered with cotton wool. As mentioned above, vanadium (II) can be oxidised by O₂ in the air. This would send it back up to a higher oxidation state; the common vanadium (V) and (IV) compounds are oxides.
  With H₂ gas also being produced by a side reaction between zinc and the acid, there is a positive pressure gradient where gas is being forced out of the flask through the cotton wool. Think of a balloon inflating, but part of its skin is permeable (that’s the cotton wool), allowing hydrogen to escape. The pressure is higher inside the flask than outside, so there is a general flow of gas out of the flask, preventing air getting in to it.
• The reaction is usually heated to speed up the process.


• These reactions converting vanadium between its oxidation states are redox reactions. If vanadium is being reduced, then something is being oxidised!
  This is what the zinc metal was added for. As with any redox reaction that occurs spontaneously, we can use cell potentials to show the steps in this reaction are feasible.


• The first step is the reduction is from vanadium (V) to vanadium (IV) or from VO₂⁺ ions to VO²⁺ ions. This is a reduction which means that the zinc metal that was added must be getting oxidised.
  The two relevant half-equations and their standard potentials^1,2 are:
• VO₂⁺ + 2H⁺ + e^- $\, \rightleftharpoons \,$ VO²⁺ + H₂O $\,$ E⁰ = +1.0 V
• Zn²⁺ + 2e^- $\, \rightleftharpoons \,$ Zn $\,$ E⁰ = -0.76 V

The lesson Predicting redox reactions explains how to use these values and what they mean. In short, the equations and E⁰ values (found in a Standard Potentials table) of these substances show you their tendency to be reduced.
As redox requires one reduction and one oxidation, the larger E⁰ - the larger reduction potential - shows the reduction process, and the lower E⁰ equation shows the oxidation process, which is the equation proceeding right to left.
With this said, H⁺ ions are present due to the acid, so the reaction for what is going on can be written:

Reduction: 2 x (VO₂⁺ + 2H⁺ + e^- $\,$→$\,$ VO²⁺ + H₂O)
Oxidation: Zn $\,$→$\,$ Zn²⁺ + 2e^-

Full reaction: 2VO₂⁺ _(aq) + 4H⁺ _(aq) + 2Zn _(s) $\,$→$\,$2VO²⁺ _(aq) + 2H₂O _(l) + 2Zn²⁺ _(aq)


• The second step is reducing vanadium (IV) to vanadium (III) by reacting VO²⁺ to V³⁺. The reduction equation² for this is:
• VO²⁺ + 2H⁺ + e^- $\, \rightleftharpoons \,$ V³⁺ + H₂O $\,$ E⁰ = +0.337 V
Again, if this reduction is to happen, an oxidation reaction needs to happen too! The zinc reduction is still a lower E⁰ value (-0.76 V) than the VO²⁺, so this will work.

Reduction: 2 x (VO²⁺ + 2H⁺ + e^- $\,$→$\,$ V³⁺ + H₂O)
Oxidation: Zn $\,$→$\,$ Zn²⁺ + 2e^-

Full reaction: 2VO²⁺ _(aq) + 4H⁺ _(aq) + Zn _(s) $\,$→$\,$2V³⁺ _(aq) + 2H₂O _(l) + 2Zn²⁺ _(aq)


• The next reaction is from vanadium (III) to vanadium (II) by the reduction of V³⁺ to V²⁺. Again, looking at the reduction half-equations we can see that if Zn metal is present then this will be oxidised as V³⁺ is reduced.
• V³⁺ + e^- $\, \rightleftharpoons \,$ V²⁺ $\,$ E⁰ = -0.255 V
• Zn $\,$→$\,$ Zn²⁺ + 2e^- $\,$ E⁰ = -0.76 V
Remember it does not matter what the signs of the two half-equations are – their actual values are just being compared to the reduction of hydrogen, which is E⁰ = 0 V by definition. It only matters that the reduction half-equation must have a higher E⁰ than the oxidation one.

In conclusion, the oxidation of zinc can reduce vanadium (V) all the way through to vanadium (II) as shown by the reduction potential values


• The redox reactions of chromium can also be predicted using their standard reduction potentials.
  Chromium’s main oxidation states are:
• +6 which is seen in the dichromate ion, Cr₂O₇^2-
• +3 from Cr³⁺
• +2 from Cr²⁺
Just like with vanadium, these oxidation states can be interconverted.


• To show the reduction of chromium (VI) to chromium (III ), just like with vanadium we can use zinc and acid, as shown in the following two half-equations²:
• Cr₂O₇^2- + 14H⁺ + 6e^- $\, \rightleftharpoons \,$ 2Cr³⁺ + 7H₂O $\,$ E⁰ = +1.36 V
• Zn $\, \rightleftharpoons \,$ Zn²⁺ + 2e^- $\,$ E⁰ = -0.76 V
Given the E⁰ values, chromium (VI) will be reduced to chromium (III) in the presence of zinc metal and acid (which is required for the chromium reduction).

Reduction: Cr₂O₇^2- + 14H⁺ + 6e^- $\, \rightleftharpoons \,$ 2Cr³⁺ + 7H₂O
Oxidation: 3 x (Zn $\, \rightleftharpoons \,$ Zn²⁺ + 2e^-)

Full reaction: Cr₂O₇^2- _(aq) + 14H⁺ _(aq) + 3Zn _(s) $\,$→$\,$ 2Cr³⁺ _(aq) + 7H₂O _(l) + 3Zn²⁺ _(aq)


• Cr³⁺ ions can be oxidised back to dichromate, Cr₂O₇^2- in basic (OH^-) conditions, followed by hydrogen peroxide and acidification.
  Starting with the chromium (III) hexaaqua species, NaOH can be added as a base to deprotonate three of the water ligands, which precipitates a chromium (III) hydroxide complex.
  This is not a ligand exchange! It’s the same as how a small amount of ammonia deprotonates H₂O ligands; we saw this in Ligand exchange reactions, except that OH^- does not act as a ligand at all. It is a strong base and only deprotonates H₂O here:

[Cr(H₂O)₆]³⁺ + 3OH^- _(aq) $\,$→$\,$ [Cr(H₂O)₃(OH)₃] _(s) + 3H₂O _(l)
Adding an excess of NaOH will then cause the precipitate to redissolve: three more deprotonations occur to give a hexahydroxychromium ion.

[Cr(H₂O)₃(OH)₃] + 3OH^- $\,$→$\,$ [Cr(OH)₆]^3- + 3H₂O
It is this [Cr(OH)₆]^3- that reacts with hydrogen peroxide to give CrO₄^2- chromate, which is a yellow coloured chromium (VI) compound.

2[Cr(OH)₆]^3- + H₂O₂ $\,$→$\,$ 2CrO₄^2- + 2OH^- + 8H₂O
Acid is then added, which converts chromate to dichromate in the equilibrium below, turning the solution from yellow to orange:

2CrO₄^2- + 2H⁺ $\, \rightleftharpoons \,$ Cr₂O₇^2- + H₂O
This reversible process means dichromate can convert back to chromate by adding base. This produces H₂O from the acidic conditions, so the equilibrium above shifts left to re-balance the sudden influx of H₂O.
To get dichromate, add acid. To get chromate, add base!


• The reduction from chromium (III) to chromium (II) can also be done by zinc:
• Cr³⁺ + e^- $\, \rightleftharpoons \,$ Cr²⁺ $\,$ E⁰ = -0.424 V
• Zn $\, \rightleftharpoons \,$ Zn²⁺ + 2e^- $\,$ E⁰ = -0.76
All that matters is that your intended reduction process has a higher (more positive, less negative) E⁰ than the intended oxidation process, which still holds true here. So zinc metal can oxidise Cr³⁺ ions to Cr²⁺ ions!


¹ Source for standard potential data: ATKINS, P. W., & DE PAULA, J. (2006). Atkins' Physical chemistry. Oxford, Oxford University Press.

² Source for vanadium and chromium standard potentials: https://chem.libretexts.org/Ancillary_Materials/Reference/Reference_Tables/Electrochemistry_Tables/P1%3A_Standard_Reduction_Potentials_by_Element