Reactions of Vanadium and Other Transition Metals

Vanadium is one of the clearest examples of why transition metals are studied as a distinct block of elements. As a chemical element, vanadium can exist in several stable oxidation states, and moving between them produces a striking sequence of colour changes that you can literally watch happen in a test tube. This lesson looks at the vanadium oxidation states in detail, then compares that behaviour with the redox chemistry of other transition metals such as iron, copper, chromium, and manganese.

Before working through this topic, it helps to be comfortable with the general electron configurations and properties covered in Introduction to transition metals and the d-block elements, since variable oxidation state is one of the defining features introduced there.

Vanadium has the electron configuration \([Ar]3d^3 4s^2\). Because the 3d and 4s electrons are close in energy, vanadium can lose different numbers of electrons depending on the reaction conditions, giving several possible vanadium oxidation numbers: plus 5, plus 4, plus 3, and plus 2. Each oxidation state exists as a differently coloured ion in aqueous solution, which is why vanadium chemistry is often used to demonstrate oxidation state changes visually in the lab.

Note that the plus 5 and plus 4 states are written as oxo-cations, \(VO_2^+\) and \(VO^{2+}\), rather than as simple \(V^{5+}\) or \(V^{4+}\) ions, because a bare 5+ or 4+ charge on such a small ion would be far too polarising to exist in water. The plus 3 and plus 2 states, by contrast, exist as simple hydrated cations, \(V^{3+}\) and \(V^{2+}\).

Starting from ammonium vanadate(V), \(NH_4VO_3\), dissolved in dilute sulfuric acid gives the yellow \(VO_2^+\) ion. Adding an excess of granulated zinc and warming gently allows the zinc, a strong reducing agent, to reduce vanadium through each oxidation state in turn. The colour of the solution changes visibly at each stage:

\(VO_2^+\) (yellow) → \(VO^{2+}\) (blue) → \(V^{3+}\) (green) → \(V^{2+}\) (violet)

The relevant half equations for each one-electron reduction step are:

\( VO_2^+ + 2H^+ + e^- \)→\( VO^{2+} + H_2O \)

\( VO^{2+} + 2H^+ + e^- \)→\( V^{3+} + H_2O \)

\( V^{3+} + e^- \)→\( V^{2+} \)

Zinc supplies electrons as it is oxidised: \( Zn \)→\( Zn^{2+} + 2e^- \). Because zinc is in excess, the reduction does not stop until vanadium reaches its lowest common oxidation state, plus 2. If the solution is left exposed to air, atmospheric oxygen slowly reoxidises \(V^{2+}\) back toward \(V^{3+}\) and \(VO^{2+}\), so these reactions are usually carried out under an inert atmosphere or with minimal air contact for a clean colour sequence.

The most common vanadium oxide formula you will meet is vanadium(V) oxide, \(V_2O_5\), a yellow-orange solid. \(V_2O_5\) is important industrially as the catalyst in the Contact process for manufacturing sulfuric acid, where it converts \(SO_2\) to \(SO_3\) by cycling between the plus 5 and plus 4 oxidation states of vanadium. This is a good example of how a transition metal's variable oxidation state enables catalysis; the general principles behind this are covered separately in Transition metal catalysts.

Vanadium is not unique in showing multiple oxidation states with characteristic colours. Several other transition metals behave similarly, and comparing them helps show this is a general feature of the d-block rather than a one-off.

• Iron: pale green \(Fe^{2+}\) is oxidised to yellow-brown \(Fe^{3+}\) by mild oxidising agents, and the reverse reduction is used in redox titrations against manganate(VII).
• Manganese: deep purple manganate(VII), \(MnO_4^-\), is reduced to almost colourless pale pink \(Mn^{2+}\) in acidic conditions, a reaction widely used as a titration indicator in its own right because the colour change is so sharp.
• Chromium: orange dichromate(VI), \(Cr_2O_7^{2-}\), is reduced to green \(Cr^{3+}\) by reducing agents such as \(Fe^{2+}\) or ethanol; \(Cr^{3+}\) can be reduced further to blue \(Cr^{2+}\) under strongly reducing conditions.
• Copper: blue \(Cu^{2+}\) reacts with iodide ions in a disproportionation-driven reaction, \( 2Cu^{2+} + 4I^- \)→\( 2CuI + I_2 \), where copper is reduced to the plus 1 state as insoluble white copper(I) iodide.

Some transition metal reactions are disproportionation reactions, where the same element is simultaneously oxidised and reduced. The copper and iodide reaction above is one example: copper(II) is reduced to copper(I) while iodide is oxidised to iodine, and it is common to see this classified as disproportionation of the overall system even though strictly only the iodide changes oxidation state on one side. A cleaner disproportionation example is copper(I) in water, where \(Cu^+\) converts partly to \(Cu^{2+}\) and partly to \(Cu\) metal, since copper(I) is thermodynamically unstable in aqueous solution.

The ability of vanadium and other transition metals to adopt multiple oxidation states, each often with its own colour, underpins several practical uses in chemistry: redox titrations (using colour change to signal an endpoint), industrial catalysis (cycling between oxidation states without being consumed), and even rechargeable battery chemistry, where vanadium redox flow batteries exploit exactly the four oxidation states described above. Recognising the oxidation state from the colour of a solution, and being able to write the correct half equation for a given electron transfer, are the key skills to take from this topic.

Related lessons

• Uk A Level Chemistry