Variable Oxidation States

Transition metals have variable oxidation states due to the incomplete d-subshell. Vanadium can have a range of oxidation states depending on the compound/ion it’s in.

You need to know the formula and colour for the following ions:

Oxidation state

Formula of ion

VO₂⁺

VO₃^–

VO²⁺

V³⁺

V²⁺

Molecular ion name

Dioxovanadium(v) ion

Vanadate (v) ion

Oxovanadium (iv) ion

Vanadium (iii) ion

Vanadium (ii) ion

Colour in solution (aq)

Yellow

Blue

Green

Violet

If you memorise the formulas of the ions and their colours you can easily work out the oxidation state as we know the oxidation state of oxygen is -2 (except in peroxides where it is -1 and when bonded to fluorine it is +2).

You also need to be able to write the standard electrode potential (reduction) equations for the reduction of vanadium from +5 to +2 by zinc in the presence of dilute HCl. Instead of memorising all the equations, if you know the formulas of the ions you can then just balance the equations.

Here you can find step by step instructions for balancing the vanadium equations. This method can be used for other equations too.

Oxidation states
+5 -> +4

VO₂⁺ -> VO²⁺

^{VO₂+ is being reduced from +5 to +4 so it must gain an e^–}

^{VO₂⁺ + e^– -> VO²⁺}

^{To balance the oxygens we add H₂O}

^{VO₂⁺ + e^– -> VO²⁺ + H₂O}

^{Now we need to balance the hydrogens and the charges by adding H⁺}

^{VO₂⁺ + e^– + 2H⁺ -> VO²⁺ + H₂O}

^{The charge on each side of the equation is 2+ and the equation is balanced. As the electrons are on the left side, this is a reduction equation.}

^{VO₂⁺ + e^– + 2H⁺ -> VO²⁺> + H₂O}^{Yellow -> Blue}

^{As this reaction occurs, the solution changes from yellow to green to blue due to the mixing of the yellow and blue solutions. This is not due to the presence of the vanadium (iii) ion.}

Oxidation states
+4 -> +3

VO²⁺ -> V³⁺

^{VO²⁺ is being reduced (its oxidation number has decreased) from +4 to +3 so it has gained an electron.}

^{VO²⁺ + e^– -> V³⁺}

^{Add H₂O to balance the oxygens}

^{VO²⁺ + e^– -> V³⁺ + H₂O}

^{Add H⁺ to balance the hydrogens and the charges}

^{VO²⁺ + 2H⁺ + e^– -> V³⁺ + H₂O
Blue -> Green}

Oxidation states
+3 -> +2

V³⁺ -> V²⁺

V³⁺ is being reduced from +3 to +2 so it has to gain an electron

^{V³⁺ + e^– -> V²⁺}

There are no oxygens and the charge on both sides is 2+ so the equation is balanced.

V³⁺ + e^– -> V²⁺Green -> Violet

Vanadium Reduction Equations from +5 to +2

+5 to +4

+4 to +3

+3 to +2

VO₂⁺ + e^– + 2H⁺ -> VO²⁺ + H₂O

Yellow -> Blue

VO²⁺ + 2H⁺ + e^– -> V³⁺ + H₂O

Blue -> Green

V³⁺ + e^– -> V²⁺

Green -> Violet

E^θ = +1.00 V

E^θ = +0.32 V

E^θ = -0.26 V

Zinc can reduce vanadium all the way from an oxidation state of +5 to +2 this is because the standard electrode potential of zinc is more negative than all of the reactions seen above:

Zn²⁺ + 2e^– -> Zn

E^θ = -0.76 V

The more negative standard electrode potential is where oxidation takes place. Zinc is oxidised and acts as a reducing agent to reduce the vanadium. This reaction takes place in acidic conditions, the H⁺ ions are provided by dilute HCl. If we calculated the EMF for each reaction we would find that they are all feasible (reduction EMF – oxidation EMF = positive = feasible).

Tollens' Reagent

Tollens’ reagent is used to distinguish between aldehydes and ketones. It contains the [Ag(NH₃)₂]⁺ complex

TO BE CONTINUED