Transition metals: Some Vanadium Chemistry

Edexcel A level Chemistry (2017)

Topic 15: Principles of transition metal chemistry

Learning Objectives related to vanadium chemistry

15/20. To know what the colours of the oxidation states of vanadium (+5, +4, +3 and +2) are in its compounds.

15/21. To understand the redox reactions for the inter-conversion of the oxidation states of vanadium (+5, +4, +3 and +2), in terms of the relevant Eo values.

Some Vanadium Chemistry

Vanadium has several stable oxidation states at room temperature.

Vanadium (II)    V²⁺          purple

Vanadium (III)   V³⁺          green

Vanadium (IV)   VO²⁺         blue

Vanadium (V)    VO₂⁺        yellow

Each of these states can exist in a redox equilibrium with the one below in acid solution.

Each redox equilibrium has its own redox potential.

You add these to an oxidation number chart or redox potential chart.

Here they are:

 V³⁺(aq)   +   e^—     ⟶      V²⁺(aq)    E^o  = —0.26v

 [VO²⁺(aq) + 2H⁺(aq)] + e^—  ⟶   V³⁺(aq) + H₂O(l)   E^o  = +0.34v

 [VO₂⁺(aq) + 2H⁺(aq)] + e^—  ⟶   [VO²⁺(aq) + H₂O(l)]   E^o  = +1.00v

Here is the oxidation number chart for the reduction of these vanadium ions:

Reduction of Vanadium(V) to Vanadium(II)

As the zinc/zinc(II) reduction potential is —0.76v then addition of zinc amalgam (zinc dissolved in mercury) or zinc powder to a solution of vanadium(V) ions will bring about a reduction of the yellow vanadium(V) eventually to purple vanadium(II).

3Zn(s)  +  2VO₂⁺(aq) + 8H⁺(aq) ⟶  2V²⁺(aq)  +  4H₂O(l) + 3Zn²⁺(aq)

yellow                            purple

Reduction of Vanadium(V) to Vanadium(III)

Adding tin powder should reduce the vanadium(V) to vanadium(III) since the tin/tin(II) reduction potential lies at —0.14v.

2Sn(s)  +  2VO₂⁺(aq) + 8H⁺(aq) ⟶  2V³⁺(aq)  +  4H₂O(l) + 2Sn²⁺(aq)

yellow                            green

Reduction of Vanadium(V) to Vanadium(IV)

Adding iodide ions to a solution of vanadium(V) will reduce the vanadium(V) down to vanadium(IV) since the iodine/iodide reduction potential lies at +0.54v.

I^—(aq)  +  VO₂⁺(aq) + 2H⁺(aq) ⟶  VO²⁺(aq) +  H₂O(l) + ½I₂(aq)

yellow                            blue                        brown

However the resultant brown colour of the product iodine will mask the blue of vanadium(IV). 

To remove this brown colour add thiosulphate ions.

   I₂(aq)      +   2S₂O₃^2—(aq) ⟶     2I^—(aq)   +     S₄O₆^2—(aq)      

brown                                       colourless

You can view the experiment to show these changes in vanadium chemistry here.