Vanadium Chemistry (Edexcel International A Level (IAL) Chemistry): Revision Note

Vanadium - Colours & Oxidation States

• Vanadium is a transition metal which has variable oxidation states

• The table below shows the important ones you need to be aware of

• The variation in oxidation states of transition metal ions is illustrated by the reaction of zinc with ammonium vanadate(V) (also known as ammonium metavanadate) under acidic conditions

• Vanadium had four common oxidation states, from +2 to +5

• Zinc is a reducing agent that is capable to reducing vanadium(V) to vanadium(II) in a sequence of steps accompanied by vibrant colour changes

The reduction of vanadate(V) ions by zinc in acidic conditions is one of the most colourful reactions in chemistry

Vanadium - Interconversions of Ions

• For vanadium, we need to consider the following standard electrode potential values

• The half equations are arranged from the most negative E^θ at the top to the most positive E^θ at the bottom

  • The strongest oxidising agent is the reactant from the electrode reaction with the most positive standard electrode potential

    • In this case, the most positive value is + 1.00 V, which means that VO₂⁺ is the strongest oxidising agent

  • The strongest reducing agent is the product from the electrode reaction with the most negative standard electrode potential

    • In this case, the most negative value is – 1.18 V, which means that V is the strongest reducing agent 

• For all of the following reactions, we will use zinc as the reducing agent

Reduction of V from +5 to +4 by Zn

• Vanadium is being reduced from an oxidation number of +5 to +4 in half equation 5

• So, the two half equations we need to consider are:

  • 2. Zn²⁺ (aq) + 2e^– $\rightleftharpoons$ Zn (s)     E^θ = – 0.76 V

  • 5. VO₂⁺ (aq) + 2H⁺ (aq) + e^– $\rightleftharpoons$ VO²⁺ (aq) + H₂O (l)     E^θ = + 1.00 V

• Half equation 5 has the most positive standard electrode potential, E^θ, value

  • Therefore, this is the reduction reaction 

  • VO₂⁺ (aq) + 2H⁺ (aq) + e^– $\rightleftharpoons$ VO²⁺ (aq) + H₂O (l) 

• Half equation 2 has the least positive standard electrode potential, E^θ, value

  • Therefore, this is the oxidation reaction and needs to be reversed

  • Zn (s) $\rightleftharpoons$ Zn²⁺ (aq) + 2e^– 

• We can obtain the overall equation by combining the reduction equation (half equation 5) and the oxidation equation (the reversed half equation 2)

  • Remember: When combining half equations, they must have the same number of electrons in both equations

    2VO2+ (aq) + 4H+ (aq) + 2e–	$\mathbf{\rightleftharpoons }$	2VO2+ (aq) + 2H2O (l)
    Zn (s)	$\rightleftharpoons$	Zn2+ (aq) + 2e–

  •  The equal number of electrons on both sides of the equation cancel out to give the overall equation: 

2VO₂⁺ (aq) + 4H⁺ (aq) + Zn (s) $\mathbf{\rightleftharpoons }$ 2VO²⁺ (aq) + Zn²⁺ (aq) + 2H₂O (l)

Reduction of V from +4 to +3 by Zn

• Vanadium is being reduced from an oxidation number of +4 to +3 in half equation 4

• So, the two half equations we need to consider are:

  • 2. Zn²⁺ (aq) + 2e^– $\rightleftharpoons$ Zn (s)     E^θ = – 0.76 V

  • 4. VO²⁺ (aq) + 2H⁺ (aq) + e^– $\rightleftharpoons$ V³⁺ (aq) + H₂O (l)     E^θ = + 0.34 V

• Half equation 4 has the most positive standard electrode potential, E^θ, value

  • Therefore, this is the reduction reaction 

  • VO²⁺ (aq) + 2H⁺ (aq) + e^– $\rightleftharpoons$ V³⁺ (aq) + H₂O (l) 

• Half equation 2 has the least positive standard electrode potential, E^θ, value

  • Therefore, this is the oxidation reaction and needs to be reversed

  • Zn (s) $\rightleftharpoons$ Zn²⁺ (aq) + 2e^– 

• We can obtain the overall equation by combining the reduction equation (half equation 5) and the oxidation equation (the reversed half equation 2)

  • Remember: When combining half equations, they must have the same number of electrons in both equations

    2VO2+ (aq) + 4H+ (aq) + 2e–	$\mathbf{\rightleftharpoons }$	2V3+ (aq) + 2H2O (l)
    Zn (s)	$\mathbf{\rightleftharpoons }$	Zn2+ (aq) + 2e–

  •  The equal number of electrons on both sides of the equation cancel out to give the overall equation: 

2VO²⁺ (aq) + 4H⁺ (aq) + Zn (s) $\mathbf{\rightleftharpoons }$ 2V³⁺ (aq) + Zn²⁺ (aq) + 2H₂O (l) 

Reduction of V from +3 to +2 by Zn

• Vanadium is being reduced from an oxidation number of +3 to +2 in half equation 3

• So, the two half equations we need to consider are:

  • 2. Zn²⁺ (aq) + 2e^– $\rightleftharpoons$ Zn (s)     E^θ = – 0.76 V

  • 3. V³⁺ (aq) + e^– $\rightleftharpoons$ V²⁺ (aq)      E^θ = – 0.26 V

• Half equation 3 has the most positive / least negative standard electrode potential, E^θ, value

  • Therefore, this is the reduction reaction 

  • V³⁺ (aq) + e^– $\rightleftharpoons$ V²⁺ (aq)

• Half equation 2 has the least positive standard electrode potential, E^θ, value

  • Therefore, this is the oxidation reaction and needs to be reversed

  • Zn (s) $\rightleftharpoons$ Zn²⁺ (aq) + 2e^– 

• We can obtain the overall equation by combining the reduction equation (half equation 5) and the oxidation equation (the reversed half equation 2)

  • Remember: When combining half equations, they must have the same number of electrons in both equations

    2V3+ (aq) + 2e–	$\mathbf{\rightleftharpoons }$	2V2+ (aq)
    Zn (s)	$\mathbf{\rightleftharpoons }$	Zn2+ (aq) + 2e–

  • The equal number of electrons on both sides of the equation cancel out to give the overall equation: 

2V³⁺ (aq) + Zn (s) $\mathbf{\rightleftharpoons }$ 2V²⁺ (aq) + Zn²⁺ (aq) 

Reduction of V from +2 to 0 by Zn

• Vanadium is being reduced from an oxidation number of +3 to +2 in half equation 1

• So, the two half equations we need to consider are:

  • 2. Zn²⁺ (aq) + 2e^– $\rightleftharpoons$ Zn (s)     E^θ = – 0.76 V

  • 1. V²⁺ (aq) + 2e^– $\rightleftharpoons$ V (s)      E^θ = – 1.18 V

• Half equation 1 has the most negative standard electrode potential, E^θ, value

  • Therefore, this is the oxidation reaction and needs to be reversed 

  • V (s) $\rightleftharpoons$ V²⁺ (aq) + 2e^– 

• Half equation 2 has the most positive / least negative standard electrode potential, E^θ, value

  • Therefore, this is the reduction reaction

  • Zn²⁺ (aq) + 2e^– $\rightleftharpoons$ Zn (s) 

• We can obtain the overall equation by combining the reduction equation (half equation 5) and the oxidation equation (the reversed half equation 2)

  • When combining these half equations, there is no need to change them as they have the same number of electrons in both equations

    V (s)	$\mathbf{\rightleftharpoons }$	V2+ (aq) + 2e–
    Zn2+ (aq) + 2e–	$\mathbf{\rightleftharpoons }$	Zn (s)

  •  The equal number of electrons on both sides of the equation cancel out to give the overall equation: 

V (s) + Zn²⁺ (aq) $\mathbf{\rightleftharpoons }$ V²⁺ (aq) + Zn (s) 

• Zn is not electron releasing with respect to V²⁺

  • This means this reaction is not thermodynamically feasible

Predicting oxidation reactions

• The same method can be used to predict whether a given oxidising agent will oxidise a vanadium species to one with a higher oxidation number