Edexcel International A Level Chemistry

Revision Notes

5.3.1 Vanadium Chemistry

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

colours-and-oxidation-states

  • 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

Colours of vanadium, downloadable AS & A Level Biology revision notes

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

6-3-1-vanadium-table-2

  • The half equations are arranged from the most negative Eθ at the top to the most positive Eθ 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 oxidising 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 rightwards harpoon over leftwards harpoon Zn (s)     Eθ = – 0.76 V
    • 5. VO2+ (aq) + 2H+ (aq) + e rightwards harpoon over leftwards harpoon VO2+ (aq) + H2O (l)     Eθ = + 1.00 V
  • Half equation 5 has the most positive standard electrode potential, Eθ, value
    • Therefore, this is the reduction reaction 
    • VO2+ (aq) + 2H+ (aq) + e rightwards harpoon over leftwards harpoon VO2+ (aq) + H2O (l) 
  • Half equation 2 has the least positive standard electrode potential, Eθ, value
    • Therefore, this is the oxidation reaction and needs to be reversed
    • Zn (s) rightwards harpoon over leftwards harpoon 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  begin mathsize 14px style bold rightwards harpoon over leftwards harpoon end style  2VO2+ (aq) + 2H2O (l) 
      Zn (s) rightwards harpoon over leftwards harpoon  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) begin mathsize 14px style bold rightwards harpoon over leftwards harpoon end style 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 rightwards harpoon over leftwards harpoon Zn (s)     Eθ = – 0.76 V
    • 4. VO2+ (aq) + 2H+ (aq) + e rightwards harpoon over leftwards harpoon V3+ (aq) + H2O (l)     Eθ = + 0.34 V
  • Half equation 4 has the most positive standard electrode potential, Eθ, value
    • Therefore, this is the reduction reaction 
    • VO2+ (aq) + 2H+ (aq) + e rightwards harpoon over leftwards harpoon V3+ (aq) + H2O (l) 
  • Half equation 2 has the least positive standard electrode potential, Eθ, value
    • Therefore, this is the oxidation reaction and needs to be reversed
    • Zn (s) rightwards harpoon over leftwards harpoon 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   begin mathsize 14px style bold rightwards harpoon over leftwards harpoon end style  2V3+ (aq) + 2H2O (l) 
      Zn (s) begin mathsize 14px style bold rightwards harpoon over leftwards harpoon end style  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) begin mathsize 14px style bold rightwards harpoon over leftwards harpoon end style 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 rightwards harpoon over leftwards harpoon Zn (s)     Eθ = – 0.76 V
    • 3. V3+ (aq) + e rightwards harpoon over leftwards harpoon V2+ (aq)      Eθ = – 0.26 V
  • Half equation 3 has the most positive / least negative standard electrode potential, Eθ, value
    • Therefore, this is the reduction reaction 
    • V3+ (aq) + e rightwards harpoon over leftwards harpoon V2+ (aq)
  • Half equation 2 has the least positive standard electrode potential, Eθ, value
    • Therefore, this is the oxidation reaction and needs to be reversed
    • Zn (s) rightwards harpoon over leftwards harpoon 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)
    • When combining these half equations, there is no need to change them as they have the same number of electrons in both equations
      V3+ (aq) + e  begin mathsize 14px style bold rightwards harpoon over leftwards harpoon end style  V2+ (aq) 
      Zn (s) begin mathsize 14px style bold rightwards harpoon over leftwards harpoon end style  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) begin mathsize 14px style bold rightwards harpoon over leftwards harpoon end style 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 rightwards harpoon over leftwards harpoon Zn (s)     Eθ = – 0.76 V
    • 1. V2+ (aq) + 2e rightwards harpoon over leftwards harpoon V (s)      Eθ = – 1.18 V
  • Half equation 1 has the most negative standard electrode potential, Eθ, value
    • Therefore, this is the oxidation reaction and needs to be reversed 
    • V (s) rightwards harpoon over leftwards harpoon V2+ (aq) + 2e 
  • Half equation 2 has the most positive / least negative standard electrode potential, Eθ, value
    • Therefore, this is the reduction reaction
    • Zn2+ (aq) + 2e rightwards harpoon over leftwards harpoon 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)  begin mathsize 14px style bold rightwards harpoon over leftwards harpoon end style  V2+ (aq) + 2e 
       Zn2+ (aq) + 2e   begin mathsize 14px style bold rightwards harpoon over leftwards harpoon end style  Zn (s)
    •  The equal number of electrons on both sides of the equation cancel out to give the overall equation: 

V (s) + Zn2+ (aq) begin mathsize 14px style bold rightwards harpoon over leftwards harpoon end style 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

Exam Tip

It is important to not get confused between the two oxo ions of vanadium VO2and VO2+

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