• Vanadium is a transition metal which has variable oxidation states
• The table below shows the important ones you need to be aware of

• Addition of zinc to the vanadium(V) in acidic solution will reduce the vanadium down through each successive oxidation state
  • ?The colour would successively change from yellow to blue to green to violet
• The ion with the V at oxidation state +5 exists as a solid compound in the form of a VO3-?ion
  • Usually as NH4VO3?known as ammonium vanadate(V)
  • It is a reasonably strong oxidising agent
  • Addition of acid to the solid will turn into the yellow solution containing the VO2+?ion.

• For vanadium we need to consider the following standard electrode potential values
• We will use zinc as our chosen oxidising agent
• The half equations are arranged from high negative EΘ?at the top to high positive?EΘ?at the bottom
  • The best reducing agent is the top right species (V2+)
  • The best oxidising agent is the bottom left species (VO2+)

Reduction from +5 to +4

• The two half equations we need to consider are 2 and 5
• Vanadium is reduced from an oxidation number of +5 to +4 in half equation 5
• The EΘ?value for half equation 2 is more negative than the?EΘ?for half equation 5
  • Zn is the best reducing agent
  • VO2+?is the best oxidising agent
• We can obtain the overall equation by reversing half equation 2 and combining with equation 5
  • When adding half equations remember to multiply them so each have the same number of electrons

2VO2+?(aq) + 4H+?(aq)?+ Zn (s) → 2VO2+?(aq) + Zn2+?(aq)?+ 2H2O (l)

Reduction from +4 to +3

• The two half equations we need to consider are 2 and 4
• Vanadium is reduced from an oxidation number of +4 to +3 in half equation 4
• The EΘ?value for half equation 2 is more negative than the?EΘ?for half equation 5
  • Zn is the best reducing agent
  • VO2+?is the best oxidising agent
• We can obtain the overall equation by reversing half equation 2 and combining with equation 4
  • When adding half equations remember to multiply them so each have the same number of electrons

2VO2+?(aq)?+ 4H+?(aq)?+ Zn (s) → 2V3+?(aq) + Zn2+?(aq)?+ 2H2O (l)

Reduction from +3 to +2

• The two half equations we need to consider are 2 and 3
• Vanadium is reduced from an oxidation number of +3 to +2 in half equation 3
• The EΘ?value for half equation 2 is more negative than the?EΘ?for half equation 3
  • Zn is the best reducing agent
  • V3+?is the best oxidising agent
• We can obtain the overall equation by reversing half equation 2 and combining with equation 3
  • When adding half equations remember to multiply them so each have the same number of electrons

2V3+?(aq)?+ Zn (s) → 2V2+?(aq) + Zn2+?(aq)

Reduction from +2 to 0

• The two half equations we need to consider are 1 and 2
• Vanadium is reduced from an oxidation number of +2 to 0 in half equation 1
• The EΘ?value for half equation 1 is more negative than the EΘ?for half equation 2
  • 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