Edexcel A Level Chemistry:复习笔记6.3.1 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

  • 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.

Interconversions of Vanadium Ions

  • 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+)

6-3-1-vanadium-table-2

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

Exam Tip

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

 

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