Variable Oxidation States

• When transition metals form ions they lose the 4s electrons before the 3d
• Iron (II) Oxidation

  • Fe2+ (green solution) can be easily oxidised to Fe3+ (brown solution) by various oxidising agents.
  • We commonly use potassium manganate (VII), although oxygen in the air will bring about the change
  • MnO4 -(aq) + 8H+ (aq) + 5Fe2+ (aq) Mn2+ (aq) + 4H2O (l) + 5Fe3+ (aq)

  
• Iron (III) Reduction

  • Fe3+ (brown solution) can be reduced to Fe2+ (green solution) by various reducing agents.
  • We commonly use potassium iodide solution.
  • The brown colour of the iodine formed can mask the colour change of the Iron.
  • 2Fe3+ (aq) + 2I- (aq)  I2 (aq) + 2Fe2+ (aq)

  
• Reducing Chromium

  • Cr3+ (green) and then Cr2+ (blue) are formed by reduction of Cr2O7 2- (orange) by the strong reducing agent zinc in (HCl) acid solution.
  • Fe2+ is a less strong reducing agent and will only reduce the dichromate to Cr3+
  • Cr2O7 2- + 14H+ + 6Fe2+  2Cr3+ + 7H2O + 6 Fe3+
  • The Fe2+ and Cr2O7 2- in acid solution reaction can be used as a quantitate redox titration. this doesn't need an indicator

  
• • When transition metals in low oxidation states are in alkaline solution they are more easily oxidised than when in acidic solution
  • It is easier to remove an electron from a negatively charged ion
• Alkaline chromium(III) can be oxidised by using oxidising agents such as hydrogen peroxide to the (yellow solution) chromate ion

  [Cr(H2O)6 ]3+ (aq) [Cr(OH)6 ]3- (aq)

  • Excess NaOH Acidified Not easy to oxidise to alkaline easier to oxidise

  Cr(OH)6 3- (aq) (H2O2 ->) CrO4 2- (aq) Green solution to yellow solution

  Reduction :H2O2 + 2e- => 2OH-

  Oxidation: [Cr(OH)6 ]3- + 2OH- => CrO4 2- + 3e- + 4H2O

  2 [Cr(OH)6 ]3- + 3H2O2 => 2CrO4 2- +2OH- + 8H2O

  
• Reduction of Cu2+ to Cu+ Cu2+

  • (blue solution) can be reduced to Cu+ by various reducing agents.
  • We commonly use potassium iodide solution
  • . 2Cu2+ (aq) + 4I- (aq) => I2 (aq) + 2CuI (s) Brown solution to White precipitate

  
• read the Disproportionation of copper(I) ions

  • Copper(I) ions when reacting with sulfuric acid will disproportionate to Cu2+ and Cu metal
  • As Eo Cu+ /Cu > Eo Cu2+/Cu+ and Ecell has a positive value of +0.37V , Cu+ disproportionates from +1 oxidation state to 0 in Cu and +2 in Cu2+