4 Variable oxidation states

Variable Oxidation States Transition elements show 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) Purple colourless 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+. 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+ Orange green The Fe2+ and Cr2O7 2- in acid solution reaction can be used as a quantitative redox titration. This does not need an indicator Oxidising Chromium When transition metals in low oxidation states are in alkaline solution they are more easily oxidised than when in acidic solution [Cr(H2O)6 ]3+ (aq) [Cr(OH)6 ]3- (aq) excessNaOH Acidified Not easy to oxidise alkaline easier to oxidise 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(OH)6 3- (aq) H2O2 CrO4 2- (aq) Green solution Reduction :H2O2 + 2e-  2OH- Oxidation: [Cr(OH)6 ]3- + 2OH- CrO4 2- + 3e- + 4H2O 2 [Cr(OH)6 ]3- + 3H2O2 2CrO4 2- +2OH- yellow + 8H2O solution. Reduction of Cu2+ to Cu+ Cu2+ (blue solution) can be reduced to Cu+ (colourless solution) by various reducing agents. We commonly use potassium iodide solution. 2Cu2+ (aq) + 2I- (aq)  I2 (aq) + 2Cu+ (aq) Disproportionation of copper(I) ions Copper(I) ions when reacting with sulphuric acid will disproportionate to Cu2+ and Cu metal 2Cu+  Cu + Cu2+ Cu+ (aq) + e−  Cu(s) Eo = +0.52 V Cu2+ (aq) + e−  Cu+ (aq) Eo = +0.15 V So Eo cell = 0.52 − 0.15 = +0.37 V 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+