Unit 5 Section 3 Variable Oxidation States

Transition metals can usually form more than one type of ion and in different ions, transition metals can have different oxidation states. These variable oxidation states are another important feature of the transition metals.

Oxidation States of Chromium:

• +6    -    Cr2O7^2-(aq)    -    Orange
• +6    -    CrO4^2-(aq)    -    Yellow
• +3    -    Cr^3+(aq)    -    Violet/Green
• +2    -    Cr^2+(aq)    -    Blue

When an alkali (OH- ions) is added to aqueous dichromate(VI) ions (Cr2O7^2-), the orange colour turns yellow, because aqueous chromate(VI) (CrO4^2-) ions are formed:

• Orange Cr2O7^2-(aq) + OH-(aq) >>> Yellow 2CrO4^2-(aq) + H+(aq)

When an acid is added to aqueous chromate(VI)  ions, the yellow colour turns orange, because aqueous dichromate(VI) ions form:

• Yellow 2CrO4^2-(aq) + H+(aq) >>> Orange Cr2O7^2-(aq) + OH-(aq)

These are opposite processes and the two ions exist in equilibrium:

• Cr2O7^2-(aq) + H2O(l) <<<>>> 2CrO4^2-(aq) + H+(aq)

The position of equilibrium depends on the pH. If H+ ions are added, the equilibrium shifts to the left so orange Cr2O7^2-(aq) ions are formed. If OH- ions are added, the H+ ions are removed and the equilibrium shifts to the right, forming yellow CrO4^2-(aq).

Dichromate(VI) ions can be reduced using a good reducing agent, such as zinc and dilute acid:

• Equation: Cr2O7^2-(aq) + 14H+(aq) + 3Zn(s) >>> 3Zn(aq) + 2Cr^3+(aq) + 7H2O(l)
• Colour:     Orange                                                                       Green
• Oxidation: +6                                         0                +2                +3

Zinc will reduce Cr^3+ further to Cr^2+:

• Equation: 2Cr^3+ (aq) + Zn(s) >>> Zn^2+(aq) + 2Cr^2+(aq)
• Colour:       Green                                                  Blue
• Oxidation: +6                  0               +2                   +2    

Unless an inert atmosphere is used Cr^2+ will revert back to Cr^3+ in air.

You can oxidise Cr^3+ to chromate(VI) ions with H2O2 in an alkaline solution:

• Equation: 2Cr^3+(aq) + 10OH-(aq) + 3H2O2(aq) >>> 2CrO4^2-(aq) + 8H2O(l)
• Colour:       Green                                                            Yellow
• Oxidation:  +3                                                                   +6

Cobalt can exist in two oxidation states: 2+ as Co^2+, and 3+ as Co^3+. It much prefers to be in the 2+ state though.

There are two ways of oxidising Co^2+ ions to Co^3+ ions. Firstly, Co^3+ can be made by oxidising Co^2+ with hydrogen peroxide in alkaline conditions:

• 2Co^2+(aq) + H2O2(aq) >>> 2Co^3+(aq) + 2OH-(aq)

Secondly, you can oxidise Co^2+ with air in an ammoniacal solution. You start with a pink solution of [Co(H2O)6]^2+ ions.

If you add a small amount of NH3(aq) to this solution, the following reaction occurs:

• [Co(H2O)6]^2+(aq) + 2NH3(aq) >>> [Co(H2O)4(OH)2](s) + 2NH4+(aq)

If you add excess of ammonia to the solution then [Co(NH3)6]^2+ ions form, producing a straw coloured solution. If these complex ions are left to stand in air, oxygen oxidises them to [Co(NH3)6]^3+ which is dark brown in colour.