Do Now Knowing what you already know… write out the reaction between hexaaquachromium(III) ion and aqueous sodium hydroxide. Knowing what you already know… write out the reaction between hexaaquachromium(III) ion and aqueous ammonia.

Chromium Chemistry

With aqueous sodium hydroxide Green/violet solution Green precipitate

With excess aqueous sodium hydroxide

Green solution Green solution But the process doesn't stop there. More hydrogen ions are removed to give ions like [Cr(H2O)2(OH)4]- and [Cr(OH)6]3-. For example: The precipitate redissolves because these ions are soluble in water. In the test-tube, the colour changes are: Green solution Green solution

How could you reverse the reaction? What is meant by amphoteric?

Amphoteric Describes acid-base reaction Reversible nature Species can react with both acids and bases

The ammonia acts as both a base and a ligand The ammonia acts as both a base and a ligand. With a small amount of ammonia, hydrogen ions are pulled off the hexaaqua ion exactly as in the hydroxide ion case to give the same neutral complex.

That precipitate dissolves to some extent if you add an excess of ammonia (especially if it is concentrated). The ammonia replaces water as a ligand to give hexaamminechromium(III) ions.

What is the oxidation state of chromium?

Changing the oxidation state… If the solution is alkaline [Cr(OH)6]3-, oxidation is easily achieved Adding oxidising agent hydrogen peroxide Solution changes from green to yellow Chromate(VI) ion, with oxidation number +6 is formed 2[Cr(OH)6]3- + 3H2O2  2CrO42- + 2OH- + 8H2O

This is then oxidised by warming it with hydrogen peroxide solution This is then oxidised by warming it with hydrogen peroxide solution. You eventually get a bright yellow solution containing chromate(VI) ions. The equation for the oxidation stage is:

Changing between them is easy: You are probably more familiar with the orange dichromate(VI) ion, Cr2O72-, than the yellow chromate(VI) ion, CrO42-. Changing between them is easy: If you add dilute sulphuric acid to the yellow solution it turns orange. If you add sodium hydroxide solution to the orange solution it turns yellow. The equilibrium reaction at the heart of the interconversion is:

Reduction of dichromate(VI) ions Dichromate(VI) ions (for example, in potassium dichromate(VI) solution) can be reduced to chromium(III) ions….. ….and then to chromium(II) ions using zinc and either dilute sulphuric acid or hydrochloric acid.

Reduction of dichromate(VI) ions Hydrogen is produced from a side reaction between the zinc and acid. This must be allowed to escape, but you need to keep air out of the reaction. Oxygen in the air rapidly re-oxidises chromium(II) to chromium(III).

An easy way of doing this is to put a bit of cotton wool in the top of the flask (or test-tube) that you are using. This allows the hydrogen to escape, but stops most of the air getting in against the flow of the hydrogen.

The equations for the two stages of the reaction are: For the reduction from +6 to +3: For the reduction from +3 to +2:

Standard Electrode Potentials Zn2+ + 2e- Zn EƟ = -0.76V Cr3+ + 3e- Cr2+ EƟ = -0.41V CrO42- + 4H2O + 3e- Cr(OH)3 + 5OH- EƟ = -0.13V H2O2 + 2e- 2OH- EƟ = +1.24V Cr2O72- + 14H+ + 6e- 2Cr3+ + 7H2O EƟ = +1.33V

Standard Electrode Potential E° values give you a way of comparing the positions of equilibrium when these elements lose electrons to form ions in solution. The more negative the E° value, the further the equilibrium lies to the left - the more readily the element loses electrons and forms ions. The more positive (or less negative) the E° value, the further the equilibrium lies to the right - the less readily the element loses electrons and forms ions.

RAD AND OAT

Standard Electrode Potentials Best reducing agent Standard Electrode Potentials Zn2+ + 2e- Zn EƟ = -0.76V Cr3+ + 3e- Cr2+ EƟ = -0.41V CrO42- + 4H2O + 3e- Cr(OH)3 + 5OH- EƟ = -0.13V H2O2 + 2e- 2OH- EƟ = +1.24V Cr2O72- + 14H+ + 6e- 2Cr3+ + 7H2O EƟ = +1.33V Best oxidising agent

Explaining oxidation from +3 to +6 2Cr(OH)3 + 3H2O2 + 4OH-  2CrO42- + 8H2O CrO42- + 4H2O + 3e- Cr(OH)3 + 5OH- EƟ = -0.13V H2O2 + 2e- 2OH- EƟ = +1.24V Cr(OH)3  more negative, so best reducing agent H2O2  less negative (positive), so best oxidising agent

Explaining reduction from +6 to +3 Cr2O72- + 14H+ + 3Zn  2Cr3+ + 7H2O + 3Zn2+ Zn2+ + 2e- Zn EƟ = -0.76V Cr2O72- + 14H+ + 6e- 2Cr3+ + 7H2O EƟ = +1.33V Zn  best reducing agent Cr2O72-  best oxidising agent

Explaining reduction from +3 to +2 2Cr3+ + Zn  2Cr2+ + Zn2+ Zn2+ + 2e- Zn EƟ = -0.76V Cr3+ + 3e- Cr2+ EƟ = -0.41V Zn  best reducing agent Cr3+  best oxidising agent

In your notes, make sure you have copied down page 129, Table A Task In your notes, make sure you have copied down page 129, Table A

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