Change of Iron(II) Ion to Iron(III) Ion
- Iron shows two oxidation numbers, that is +2 and +3.
- The aqueous solution of iron(II) ion Fe2+ is light green in colour. The aqueous solution of iron(III) ion Fe3+ is brown in colour.
- The change of iron(II) ion for iron(III) ion is an oxidation process. This can be done by mixing an oxidation agent.
Example
Procedure:
- 2 cm³ of iron(II) sulphate solution is poured into a test tube.
- Bromine water is added drop by drop into the solution until no further changes are observed.
- The mixture is then shaken and warmed gently.
- The observation is recorded.
Observation:
- The brown colour of bromine water turns colourless.
- The colour of the solution changes from light green to yellowish brown.
Half Equations:
Ionic Equation
Explanation:
- The light green colour of iron(II) sulphate solution turns brown because iron(II) ions Fe2+ are oxidise
- to become iron(III) ion, Fe3+.
- Iron(II) ion, Fe2+ undergoes oxidation by releasing an electron to form iron(III) ion, Fe3+.
- The brown colour of bromine water turns colourless because bromine molecules are reduced to become bromide ions.
- Bromine molecules receive electrons and undergo reduction to form bromide ion, Br–.
Oxidising agent: Bromine water
Reducing agent: Iron(II) ions Fe2+
Confirmation Test
2 cm³ of the solution of the product is filled into a test tube.
Test 1: Dilute sodium hydroxide solution (NaOH) is then added to the test tube until excess.
Result: Brown precipitate formed. The precipitate does not dissolve in excess sodium hydroxide solution.
Test 2: Dilute ammonium hydroxide solution (NH4OH)/ammonia aqueous (NH3) is then added into the test tube until excess
Result: Brown precipitate formed. The precipitate does not dissolve in excess ammonium hydroxide solution /ammonia aqueous.
Test 3: 2cm³ of potassium thiocyanate is added to the test tube.
Result: Red blood solution formed.
Other oxidation agents that get to replace bromine water
- Chlorine water
Half EquationsCl2 + 2e → 2Cl– - Acidic potassium manganate (VII)
Half EquationsMnO4– + 8H+ + 5e → Mn2+ + 4H2O - Potassium dichromate (VI)
Half EquationsCr2O72- + 14H+ + 6e → 2Cr3+ + 7H2O - Hydrogen peroxide
Half EquationsH2O2 + 2H+ + 2e → 2H2O - Concentrated nitric acid
Half EquationsNO3– + 4H+ + 3e → NO + 2H2O
Change of Iron(III) Ion to Iron(II) Ions
- The change of iron(III) ion for iron(II) ion is a reduction process. This can be done by mixing a reducing agent.
Example
Procedure:
- 2 cm³ of iron(III) sulphate solution is poured into a test tube.
- Half spatula of zinc powder is added to the solution.
- The mixture is then shaken and warmed gently.
- The observation is recorded.
Observation:
- Zinc powder dissolves.
- The brown coloured of iron(III) sulphate solution turn light green.
Half Equations:
Ionic Equation
Explanation:
- The brown colour of iron(III) sulphate solution turns light green shows that iron(III) ions, Fe3+ are
- reduced to iron(II) ion, Fe2+.
- Iron(III) ion, Fe3+ undergoes reduction by receiving electron to form iron(II) ion, Fe2+
- Zinc powder added is oxidised to form zinc ion, Zn2+.
Oxidising agent: Iron(III) ions Fe3+
Reducing agent: Zinc powder
Confirmation Test
2 cm³ of the solution of the product is filled into a test tube.
Test 1: Dilute sodium hydroxide solution (NaOH) is then added to the test tube until excess.
Result: Dirty green precipitate formed. The precipitate does not dissolve in excess sodium hydroxide solution.
Test 2: Dilute ammonium hydroxide solution (NH4OH)/ammonia aqueous (NH3) is then added into the test tube until excess
Result: Green precipitate formed. The precipitate does not dissolve in excess ammonium hydroxide solution /ammonia aqueous.
Test 3: 2cm³ of potassium thiocyanate is added to the test tube.
Result: No change observed
Other oxidation agents that get to replace bromine water
- Magnesium
Half EquationsMg → Mg2+ + 2e - Sulphur dioxide gas
Half EquationsSO2 + 2H2O → SO42- + 4H+ + 2e - Hydrogen sulphide gas
Half EquationsH2S → 2H+ + S + 2e - Solution of sodium sulphide, Na2SO3
Half EquationsSO32- + H2O → SO42- + 2H+ + 2e - Solution of tin(II) Chloride
Half EquationsSn2+ → Sn4+ + 2e