Change of Iron(II) Ion to Iron(III) Ion and Vice Versa

Example

Procedure:

1. 2 cm³ of iron(II) sulphate solution is poured into a test tube.
2. Bromine water is added drop by drop into the solution until no further changes are observed.
3. The mixture is then shaken and warmed gently.
4. The observation is recorded.

Observation:

1. The brown colour of bromine water turns colourless. 
2. The colour of the solution changes from light green to yellowish brown.

Half Equations:

Ionic Equation

Explanation:

1. The light green colour of iron(II) sulphate solution turns brown because iron(II) ions Fe²⁺ are oxidise
2. to become iron(III) ion, Fe³⁺.
3. Iron(II) ion, Fe²⁺ undergoes oxidation by releasing an electron to form iron(III) ion, Fe³⁺.
4. The brown colour of bromine water turns colourless because bromine molecules are reduced to become bromide ions. 
5. 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

1. Chlorine water
   Half Equations
   
   Cl₂ + 2e → 2Cl^–
2. Acidic potassium manganate (VII)
   Half Equations
   
   MnO₄^– + 8H⁺  +  5e   →      Mn²⁺ +      4H₂O
3. Potassium dichromate (VI)
   Half Equations
   
   Cr₂O₇^2- +   14H⁺ +  6e   →      2Cr³⁺ +     7H₂O
4. Hydrogen peroxide
   Half Equations
   
   H₂O₂ +     2H⁺ +   2e  →     2H₂O
5. Concentrated nitric acid
   Half Equations
   
   NO₃^– +    4H⁺ +   3e  →      NO +   2H₂O