Halogen Displacement Reactions (Edexcel A Level Chemistry)

• The reactivity of halogens is also shown by their displacement reactions with other halide ions in solutions
• A more reactive halogen can displace a less reactive halogen from a halide solution of the less reactive halogen
• Eg. The addition of chlorine water to a solution of bromine water:

Cl₂ (aq) + 2NaBr (aq) → 2NaCl (aq) + Br₂ (aq)

• The chlorine has displaced the bromine from solution as it is more reactive which can be summarised in the following ionic equation by removing the sodium spectator ions:

Cl₂ (aq) + 2Br^- (aq) → 2Cl^- (aq) + Br₂ (aq)

• Chlorine can also displace iodine from a solution of iodide ions:

Cl₂ (aq) + 2I^- (aq) → 2Cl^- (aq) + I₂ (aq)

Chlorine with Bromides & Iodides

If you add chlorine solution to colourless potassium bromide or potassium iodide solution a displacement reaction occurs:

• The solution becomes yellow-orange as bromine is formed
• The solution becomes brown as iodine is formed
• If an organic solvent is added, such as cyclohexane, the following observations are seen:
  • The organic layer will appear yellow-orange as bromine is formed
  • The organic layer will appear purple as iodine is formed 
  • The organic solvent is useful as the the halogens are more soluble in this layer which helps observe the colour changes more easily
• Chlorine is above bromine and iodine in Group 7 so it is more reactive
• Chlorine will displace bromine or iodine from an aqueous solution of the metal halide:

Cl₂ (aq) + 2KBr (aq) → 2KCl (aq) + Br₂ (aq)

chlorine + potassium bromide →  potassium chloride + bromine

Cl₂ (aq) + 2KI (aq) → 2KCl (aq) + I₂ (aq)

chlorine + potassium iodide →  potassium chloride + iodine

Bromine with Iodides

• Bromine is above iodine in Group 7 so it is more reactive
• Bromine will displace iodine from an aqueous solution of the metal iodide

bromine + potassium iodide →  potassium bromide + iodine

Br₂ (aq) + 2KI (aq) → 2KBr (aq) + I₂ (aq)

• The reaction will turn brown as iodine is formed 
• If an organic solvent is added the organic layer will appear purple as iodine is formed 
• We can show that this is a redox reaction by looking at the changes in oxidation numbers:
  • I = -1 → 0 so the iodine has been oxidised 
  • Br = 0 → -1 so the bromine has been reduced 
  • No change in oxidation number for the potassium
• Rather than writing the full equation, we can also write the ionic equation by removing the potassium spectator ion

Br₂ (aq) + 2I^- (aq) → 2Br^- (aq) + I₂ (aq)