Suggested Practicals

Investigating Halogen Displacement

If you want to know which wrestler is the strongest, you put them in a ring together to see who wins. A displacement reaction works on the exact same principle: a more reactive element will "kick out" and take the place of a less reactive element in a compound.

To observe this practically, you can test different halogens against various halide solutions. You must wear safety goggles and work in a fume cupboard, as both chlorine and bromine release toxic vapours.

1. Place a small volume (approximately $1-2\text{ cm}^3$) of a colourless metal halide salt solution (like potassium chloride, potassium bromide, or potassium iodide) into a spotting tile or test tube.
2. Add a few drops of an aqueous halogen solution (chlorine water, bromine water, or iodine solution) to the same well.
3. Observe the mixture carefully and record any colour changes.

Observations and Colour Changes

Metal halide solutions are inherently colourless. If a reaction occurs, the newly formed elemental halogen will change the colour of the solution. If no reaction occurs, the mixture simply remains the colour of the added halogen water.

Sometimes, distinguishing between a dark orange bromine solution and a pale brown iodine solution can be tricky. To clarify the results, you can add a highly flammable organic solvent called cyclohexane, stopper the tube, and shake it. The halogens dissolve much more easily in this top organic layer, producing distinct diagnostic colours:

• Chlorine: Pale green / colourless
• Bromine: Orange
• Iodine: Purple / Lilac / Pink

Explaining Displacement as Redox

Every time you see a halogen displacement, you are watching a transfer of electrons. These are redox reactions, meaning both oxidation (loss of electrons) and reduction (gain of electrons) happen simultaneously. You can remember this using the acronym OIL RIG.

• The more reactive halogen acts as the oxidizing agent. It pulls electrons away from the halide, gaining them itself and becoming reduced.
• The less reactive halide ion acts as the reducing agent. It loses its extra electron to the stronger halogen, meaning it is oxidized back into its elemental form.
• The metal ions attached to the halide (like potassium or sodium) are spectator ions. They do not gain or lose electrons, so we omit them from the net ionic equation.

Worked Example

Write the full balanced equation, the net ionic equation, and the half-equations to explain the redox reaction between aqueous bromine and sodium iodide.

Step 1: Write the full balanced symbol equation.

• $Br_2(aq) + 2NaI(aq) \rightarrow 2NaBr(aq) + I_2(aq)$

Step 2: Identify and remove the spectator ions to construct the net ionic equation.

• The sodium ions ($Na^+$) do not participate in the electron transfer.
• $Br_2(aq) + 2I^-(aq) \rightarrow 2Br^-(aq) + I_2(aq)$

Step 3: Split the net ionic equation into two half-equations showing electron transfer.

• Reduction (gain of electrons): $Br_2(aq) + 2e^- \rightarrow 2Br^-(aq)$
• Oxidation (loss of electrons): $2I^-(aq) \rightarrow I_2(aq) + 2e^-$