6.12 - Halogen displacement reactions as redox

From displacement to ions

A halogen displacement reaction involves a halogen molecule and halide ions in aqueous solution. A more reactive halogen displaces a less reactive halogen from its halide ions.

For example, bromine is more reactive than iodine, so bromine reacts with iodide ions:

bromine + potassium iodide -> potassium bromide + iodine
Br2(aq) + 2KI(aq) -> 2KBr(aq) + I2(aq)

The potassium ions are present before and after the reaction. They are spectator ions, so they are not the particles doing the electron transfer.

The net ionic equation keeps only the reacting particles:

Br2(aq) + 2I-(aq) -> 2Br-(aq) + I2(aq)

This equation shows the key swap. Bromine molecules become bromide ions, and iodide ions become iodine molecules.

Oxidation and reduction

Edexcel 6.12 asks for redox in terms of electron gain and electron loss. The useful memory aid is:

OIL = Oxidation Is Loss of electrons
RIG = Reduction Is Gain of electrons

In a halogen displacement reaction, electrons move from the less reactive halide ions to the more reactive halogen molecules.

The more reactive halogen molecule gains electrons to form halide ions, so it is reduced:

X2 + 2e- -> 2X-

The less reactive halide ions lose electrons to form a halogen molecule, so they are oxidised:

2Y- -> Y2 + 2e-

Here X is the more reactive halogen and Y− is the halide ion of the less reactive halogen.

Accounting for the electrons

Use the ionic equation to decide what has gained or lost electrons:

Br2(aq) + 2I-(aq) -> 2Br-(aq) + I2(aq)

Bromine starts as $\ce{Br2}$ and ends as $\ce{Br-}$ ions. To become negative ions, bromine must gain electrons:

Br2(aq) + 2e- -> 2Br-(aq)

Bromine is reduced because it gains electrons.