Chlorine gas is bubbled through a colourless solution of potassium bromide. (a) State the colour change seen in the solution during the reaction. [1] (b) Na...

This question tests the halogen displacement rule: a more reactive halogen removes a less reactive halogen from a solution of its ions. Chlorine is above bromine in Group VII, so it is the more reactive of the two.

(a) The potassium bromide solution starts colourless. As chlorine displaces bromine, free bromine (Br₂) dissolves in the water and colours it, so the colour change is colourless to orange / yellow-brown [1]. The orange-brown colour is the visible sign that a halogen has been set free.

(b) One element (bromine) is pushed out of its compound by a more reactive element (chlorine), so this is a displacement reaction [1]. Because electrons are also transferred, redox is accepted.

(c) The balanced ionic equation, showing only the species that change, is:

Cl₂ + 2Br^- → 2Cl^- + Br₂ [1 for correct species; 1 for balancing with charges = 2]

Check: 2 Cl and 2 Br on each side, and total charge is 2- on both sides. The potassium ions (K⁺) are spectator ions, so they are left out of the ionic equation.

(d) Chlorine reacts this way because it is more reactive than bromine, so it is able to take the electrons from the bromide ions and displace bromine from solution [1]. Reactivity falls down Group VII, so the halogen that is higher up always displaces the one below it.

Exam tip: for the ionic equation, cancel the spectator ion (K⁺) and always check the charges balance, not just the atoms.