Displacement as Redox

$\mathrm{Cl_2(aq) + 2NaBr(aq) \rightarrow 2NaCl(aq) + Br_2(aq)}$

The sodium ions are present before and after the reaction but do not change. Leaving them out makes it easier to see the reacting particles:

$\mathrm{Cl_2(aq) + 2Br^-(aq) \rightarrow 2Cl^-(aq) + Br_2(aq)}$

Following the electrons

Each bromide ion loses one electron. Two bromide ions therefore lose two electrons and form one bromine molecule:

$\mathrm{2Br^- \rightarrow Br_2 + 2e^-}$

Bromide ions are oxidised because oxidation is loss of electrons. The chlorine molecule gains those two electrons, forming two chloride ions:

$\mathrm{Cl_2 + 2e^- \rightarrow 2Cl^-}$

Chlorine is reduced because reduction is gain of electrons. The equations are called half-equations: one shows the electron loss and the other shows the electron gain. Together they account for the two electrons transferred, which cancel when the half-equations are combined.

Why this is redox

Oxidation and reduction occur together here: electrons lost by bromide ions are gained by chlorine. That makes the displacement a redox reaction. The same reasoning applies when a more reactive halogen displaces another halogen from a halide solution: identify which halide ions lose electrons, and which halogen molecules gain them.