Which will liberate bromine from a solution of potassium bromide?

Since, $E^{\circ}$ of $\mathrm{Cl}_2$ is higher than that of $\mathrm{Br}_2$, therefore $\mathrm{Cl}_2$ has a higher tendency to accept electrons than $\mathrm{Br}_2$. Conversely, $\mathrm{Br}^{-}$ion has a higher tendency to lose electrons than $\mathrm{Cl}^{-}$ion. Therefore, the following reaction will occur $\mathrm{KBr} \longrightarrow \mathrm{K}^{+}+\mathrm{Br}^{-}$ $2 \mathrm{Br}^{-}+\mathrm{Cl}_2 \longrightarrow 2 \mathrm{Cl}^{-}+\mathrm{Br}_2$ Thus, $\mathrm{Br}^{-}$ions are oxidised to $\mathrm{Br}_2$ by $\mathrm{Cl}_2$, which itself is reduced to $\mathrm{Cl}^{-}$ions.