$ C{{l}_{2}}+2B{{r}^{-}}\,\to \,2C{{l}^{-}}+B{{r}_{2}} $

$ 1.\,C{{l}^{-}} $

$ 2.\,B{{r}^{-}} $

$ 3.\,B{{r}_{2}} $

$ 4.\,C{{l}_{2}} $

Solution:  $ 4.\,C{{l}_{2}} $

Br is the reducing agent because it loses an electron when it is oxidised from Br^– to Br₂. Cl₂ loses one electron as it is reduced from Cl₂ to 2 Cl^–, making Cl₂ the oxidising agent.