Answer:
At anode - [tex]{Br_2}_{(l)}[/tex]
At cathode - [tex]{H_2}_{(g)}[/tex]
Explanation:
Electrolysis of NaBr:
Water will exist as:
[tex]H_2O\rightleftharpoons H^++OH^-[/tex]
The salt, NaBr will dissociate as:
[tex]NaBr\rightarrow Na^++Br^-[/tex]
At the anode, oxidation takes place, as shown below.
[tex]{Br^-}_{(aq)}\rightarrow {Br_2}_{(l)}+2e^-[/tex]
At the cathode, reduction takes place, as shown below.
[tex]2{H^+}_{(aq)}+2e^-\rightarrow {H_2}_{(g)}[/tex]