Answer:
Since \[E{}^\circ \]of \[B{{r}_{2}}\] is higher than that of \[{{I}_{2}},\] therefore, \[B{{r}_{2}}\] has a higher tendency to accept electrons that \[{{I}_{2}}.\] Conversely, \[{{I}^{-}}\] ion has a higher tendency to lose electrons than \[B{{r}^{-}}\] ion. Therefore, the following reaction will occur:
\[\begin{align} & \,\,\,\,\,\,\,\,\,\,\,2{{I}^{-}}\xrightarrow{\,}\,{{I}_{2}}+2{{e}^{-}} \\ & B{{r}_{2}}+2{{e}^{-}}\xrightarrow{\,}\,2B{{r}^{-}} \\ & \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ & 2{{I}^{-}}+B{{r}_{2}}\xrightarrow{\,}\,{{I}_{2}}+2\,B{{r}^{-}} \\ \end{align}\]
In other words \[{{I}^{-}}\] ion will be oxidised to \[{{I}_{2}}\] while \[B{{r}_{2}}\] will be reduced to \[B{{r}^{-}}\] ions.
You need to login to perform this action.
You will be redirected in
3 sec