question_answer

\[{{I}_{2}}\]and\[B{{r}_{2}}\] are added to a solution containing \[B{{r}^{-}}\] and \[{{I}^{-}}\] ions. What reaction will occur if, \[{{I}_{2}}+2{{e}^{-}}\xrightarrow{}2{{I}^{-}};{{E}^{{}^\circ }}=+0.54V\,and\]\[B{{r}_{2}}+2{{e}^{-}}\xrightarrow{}2B{{r}^{-}};{{E}^{{}^\circ }}=+1.09V?\]

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.