• # 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?$

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 will be redirected in 3 sec 