Answer:
\[B{{r}_{2}}\]is
stronger oxidising agent than \[{{I}_{2}}\] , it oxidises\[{{S}_{2}}O_{3}^{2-}\]
to \[SO_{4}^{2-}\]i.e., from +2
state to +6 state of sulphur. However, \[{{I}_{2}}\] being weaker oxidising
agent, oxidizes \[{{S}_{2}}O_{3}^{2-}\]to\[{{S}_{4}}O_{6}^{2-}\] ion i.e., from
+2 to +2.5 state of sulphur.
You need to login to perform this action.
You will be redirected in
3 sec