| Given that \[E_{{{O}_{2}}/{{H}_{2}}O}^{\Theta }=+1.23\,V;E_{{{S}_{2}}O_{8}^{2\,-}}^{\Theta }=2.05V\] |
| \[E_{B{{r}_{2}}/B{{r}^{\,-}}}^{\Theta }=+1.09\,V;E_{A{{u}^{3+}}/Au}^{\Theta }=+1.4V\] |
| The strongest oxidizing agent is: |
A) \[{{S}_{2}}O_{8}^{2-}\]
B) \[{{O}_{2}}\]
C) \[A{{u}^{3+}}\]
D) \[B{{r}_{2}}\]
Correct Answer: A
Solution :
Strongest oxidising agent \[\propto \]lowest position in electrochemical series (ECS) So, Ans. \[={{S}_{2}}O_{8}^{2-}.\]
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