QJEEM19Ch4736: JEE Main 2019 Chemistry - Class 12 JEEM19Ch4736 Solution: Given that E Θ O 2 / H 2 O = 1.23 V ; E Θ S 2 O 8 2 − / S O 4 2 …

Which electrode have higher value of standard reduction potential (SRP), that electrode will be strongest oxidizing agent.

Tendency to gain electrone is called standard reduction potential. When tendency to gain electron is more then that electrode will have more oxidizing power.

Here given SRP's are :

$${E^{\mathrm{\Theta }}}_{O_2/H_{2}O}=1.23V$$ ;

$${E^{\mathrm{\Theta }}}_{S_2{O}_8^{2-}/S{O}_4^{2-}}=2.05V$$

$${E^{\mathrm{\Theta }}}_{B{r}_2/B{r}^{-}}=1.09V$$

$${E^{\mathrm{\Theta }}}_{A{u}^{3+}/Au}=1.4V$$

You can see $${E^{\mathrm{\Theta }}}_{S_2{O}_8^{2-}/S{O}_4^{2-}}=2.05$$,

have highest SRP so $$S_2{O}_8^{2-}$$ is the strongest oxidizing agent.