Question

Reduction potentials of some ions are given below. Arrange them in decreasing order of oxidizing power.

Ion	$Cl{O}_4^{-}$	$I{O}_4^{-}$	$Br{O}_4^{-}$
Reduction potential $E^{\circ }$/V	$E^{\circ }$= 1.19V	$E^{\circ }$= 1.65V	$E^{\circ }$= 1.74V

• A. $Cl{O}_4^{-}$ > $I{O}_4^{-}$ > $Br{O}_4^{-}$
• B. $I{O}_4^{-}$ > $Br{O}_4^{-}$ > $Cl{O}_4^{-}$
• C. $Br{O}_4^{-}$ > $I{O}_4^{-}$ > $Cl{O}_4^{-}$
• D. $Br{O}_4^{-}$ > $Cl{O}_4^{-}$ > $I{O}_4^{-}$

Answer 1

The correct option is C $Br{O}_4^{-}$ > $I{O}_4^{-}$ > $Cl{O}_4^{-}$

The reduction potential of the substance is the ability of the substance to be reduced.So as the reduction potential increases reducing ability of the substance increases.It means oxidizing power of the substance (The ability of the substance to make the other substance to lose the electrons increases which noting but oxidizing power) increases.

So the decreasing order of oxidizing power is $Br{O}_4^{-}$ > $I{O}_4^{-}$ > $Cl{O}_4^{-}$.

Hence option $C$ is correct.