Ncert Exercises & Solutions

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

Ion	${ClO}_{4}^{-}$	${IO}_{4}^{-}$	${BrO}_{4}^{-}$
Reduction potential E^-/V	= 1.19V	= 1.65V	= 1.74V

$1 . {ClO}_{4}^{-} > {IO}_{4}^{-} > {BrO}_{4}^{-}$
$2 . {IO}_{4}^{-} > {BrO}_{4}^{-} > {ClO}_{4}^{-}$
$3 . {BrO}_{4}^{-} > {IO}_{4}^{-} > {ClO}_{4}^{-}$
$4 . {BrO}_{4}^{-} > {ClO}_{4}^{-} > {IO}_{4}^{-}$

Hint: As the reduction potential increases, the oxidising power increases

Greater the SRP value of species higher will be its oxidizing power.

Species

$E_{cell}^{°}$

${ClO}_{4}^{-}$

${IO}_{4}^{-}$

${BrO}_{4}^{-}$

1.19 V

1.65 V

1.74 V

SRP of species increases.

Oxidizing power increase.

Here, SRP = standard reduction potential.

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