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Introduction to d and f Block Elements

Which of the ...

Question

Which of the following is the strongest oxidising agent?

B_{r} O_{3}^{-} / B_{r}^{2 +}, E^{0} = + 1.50

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F_{e}^{3 +} / F_{e}^{2 +}, E^{0} = + 0.76

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M n O_{4}^{-} / M_{n}^{2 +}, E^{0} = + 1. 52

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C r_{2} O_{7}^{2 -} / C_{r}^{3 +}, E^{0} = + 1.33

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Solution

The correct option is C $M n O_{4}^{-} / M_{n}^{2 +}, E^{0} = + 1. 52$
Seeing the oxidation state of atoms of the provided options it is clear that the given E^o are there respective reduction potential. Clearly higher reduction potential implies greater oxidising nature because the atoms will easily get reduced and therefore will easily oxidise other atoms.Hence option (C) has greatest oxidising nature.

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Similar questions

[F e {(C N)}_{6}]^{4 -} \to [F e {(C N)}_{6}]^{3 -} + e^{- 1}, E^{0} = - 0.35 V

F e^{2 +} \to F e^{3 +} + e^{- 1}, E^{0} = - 0.77 V

E^{0}

I . M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O E^{0} = 1.5 V

I I . S n^{2 +} \to S n^{4 +} + 2 e^{-} E^{0} = - 0.15 V

I I I . C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 C r^{2 +} + 7 H_{2} O E^{0} = 1.33 V

I V . C e^{3 +} \to C e^{4 +} + e^{-} E^{0} = - 1.61 V

F_{2} (g) + 2 e^{-} \to 2 F^{-} (a q); E^{0} = + 2.85 V

C l_{2} (g) + 2 e^{-} \to 2 C l^{-} (a q); E^{0} = + 1.36 V

B r_{2} (l) + 2 e^{-} \to 2 B r^{-} (a q); E^{0} = + 1.06 V

I_{2} (s) + 2 e^{-} \to 2 I^{-} (a q); E^{0} = + 0.53 V

E^{0}

I . M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O E^{0} = 1.5 V

I I . S n^{2 +} \to S n^{4 +} + 2 e^{-} E^{0} = - 0.15 V

I I I . C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 C r^{2 +} + 7 H_{2} O E^{0} = 1.33 V

I V . C e^{3 +} \to C e^{4 +} + e^{-} E^{0} = - 1.61 V

Given standard electrode potentials:

$F e^{3 +} + 3 e^{-} \to F e; E^{0} = - 0.036 V$

$F e^{2 +} + 2 e^{-} \to F e; E^{0} = - 0.440 V$

The standard electrode potential $E^{o}$ for $F e^{3 +} + e^{-} \to F e^{2 +}$ is:

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