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Reactivity Towards Air

Arrange the f...

Question

Arrange the following substance in order of increasing strength as an oxidising agent, and account for the trend :
(a) $M n^{2 +} (b) M n O_{2} (c) M n O_{4}^{-}$

M n^{2 +} < M n O_{2} < M n O_{4}^{-}

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M n O_{2} < M n^{2 +} < M n O_{4}^{-}

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M n^{2 +} < M n O_{4}^{-} < M n O_{2}

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M n O_{4}^{-} < M n^{2 +} < M n O_{2}

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Solution

The correct option is A $M n^{2 +} < M n O_{2} < M n O_{4}^{-}$
More the oxidation state of the Mn, more will be oxidizing power.
As $M n^{2 +}$ has $+ 2$ oxidation state, $M n O_{2}$ has $+ 4$ oxidation state of $M n$ whereas $M n O_{4}^{-}$ has $+ 7$ oxidation state of the $M n$ .
So, as $M n O_{4}^{-}$ has maximum oxidation state of $M n$ , it is the strongest oxidizing agent and the order of increasing strength as an oxidizing agent is as follows:
$M n^{2 +} < M n O_{2} < M n O_{4}^{-}$

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Similar questions

M n O_{4}^{-} / M n O_{2}

E_{M n O_{4}^{-} / M n^{2 +}}^{o} = 1.51 V

E_{M n O_{2} / M n^{2 +}}^{o} = 1.23 V

M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O; E^{O} = 1.51 V; Δ G_{1}^{0} = - 5 \times 1.51 \times F

M n O_{2} + 4 H^{+} + 2 e^{-} \to M n^{2 +} + 2 H_{2} O; E^{O} = 1.23 V; Δ G_{2}^{0} = - 5 \times 1.23 \times F

E_{M n O_{4}^{-} | M n O_{2}}^{\circ}

E^{o}

M n^{2 +} + 2 H_{2} O \to M n O_{2} + 4 H^{+} + 2 e^{-}

E^{o} = - 1.23 V

M n O_{4}^{-} + 4 H^{+} + 3 e^{-} \to M n O_{2} + 2 H_{2} O

E^{o} = + 1.70 V

k M n O_{4}

M n O_{4}^{2 -}, M n O_{2}, M n_{2} O_{3},

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100

H_{2} O_{2}

100

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100

H_{2} O_{2}

V

0.01 M

K M n O_{4}

M n O_{4}^{⊝}

M n O_{2}

V

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