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Valency and Oxidation State

0.1 mol of Mn...

Question

0.1 mol of $M n O_{4}^{(-)}$ (in acidic medium) can:

Oxidise 0.5 mol of

F e^{(2 +)}

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Oxidise 0.166 mol of

F e C_{2} O_{4}

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Oxidise 0.25 mol of

C_{2} O_{4}^{(2 -)}

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Oxidise 0.083 mol of

C r_{2} O_{7}^{(2 -)}

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Solution

The correct options are
A Oxidise 0.5 mol of $F e^{(2 +)}$
B Oxidise 0.166 mol of $F e C_{2} O_{4}$
C Oxidise 0.25 mol of $C_{2} O_{4}^{(2 -)}$
In acidic medium, $+ 7 M n O_{4}^{-} \to + 2 M n; Δ O . N . = 5$
$no. of eq. = 0.1 \times 5 = 0.5 e q .$

a) 0.5 mol of $F e^{(2 +)}$
$F e^{(2 +)} \to F e^{(3 +)} + e^{-}; Δ O . N . = 1$
$no. of eq. = 0.5 e q .$

b) 0.166 mol of $F e C_{2} O_{4}$
$F e^{(2 +)} \to F e^{(3 +)} + e^{-}; Δ O . N . = 1$
${+ 3 C}_{2} O_{4}^{(2 -)} \to 2 + 4 C O_{2} + 2 e^{-}; Δ O . N . = 2$
$Total Δ O . N . = 3$
$no. of eq. = 0.5 e q .$

c) 0.25 mol of $C_{2} O_{4}^{(2 -)}$
${+ 3 C}_{2} O_{4}^{(2 -)} \to 2 + 4 C O_{2} + 2 e^{-}; Δ O . N . = 2$
$no. of eq. = 0.5 e q .$

d) 0.083 mol of $C r_{2} O_{7}^{(2 -)}$
$+ 6 C r_{2} O_{7}^{(2 -)} \to 2 + 3 C r; Δ O . N . = 6$
Here $C r$ is in its highest oxidation state so can not be oxidised further.

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