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Valency and Oxidation States

4 mol of a so...

Question

$4$ mol of a solution containing $A^{n +}$ requires $1.6$ mol of $M n O_{4}^{⊝}$ for the oxidation of $A^{n +}$ to $A O_{3}^{⊝}$ in acidic medium. The value of $n$ is :

1

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Solution

The correct option is C $3$
$A^{n +} x = + n \to A O_{3}^{-} x = + 5$ $(u n d e r g o e s o x i d a t i o n)$
$∴$ Number of $e^{-}$ released in oxidation $= (5 - n)$
$∴$ Equivalent of $M n O_{4}^{⊝} = 4 \times (5 - n)$
Equivalents of $M n O_{4}^{⊝} = 1.6 \times 5$ (Mn(VII) is reduced to Mn(II) in acidic medium, therefore, oxidation state changes by 5).
$∴ 4 \times (5 - n) = 1.6 \times 5$
$∴ n = 3$

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