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Oxidising Reagent and Reducing Reagent

Given a react...

Question

Given a reaction:
$M^{x +} + M n O_{4}^{-} ⟶ M O_{3}^{-} + {M n}^{2 +} + \frac{1}{2} O_{2}$
If $1$ mol of $M n O_{4}^{-}$ oxidises $1.67$ mol of $M^{x +}$ to $M O_{3}^{-}$ , then the value of $x$ in this reaction is:

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Solution

$5 e^{-} + M n O_{4}^{-} ⟶ {M n}^{2 +}$
$M^{x +} ⟶ M O_{3}^{-} + (5 - x) e^{-}$

Equivalent of $M n O_{4}^{-} =$ Equivalent of $M^{x +}$
$1$ mol $\times 5 = 1.67$ mol $\times (5 - x)$
$∴ x = 2$

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