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Reaction with Metals

Mx+ +MnO4-1→ ...

Question

$M^{x +} + M n O_{4}^{- 1} \to M O_{3}^{- 1} + M n^{+ 2}$

If 1 mol of $M n O_{4}^{- 1}$ oxidises 1.67 moles of $M^{n +}$ to $M O_{3}^{- 1}$ , then the value of x in the reaction is __________.

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Solution

$M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{+ 2} + 4 H_{2} O$

$M^{x +} + 3 H_{2} O ⟶ M O_{3}^{- 1} + 6 H^{+} + (5 - x) e^{-}$

geq of $M n O_{4}^{- 1} =$ geq of $M^{x +}$

$1 \times 5 = [1.67 \times (5 - x)]$

$\to x = 2$

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