Determine equivalent weight of underlined species taking its molecular weight as $M$ .

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

M n O_{2} = \frac{5 M}{6}; M n O_{4}^{-} = \frac{M}{3}

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M n O_{2} = \frac{6 M}{5}; M n O_{4}^{-} = \frac{M}{3}

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M n O_{2} = \frac{M}{3}; M n O_{4}^{-} = \frac{M}{2}

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None of these

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Solution

The correct option is A $M n O_{2} = \frac{5 M}{6}; M n O_{4}^{-} = \frac{M}{3}$
$M n O_{2} ⟶ M n O_{4}^{-} + {M n}^{2 +}$

Equivalent weight of $M n O_{2}$
For reduction-

$M n O_{2} = \frac{M}{2}$ ( $M n O_{2}$ to $M n^{+ 2}$ )

For oxidation-

$M n O_{2} = M / 3$ ( $M n O_{2}$ to $M n O_{4}^{-}$ )

So, equivalent weight of $M n O_{2}$ should be = $M / 2 + M / 3 = 5 M / 6$

Equivalent weight of $M n O_{4}^{-}$ = $\frac{M}{3}$

Suggest Corrections

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4 H_{3} {P O}_{3} - ------- - Δ - \to 3 H_{3} {P O}_{4} + {P H}_{3}

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

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