In the redox ionic equation, the equivalent weight of potassium permaganate of molar mass $M$ $g m o l^{- 1}$ is:

$2 M n O_{4}^{-} + B r^{-} + H_{2} O \to 2 M n O_{2} + B r O_{3}^{-} + 2 O H^{-}$

\frac{M}{5}

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\frac{M}{3}

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\frac{M}{6}

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\frac{M}{2}

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Solution

The correct option is A $\frac{M}{3}$
The oxidation number of bromine changes from $+ 7$ ( in $M n O_{4}^{-}$ ) to $+ 4$ (in $M n O_{2}$ ).

Total change in the oxidation number is 6.

Hence, the equivalent weight of bromine will be $\frac{M}{3}$ .

Hence, option $B$ is the correct answer.

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Similar questions

M n O_{4}^{-} (a q) + B r^{-} (a q) \to M n O_{2} (s) + B r O_{3}^{-} (a q)

M n O_{4}^{-} (a q) + B r^{-} (a q) \to M n O_{2} (s) + B r O_{3}^{-} (a q)

A s O_{2}^{2 -} + B r O_{3}^{-} + H^{+} ⟶ B r^{-} + A s O_{4}^{-} + H_{2} O

B r O_{3}^{-} = \frac{M}{x}

x

For the reaction

$3 {B r}_{2}$ + $6 {O H}^{Θ}$ $\to$ $5 {B r}^{Θ}$ + $B r {O_{3}}^{Θ}$ + $3 H_{2} O$

Equivalent weight of ${B r}_{2}$ (Molecular weight M) is

B r_{2} + O H^{-} \to B r^{-} + B r O_{3}^{-} + H_{2} O

B r_{2}

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