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Electrolysis and Electrolytes

When BrO 3 ...

Question

When ${B r O}_{3}^{-}$ ion reacts with ${B r}^{-}$ in acid medium, ${B r}_{2}$ is liberated. The equivalent weight of ${B r}_{2}$ in this reaction is:

A

\frac{5 M}{8}

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B

\frac{5 M}{3}

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C

\frac{3 M}{5}

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D

\frac{4 M}{6}

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Solution

The correct option is C $\frac{3 M}{5}$
The reaction is as follows,
${B r O}_{3}^{-}$ + $5 B r^{-}$ + $6 H^{+} \to$ $3 B r_{2}$ + $3 H_{2} O$
reduction $n_{1}$ = 5
So, n factor is 5
Formula of equivalent weight = $\frac{M a s s o f e l e m e n t}{n}$
equivalent weight of $3 r_{2}$ = $\frac{3 M}{n}$
= $\frac{3 M}{5}$ ( 5 stands of 'n' factor)

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