For the reaction $3 \mathrm{Br}_{2}+6 \mathrm{OH}^{\ominus} \longrightarrow 5…

For the reaction
$3 \mathrm{Br}_{2}+6 \mathrm{OH}^{\ominus} \longrightarrow 5 \mathrm{Br}^{\ominus}+\mathrm{BrO}_{3}^{\ominus}+3 \mathrm{H}_{2} \mathrm{O}$
Equivalent weight of $\mathrm{Br}_{2}$ (molecular weight $M$ ) is
  1. $\frac{M}{2}$
  2. $\frac{M}{10}$
  3. $\left(\frac{M}{2}+\frac{M}{10}ight)$
  4. $\left(\frac{M}{6}ight)$

Solution

$\mathrm{Br}_{2}$ disproportionates (simultaneous oxidation and reduction), its equivalent weight is the sum of equivalent weights of the two half reactions.
$2 e^{-}+\mathrm{Br}_{2} \longrightarrow 2 \mathrm{Br}^{\ominus}(x=2)$ (reduction)
$\mathrm{Br}_{2} \longrightarrow 2 \mathrm{BrO}_{3}^{\ominus}+10 \mathrm{e}^{-}\left(' n^{\prime}=10ight)$ (oxidation)
$2 x=0 \quad \quad2 x-12=-2$
$\quad \quad \quad \quad \quad 2 x=10$
$\therefore E w=\frac{M}{2}+\frac{M}{10}=\left(\frac{80 \times 2}{2}+\frac{80 \times 2}{10}ight)=96$ .

Asked in: JEE-TOPICTESTS-CHEMISTRY

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