In the reaction, ${C l O}_{3}^{-} + {6 H}^{+} + {6 e}^{-} ⟶ {C l}^{-} + 3 H_{2} O$ . The equivalent weight of ${C l O}_{3}^{-}$ is :

Formula weight / 6

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Formula weight / 3

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Formula weight / 4

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Formula weight / 5

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Solution

The correct option is A Formula weight / 6
$C l O_{3}^{-} + 6 H^{+} + 6 e^{-} \to C l^{-} + 3 H_{2} O$

Cl oxidation in $C l O_{3}^{-} = + 5$
Cl oxidation state in $C l^{- 1} = - 1$
Valency factor $= 6$

$∴ E q u i v a l e n t w e i g h t = \frac{1}{6} \times$ formula weight

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