What is the equivalent mass of $S_{2} O_{3}^{2 -}$ as per the following disproportionation reaction
$S_{2} O_{3}^{2 -} ⟶ S_{4} O_{6}^{2 -} + S_{2}^{2 -}$
where the charge above species represents charge on the ion and not oxidation state?

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132

130.6

113.15

answer is C.

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Detailed Solution

$6 e^{-} + S_{2} O_{3}^{2 -} ⟶ S_{2}^{2 -}$

$n = 6$

$2 S_{2} O_{3}^{2 -} \to S_{4} O_{6}^{2 -} + 2 e^{-}$

$n_{2} = \frac{2}{2} = 1$

$n_{eff} = \frac{1 \times 6}{1 + 6} = \frac{6}{7}$

$E = \frac{112}{6} \times 7 = 130.6$

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