In the reaction I_2 + 2S_2O_3^{2-} to 2I^{-} | vedclass

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(B) The given reaction is: $I_2 + 2S_2O_3^{2-} 	o 2I^{-} + S_4O_6^{2-}$.In this reaction,the oxidation state of iodine changes from $0$ in $I_2$ to $

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$1/2$ the molecular weight

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(B) The given reaction is: $I_2 + 2S_2O_3^{2-} 	o 2I^{-} + S_4O_6^{2-}$.In this reaction,the oxidation state of iodine changes from $0$ in $I_2$ to $-1$ in $I^{-}$.The change in oxidation state per atom of iodine is $1$. Since there are $2$ atoms of iodine in $I_2$,the total change in oxidation state is $2 	imes 1 = 2$.The $n$-factor for $I_2$ is $2$.The equivalent weight is calculated as: $	ext{Equivalent weight} = \frac{	ext{Molecular weight}}{n	ext{-factor}} = \frac{	ext{Molecular weight}}{2}$.

In the reaction $I_2 + 2S_2O_3^{2-} \to 2I^{-} + S_4O_6^{2-}$,the equivalent weight of iodine will be equal to

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