Question

In the reaction,
$I_2+2S_2{O}_3^{2-}\to 2I^{-}+S_4{O}_6^{2-}$
equivalent weight of iodine will be equal to

• A. $4/6$ of molecular weight
• B. Molecular weight
• C. $2/9$ of molecular weight
• D. Twice the molecular weight

Answer 1

The correct option is B Molecular weight

$I_2+2S_2{O}_3^{2-}\to 2I^{-}+S_4{O}_6^{2-}$
In the above reaction $I_2$ is converted to $I^{-}$ where the oxidation state changed from 0 to -1
So equivalent weight of iodine will be equal to molecular weight/1.
So equivalent weight of iodine will be equal to molecular weight.
Hence option B is correct.