In alkaline medium,what will be the equivalen | vedclass

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(B) The chemical reaction for the conversion of $I_2$ to iodate ion $(IO_3^-)$ in an alkaline medium is: $I_2 + 12OH^- 	o 2IO_3^- + 6H_2O + 10e^-$In

Vedclass · Accepted Answer

$25.4$

Vedclass · Answer

(B) The chemical reaction for the conversion of $I_2$ to iodate ion $(IO_3^-)$ in an alkaline medium is: $I_2 + 12OH^- 	o 2IO_3^- + 6H_2O + 10e^-$In this reaction,the oxidation state of iodine changes from $0$ in $I_2$ to $+5$ in $IO_3^-$. Since there are two iodine atoms in $I_2$,the total change in oxidation number is $2 	imes (5 - 0) = 10$. Thus,the $n$-factor is $10$. The equivalent weight $(E)$ is calculated as: $E = \frac{	ext{Molar mass of } I_2}{n	ext{-factor}} = \frac{254}{10} = 25.4$.

In alkaline medium,what will be the equivalent weight of $I_2$ on its conversion to iodate ion $(IO_3^-)$?

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