The equivalent weight of KIO_3 in the given r | vedclass

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(C) In the given reaction: $2 Cr(OH)_3 + 4 OH^{-} + KIO_3 \rightarrow 2 CrO_4^{2-} + 5 H_2O + KI$The oxidation state of iodine in $KIO_3$ changes from

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$M / 6$

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(C) In the given reaction: $2 Cr(OH)_3 + 4 OH^{-} + KIO_3 ightarrow 2 CrO_4^{2-} + 5 H_2O + KI$The oxidation state of iodine in $KIO_3$ changes from $+5$ to $-1$ in $KI$.The change in oxidation state of iodine is $|5 - (-1)| = 6$.The equivalent weight of an oxidizing agent is given by the formula: $	ext{Equivalent weight} = \frac{	ext{Molecular mass}}{	ext{Change in oxidation state}}$.Therefore, the equivalent weight of $KIO_3 = M / 6$.

The equivalent weight of $KIO_3$ in the given reaction is ($M=$ molecular mass) :
$2 Cr(OH)_3 + 4 OH^{-} + KIO_3 \rightarrow 2 CrO_4^{2-} + 5 H_2O + KI$

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The equivalent weight of $KIO_3$ in the given reaction is ($M=$ molecular mass) :$2 Cr(OH)_3 + 4 OH^{-} + KIO_3 \rightarrow 2 CrO_4^{2-} + 5 H_2O + KI$