The equivalent weight of KIO_3 in the reactio | vedclass

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Question

(B) In the given reaction: $2Cr(OH)_3 + 4OH^{-} + KIO_3 	o 2CrO_4^{2-} + 5H_2O + KI$First,determine the change in oxidation state for the iodine atom

Vedclass · Accepted Answer

$\frac{Molecular 	ext{ } weight}{6}$

Vedclass · Answer

(B) In the given reaction: $2Cr(OH)_3 + 4OH^{-} + KIO_3 	o 2CrO_4^{2-} + 5H_2O + KI$First,determine the change in oxidation state for the iodine atom in $KIO_3$.In $KIO_3$,the oxidation state of $I$ is $+5$.In $KI$,the oxidation state of $I$ is $-1$.The change in oxidation state is $|5 - (-1)| = 6$.However,looking at the stoichiometry,$1$ mole of $KIO_3$ produces $1$ mole of $KI$.The $n$-factor for $KIO_3$ is $6$.Wait,let us re-evaluate the reaction: $I^{+5} + 6e^{-} 	o I^{-1}$. The $n$-factor is $6$.Therefore,the equivalent weight is $\frac{Molecular 	ext{ } weight}{6}$.Correcting the provided option: The correct answer is $B$.

The equivalent weight of $KIO_3$ in the reaction $2Cr(OH)_3 + 4OH^{-} + KIO_3 \to 2CrO_4^{2-} + 5H_2O + KI$ is

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