What is the equivalent mass of IO_4^- when it | vedclass

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(B) The equivalent mass is calculated using the formula: $	ext{Equivalent mass} = \frac{	ext{Molecular weight}}{	ext{n-factor}}$.In the reduction o

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$M/7$

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(B) The equivalent mass is calculated using the formula: $	ext{Equivalent mass} = \frac{	ext{Molecular weight}}{	ext{n-factor}}$.In the reduction of $IO_4^-$ to $I_2$,the oxidation state of Iodine changes from $+7$ in $IO_4^-$ to $0$ in $I_2$.The balanced half-reaction is: $2IO_4^- + 16H^+ + 14e^- ightarrow I_2 + 8H_2O$.The total change in oxidation number for $2$ moles of $I$ atoms is $2 	imes (7 - 0) = 14$.Therefore,the n-factor per mole of $IO_4^-$ is $14/2 = 7$.Thus,the equivalent mass $= M/7$.

What is the equivalent mass of $IO_4^-$ when it is converted into $I_2$ in an acidic medium?

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