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  4. Equivalent weight of I 2 in the reaction, I 2+5 Cl 2+6 H 2 O → 2 HIO 3+10 HCl, is

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Q. Equivalent weight of $I _{2}$ in the reaction, $I _{2}+5 Cl _{2}+6 H _{2} O \to 2 HIO _{3}+10 HCl$, is

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Solution:

$\overset{0}{I_{2}}+5Cl_{2}+6H_{2}O \to 2 \overset{+5}{HIO_{3}}+10HCl$

Change in oxidation number = $5$

Equivalent weight $=\frac{\text { Atomic weight }}{\text { Change in oxidation number }}=\frac{M}{5}$