Question

What is the equivalent mass $I{O}_4^{-}$ when it is converted into $I_2$ in acid medium ?

• A. M/6
• B. M/ 7
• C. M/5
• D. M/4

Answer 1

The correct option is B M/ 7

Given,

${I{O}_H}^{-}$ is converting to $I_2$ in acidic medium

The initial reaction is

${I{O}_4}^{-}⟶I_2$

Now by balancing $I$ atoms

${2I{O}_4}^{-}⟶I_2$

Then balancing $O$ atoms

${2I{O}_4}^{-}⟶I_2+8H_{2}O$

Balancing $H^{+}$ atoms

${2I{O}_4}^{-}+16H^{+}⟶I_2+8H_{2}O$

Finally balancing the charge

${2I{O}_4}^{-}+16H^{+}+14e^{-}⟶I_2+8H_{2}O$

Oxidation state of $I$ here is $+7$ Oxidation state of $I$ here is $O$

$\therefore$ No. of $e^{-}$ transferred for $1$ mole=$7e^{-}$

$\Rightarrow$ Equivalent weight=$\frac{Molarmass}{No.of{e}^{-}transferred}$

$=\frac{M}{7}$