Question

In the reaction, $2S_2{O}_3^{2-}+I_2\to S_4{O}_6^{2-}$, the eq. wt. of ${Na}_2{S}_2{O}_3$ is equal to its:

• A. $Mol.wt.$
• B. $Mol.Wt./2$
• C. $2\times Mol.wt.$
• D. $Mol.wt./6$

Answer 1

Answer: (C)

$2\times Mol.wt.$

$2S_2{O}_3^{2-}+I_2⟶S_4{O}_6^{2-}+2I^{-}$

Oxidation state of $S$ in $S_2{O}_3^{2-}$ is $+2$

Oxidation state of $S$ in $S_4{O}_6^{2-}$ is $+\frac{5}{2}$

$\therefore$ Change in oxidation state $=\frac{5}{2}-2$

$=\frac{1}{2}$

$\therefore$ Equivalent weight $=\frac{Mol.wt}{1/2}$

$=2\times Mol.wt$

Hence, option $C$ is correct.