Question

In the reaction, $2S_2{O}_3^{2-}+I_2\to S_4{O}_4^{2-}+2I^{-}$, the eq.wt of ${Na}_2{S}_2{O}_3$ is equal to its:

• A. molecular weight
• B. molecular weight/2
• C. 2$\times$ molecular weight
• D. molecular weight/6

Answer 1

The correct option is B molecular weight/2

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

Oxidation state of $S$ in $S_4{O}_4^{2-}=+1$

Chnage in oxidation state of $S$=$1$

Therefore, change in oxidation state of $S_2{O}_3^{2-}$=$2$

$\therefore$ Equivalent weight= $\frac{Molecularweight}{2}$