Question

The equivalent weight of ${Na}_2{S}_2{O}_3$ in the following reaction is

$2{Na}_2{S}_2{O}_3+I_2⟶{Na}_2{S}_4{O}_6+2NaI$

• A. $M$
• B. $M/8$
• C. $M/0.5$
• D. $M/2$

Answer 1

The correct option is A $M$

$2S_2{O}_3^{2-}⟶S_4{O}_6^{2-}+2e^{-}$

In the given reaction, the oxidation state of S changes from +2 to +2.5. Since two S atoms are present in each formula unit so change for each formula unit is 1.

$E_{{Na}_2{S}_2{O}_3}=\frac{\text{Molecular weight}}{\text{n-factor}}=\frac{M}{1}=M$