Q. $Na_{2}S_{2}O_{3}$ is reduced by $I_{2}$ to
A) $Na_{2}S$
B) $Na_{2}SO_{4}$
C) $NaHSO_{3}$
D) $Na_{2}S_{4}O_{6}$
Explanation
(d)
$Na_{2}S_{2}O_{3}\ r\ 2Na^{+} + S_{2}O_{3}^{2 -}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ S_{2}O_{3}^{2 -} + I_{2}\ \longrightarrow S_{4}O_{6}^{2 -} + I^{-}$
$i.e.,\ \ \ \ \ \ \ \ \ \ \ 2Na_{2}S_{2}O_{3} + I_{2}\ \longrightarrow Na_{2}S_{4}O_{6} + 2NaI$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.