Question

In balancing the half reaction: $S_2{O}_3^{2-}\to S\left(s\right)$, the number of electrons that must be added is :

• A. $2$ on the right
• B. $2$ on the left
• C. $3$ on the right
• D. $4$ on the left

Answer 1

Answer: (D)

$4$ on the left

$S_2{O}_3^{2-}\to S\left(s\right)$
+2 0
As change in oxidation state of each S is $2$,
$4e+S_2^{2+}\to 2S^0$
Hence, the number of electrons that must be added on the left is $4$.