Question

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

Answer 1

Given the half reaction:
$S_2{O}_3^{2-}\to S\left(s\right)$
+2 0
The change in oxidation state of each $S$ is $2$.
$4e+S_2^{2+}\to 2S^0$
Therefore, number of electrons added on the left is $4$.