Question

Determine the change in oxidation number of sulphur in $H_{2}S$ and $S{O}_2$ respectively in the following reaction.
$2H_{2}S+S{O}_2\to 2H_{2}O+3S$.

• A. $0,+2$
• B. $+2,-4$
• C. $-2,+2$
• D. $+4,0$

Answer 1

The correct option is B $+2,-4$

Oxidation state of $S$ changes from -2 in $H_{2}S$ and +4 in $S{O}_2$ to 0 in $S$.

Change in oxidation number of sulphur in $H_{2}S$ $=+2$

Change in oxidation number of sulphur in $S{O}_2$ $=-4$