# Question 3a748

Jun 13, 2017

The compound oxidized is $\text{H"_2"S}$.

#### Explanation:

We start by identifying the oxidation number of every atom in the equation.

stackrelcolor(blue)("+1")("H")_2stackrelcolor(blue)("-2")("S") + stackrelcolor(blue)(0)("Cl")_2 → stackrelcolor(blue)(0)("S") + stackrelcolor(blue)("+1")("H")stackrelcolor(blue)("-1")("Cl")#

We see that the oxidation number of $\text{S}$ has increased from -2 in $\text{H"_2"S}$ to 0 in $\text{S}$.

Also, the oxidation number of $\text{Cl}$ has decreased from 0 in ${\text{Cl}}_{2}$ to -1 in $\text{HCl}$.

Since oxidation is an increase in oxidation number, the $\text{H"_2"S}$ has been oxidized.