Question

Question #3a748

Answer 1

The compound oxidized is `"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 `"S"` has increased from -2 in `"H"_2"S"` to 0 in `"S"`.

Also, the oxidation number of `"Cl"` has decreased from 0 in `"Cl"_2` to -1 in `"HCl"`.

Since oxidation is an increase in oxidation number, the `"H"_2"S"` has been oxidized.