Question #3a748

1 Answer
Jun 13, 2017

Answer:

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.