If you need a refresher on determining oxidation numbers: http://www.wikihow.com/Find-Oxidation-Numbers
I'm just going to do the whole problem in case someone else doesn't understand why Cl2 is being reduced as well.
EQ: H2S(aq) + Cl2(g) --> S(s) + Cl^(-)(aq)
First, let's determine all of the oxidation numbers.
H usually has a +1 oxidation number unless it's in the form of hydrogen gas, in which case it's oxidation number is 0; or it makes up a compound called a hydride, where it has an oxidation number of -1.
Thus, for H: 2(+1)= +2. In regards to S, the compound has no overall charge so the sum of the charges of the atoms must equal 0. So, +2 + ? = 0 so ? = -2
This gas is elemental, so it's oxidation number is 0.
This solid is also elemental, so it's oxidation number is 0.
This is an ion, so it's oxidation number is its charge: -1.
Putting everything all together:
H2S(aq) + Cl2(g) --> S(s) + Cl^(-)(aq)
2(+1)&(-2) (0) (0) (-1)
Second, let's determine what is oxidized and what is reduced.
1. Cl2(g) goes from an oxidation number of 0 to -1. Therefore, it must gain an electron, so it is reduced.
2. The S atom in H2S goes from an oxidation number of -2 to 0, so it must lose two electrons and is oxidized.
I think that your question is why H2S, the compound, whose overall charge is 0, is what is oxidized in this equation, and I believe the answer is that it's really Sulfur which is oxidized, but it is a part of the compound H2S so we say that H2S is oxidized.
Hope this helps.