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### Blancing redox

Posted: Sat Feb 04, 2017 4:44 pm
For the equation of H2S (aq) + Cl2 (g) ===> S (s) +Cl- (aq), why is H2S being oxidized? I am trying to write the two balanced half reactions. I know that CL2 is being reduced because it gains a - charge and therefore gains electrons. I am having trouble figuring out the charges on H2S and S because they appear to be both neutral.

### Re: Blancing redox

Posted: Sat Feb 04, 2017 5:53 pm
H2S and S may look neutral, but to write a redox reaction we look at each element individually to determine charge. So while the compound H2S is neutral, the sulfur atom is not and in fact has a -2 charge to balance out the 2 hydrogen atoms with +1 charges within the molecule. The sulfur is being oxidized, going from a -2 to a 0 oxidation state. So the half reaction for H2S would look something like $H_{2}S\rightarrow S+2e^{-}+2H^+$. I hope this helped!

### Re: Blancing redox  [ENDORSED]

Posted: Sat Feb 04, 2017 5:55 pm
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.

H2S(aq):
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
Cl2(g):
This gas is elemental, so it's oxidation number is 0.
S(s):
This solid is also elemental, so it's oxidation number is 0.
Cl^(-)(aq):
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.