Blancing redox  [ENDORSED]

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Parsia Vazirnia 2L
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Blancing redox

Postby Parsia Vazirnia 2L » 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.

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Re: Blancing redox

Postby Rachel_Prescott_2M » 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 . I hope this helped!

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Re: Blancing redox  [ENDORSED]

Postby Natalie_Boyd_1C » Sat Feb 04, 2017 5:55 pm

If you need a refresher on determining 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.

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