[Firewall]

Here is the question that I am having problems with.

Balance the following oxidation-reduction reaction using the half-reaction method. It occurs in an acidic solution.

S₈ (s) + O₂ (g) ---> SO₄^2- (aq)

It is easy to see where the oxidation and reduction occurs. Sulfer is oxidized to a 6+ oxidation state; oxygen is reduced to 2-. What I cannot figure out is how to separate that equation into two half reactions. WTF? I even asked my professor, and I didn't come away feeling any better.

She said that you could do the following:

S₈ (s) ---> SO₄^2- (aq)

O₂ (g) ---> SO₄^2- (aq)

Evidently, you can ignore the sulfur in the last half reaction because it is going to be taken care of in the other one. But what about the gaseous oxygen in that same reaction? What are you supposed to do with it? My textbook says that you are supposed to balance oxygens with H₂0 when the net reaction takes place in an acidic solution. However, if you balance the oxygen with water, then you don't have any reason to manipulate the O₂, right?

Here is the answer from the back of my book:

S₈ (s) + 12 O₂ (g) + 8 H₂0 (l)  ---> 8 SO₄^2- (aq) + 16 H⁺ (aq)

Any help would be appreciated.

[Firewall]

Wow.

Just wow.

After many hours of toil, I finally discovered what I originally failed to notice. It turned out to be really easy and entirely conventional.



The thought occurred to me: if the gaseous oxygen is only reacting with the sulfur, then where the f&#$ does the acid come into play? I realized that the oxygen is actually reacting the hydrogen ions in solution to produce water, and you are allowed to add both of those to either side. It all makes sense now: oxygen is being reduced to water. That is the other half of the net equation. It only took me all day to figure it out.