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Name: Amy P.
Status: educator
Age: 20s
Location: N/A
Country: N/A
Date: 11/27/2004

Can you explain (in understandable terms) the stoichiometry when Persulfate oxidizes VOCs (particular benzene or BTEX)? I'm working on a project in AP Chemistry involving the oxidation of separated phase hydrocarbons (i.e. free product or gasoline).

"Persulfate" is usually Ammonium Peroxydisulfate: (NH4)2 S2O8. If there's strong acid present, it's equivalent to H2(S2O8). It's about like H2O2 is to H2O: take two, steal a hydrogen off each, stick them together.


In action, it splits at the center O - O bond, takes up one electron each side, then finds H+ ions to become 2(H2SO4). There may be pathways distinct from that, but the result is the same, so that's a clear way to think about the stoichiometry when persulfate oxidizes.

S2O8(2-) + 2 e- -> 2 SO4(2-)

total charge is unchanged, (4-) on both sides.

it is nominally interchangeable with H2O2:
      H2S2O8 <--> 2 HSO4*
HSO4* + OH-  <-->  HSO4-  + HO*
      2 HO*  <--> H2O2
So it steals the same number of electrons as H2O2 would.

Reacting with Benzene:

Lacking water, a direct reaction might happen:

C6H6 + 1/2 H2S2O8 -> C6H5-SO3 -H + H2O

( benzene sulfonic acid )

Here one must assume that an OH group came off a sulfur to form H2O with the H abstracted from the benzene. But most often the oxygens seem to stay on the sulfur while grabbing an electron to themselves.

With water, perhaps the first oxidation will be to phenol:

C6H6 + 1/2 S2O8(2-) +H2O -> C6H5-OH + HSO4-

( phenol )

Existence of the first hydroxyl group probably makes the second oxidation easier:

C6H5OH + 1/2 S2O8(2-) +H2O -> C6H4(OH)2 + HSO4-

( benzene di-ol, probably either para- or ortho-, also called catechol )

The oxidation number of C's and H's in Benzene is nominally 0. The C's and H's are equally electronegative as each other. The C-H bond is a very covalent bond, with no discernible polar character. So one cannot assign (+) to one and (-) to the other. Likewise when one hydrogen is arbitrarily removed, forming a benzyl radical I would assign a nominal oxidation number of 0. (If you wish you could figure an H+ left, leaving a C6H5- behind. Change all your reaction-balancing likewise, and stoichiometry will be clear enough.)

Adding -OH to form phenol, O is -2, the H attached to the highly electronegative Oxygen is +1, so the -OH group is net (-1). Think of it as a pseudo-halogen. For the time being, the H is staying on the O, so they carry around the same oxidation number as a Fluorine would. When a hydroxyl attaches to the benzyl group C6H5, the carbon it bonds to must become +1. The benzene ring electrons might be considered to distribute the single unit of oxidation to the C6H5 group as a whole. Which doesn't really change our figuring.

Catechol has two hydroxyls. So the ring has a +2 oxidation number, 2 ox. units to distribute broadly or place where it likes.

I think this can be followed consistently all the way to CO2 + H2O. By that point we have accumulated +4 oxidation units for each carbon atom.

Hope your project works out well.

Jim Swenson

Persulfate, perchlorate, and the other "per--" oxidizing agents react by generating free radicals, that is, a chemical species with an unpaired electron in the valence shell. Free radicals are very reactive and undergo numerous chemical reactions because that free electron "wants" to pair up with almost anything it can find. As a consequence, the reaction products tend to be complex mixtures. There are some exceptions where the free radical initiates a polymerization chain reaction, such as the production of polystyrene from styrene initiated by a free radical such as a persulfate. However, for hydrocarbons such as BTEX (the acronym BTEX stands for Benzene, Toluene, Ethyl benzene, Xylene) the products are likely to be complex mixtures depending upon reaction conditions.

Vince Calder

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