Hi Jordan,

Below is a message from Anastasia:

Hi Jordan,

For some reason I can’t open the question edit that you added later but hopefully I cover it with the following summary.

OK, lots going on here. Sulfur can be challenging as it can have oxidation numbers of -2 to +6. On the plus side, some of the details that you might be concerned about here go beyond what is explained on page 355 of the classroom companion and this means that some aspects will be beyond the scope of the MCAT.

HOWEVER, there are things that I would want you to recognize as I would count them as fair game content:

First let’s look at this ‘quantitatively’ and assign oxidation numbers where we can:

1. If you are given the sulfite ion (SO32- , far right of your image) you definitely could determine the oxidation number for sulfur.
a. Since the overall charge of the ion is -2, the sum of the oxidation numbers for all the atoms in the species will sum to -2 as well. Let’s say that X is the oxidation number for sulfur in this ion then X + 3(-2) = -2 where 3(-2) is the term representing the product of the oxidation state of oxygen (it’s not a peroxide so I am pretty safe assuming -2) and the number of oxygen atoms in the species which happens to be 3. Solving for X yields an oxidation number for sulfur of +4. Great, a match!
2. What you probably noticed and I assume why you are asking the question, is that if you try to treat the thiosulfate ion (S2O32-) with two sulfur atoms in differing oxidation states the same way as the sulfite ion, everything breaks down; you get +2 for each of them which does not jibe with the scheme. You are unlikely then, without a bunch of explanation in the corresponding passage, to be asked to determine the specific oxidation numbers for two of the same atoms (in the same species) in differing oxidation states as in thiosulfate.

Now let’s look at this ‘qualitatively’:

It’s pretty important to recognize the oxidation of thiols to disulfides (and of course, the reverse, the reduction of disulfides to thiols) because of the structural role that disulfide bridges between cysteine residues play in the overall structure of many proteins so lets use them as an example:

R-SH → R-S-S-R (oxidation)
R-SH ← R-S-S-R (reduction)

It’s key to notice that it is the sulfur atom which changes oxidation state, not the carbon. We can draw on two tools that we already know to solidify which process is happening in either direction for any sulfur atom. The following should help in identifying what has been oxidized and what has been reduced.

1. Often in organic chemistry, we don’t calculate the oxidation number for carbon we just say something like...”if the carbon of interest forms more bonds to hydrogen that carbon is being reduced”. Let’s apply that to the oxidation state of sulfur in the conversion of the disulfide backwards to the thiol and we can say...more bonds to hydrogen must be reduction. Now look at the red sulfur atom in your screen shot going right to left and apply this same idea and with a change in oxidation number from -1 to -2 it indeed has been reduced. Now going now right to left, there are fewer bonds between the red sulfur atom and hydrogen so it is being the opposite of reduced, it is being oxidized with a corresponding change in oxidation number from -2 to -1.

2. Recall that we can also say... ”if the carbon of interest forms more bonds to oxygen that carbon is being oxidized”. Now have a look at the sulfur atom in blue in the screen shot on the left and right hand sides of the arrow. There are no bonds to hydrogen but there are more bonds to oxygen and other sulfur atoms (remember that sulfur can often be treated as analogous to oxygen, thinking of where it is on the periodic table) on the left hand side than the right; therefore going left to right, that blue sulfur atom must be reduced. This is confirmed by a change in oxidation number of +5 to +4.

If the question is asking about oxidizing and reducing agents remember that the oxidizing agent is itself reduced and the reducing agent is itself oxidized so identify them that way.

I have a feeling that this explanation might be overkill but I hope it helps!

Anastasia - Instructor