Half-hearted Reaction Balance

Chm 1311 Lecture for 24 June 1999

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Caesar
was
Right!
The example I used in class today, oxidation of HSO3- by NO3-, has been treated in a web page for an earlier class. Remember your BACK button to return here after reading about the hard and easy ways of balancing redox reactions.

Now that web page from which you've returned paid insufficient attention to a not-inconsiderable detail: SO32- is not a proper species for an acid solution. Since HSO3- is not one of the strong acids, it must be a weak one...actually very weak! So in the presence of H+ (or H3O+, if you insist), SO32- will eagerly snatch up aqueous protons and turn itself into HSO3-.

The same thing is true of SO42-; in acid solution, it's found almost exclusively as HSO4- since that too is a weak acid!

So the result the other page touted:

2 NO3- + 3 SO32- + 2 H+  arrow right 2 NO + 3 SO42- + H2O

Fixing
the
Species
is certainly balanced, but uses the wrong form of the oxyanions. Well, since HNO3 is a strong acid, NO3- has no motivation to recombine with H+; so that anion is in the correct form. To fix the others, we just add the same number of H+ ions to both sides (3) in order to recombine with those weak acids:

2 NO3- + 3 H+ + 3 SO32- + 2 H+  arrow right 2 NO + 3 H+ + 3 SO42- + H2O

and then let the weak acid form to give the correct balanced equation:

2 NO3- + 3 HSO3- + 2 H+  arrow right 2 NO + 3 HSO4- + H2O

In a lesser offense, that prior webpage called NO by its ancient name, nitric oxide, which is known today as nitrogen monoxide.

Common
Oxidation
States
Some atoms are really boring in their oxidation numbers. The noble gases come to mind; they exist only (with bizarre exceptions) as monatomic elemental gases. So Group VIII element oxidation states come only as zero.

Even the alkali metals beat that out since they are zero as elements but +1 as ions.

But think about the rich possibilities for the oxidation states of the compounds we know about built from the (non-fluorine) halogens:

SpeciesCl- Cl2ClO- ClO2- ClO3- ClO4-
Cl oxidation state-10+1+3 +5+7

Notice that if ClO4- were to engage in a redox reaction, he'd have nowhere to go but down; that is, he can only be reduced. That makes him the most vigourously oxidizing species in this list! But even chlorine gas, Cl2, can be reduced (to Cl-) and is used as an oxidizing agent (in municipal water supplies to kill bacteria). But Cl- can't be reduced any further; so it cannot oxidize anything.

Does the fact that Cl- can only be oxidized make it a good reducing agent? No, because while it can be oxidized, it can only be oxidized by something that wants its electrons even more than it does. So Cl- isn't a good reducing agent.

What would be a good reducing agent? Something that didn't like to hang onto electrons. A metal would be a good candidate, and an alkali metal would be best of all. Unfortunately, we wouldn't want to carry out such a reduction in water, since the alkali metal would the reduce the hydrogen of the water to elemental dihydrogen gas with spectacular results.

Sulfur has a wide range of oxidation states as well. Here they are for some of the sulfur compounds we've already run across:

SpeciesS- S(8)S2O32- SO2SO3
S oxidation state-2 0246

Notice that those last two are anhydrides of H2SO3 and H2SO4, sulfurous and sulfuric acids respectively. Remember that an anhydride becomes its acid merely by the addition of water, neutral H2O. Since the water brings with it no charges, the oxidation number for the sulfur in its acid is the same as it is in the corresponding anhydride! So since SO3 is a highly oxidizing species, so is sulfuric acid! (Even SO2 is used to bleach fruit . . . so maraschino cherries can be stained a uniform red . . . sounds like Alice in Wonderland, doesn't it?)

You can actually use that fact to go hunting for anhydrides. For example, what's the anhydride of nitric acid, HNO3? The N in the acid bears a +5 oxidation number; so must the anhydride. That leads us to N2O5 as a likely candidate; so we try to balance the (non-redox since no oxidation numbers change) anhydride reaction:

N2O5 + H2O  arrow right 2 HNO3

Bingo.

Even lowly carbon takes on several oxidation states, exemplified (but not exhaused) by the following:

SpeciesCH4 CH3OH CH2O HCO2H CO2
Namemethane methanolmethanalmethanoic acidcarbon dioxide
Oxidation Number -4-20 +2+4

This tells you that the bacteria (acetobacter) that attack wine (ethanol) will oxidize it to ethanoic (acetic) acid. But it also tells you that's what your liver does when it's trying to counter your binge; worse still, it does it in stages! That means that at some point, it has only managed to oxidize the ethanol to ethanal otherwise known as acetaldehyde, cousin to embalming fluid, formaldehyde. So when folks speak of being pickled by too much alcohol, they are literally correct. If you must drink, do so in a moderation that bespeaks an understanding of the chemistry to which you're forcing your liver. It's the embalming fluid in your brain that gives it an entirely justified hangover.
It becomes a little dicey talking of the oxidation number of carbon in organic molecules because each of the carbons in some molecules might be in a different oxidation state depending upon to what it is bound. But if all one wished was the average oxidation number of all the carbons, that wouldn't be a problem; we'd just add up the numbers of all of the other species, correct for ionic charge (if any), negate, and divide by the number of carbons.

So HOCH2CH2CO2H (3-hydroxypropanoic acid) has three carbons with oxidation numbers of -1 (HOCH2), -2 (CH2), and +3 (CO2H), respectively, giving an average of zero! Taking that average directly, the 3 oxygens yield -6 while the 6 hydrogens yield +6; again the average carbon is in the zero oxidation state even though no one of the carbon atoms is in that state.

Fortunately, all the redox arithmetic works out just as well with the average! Not a problem, or as the Russians say, "Nye problema."
Molar
Concentrations
Earlier, I said the advantage to solution chemistry lay in being able to dilute and deliver, effectively. Delivery is easy since liquids are so easy to control. Dilute means to reduce the concentration so that the delivery is as finely-tuned as you'd like.

Of course, all of these advantages are lost if we can't tell how many moles we're delivering! So we must be able to convert between (easily measured) solution volumes and the moles of solute that they represent. If we know how many moles there are in some standard volume measure, like the decimeter3, dm3, more commonly called the liter (a bit over a quart), then scaling those moles by the volume we have tells us how many moles are in our volume.

That leads us to define MOLARITY as the number of moles of solute in a liter of the solution at hand. Seawater, for example, is 0.6 M in NaCl, where "M" is read "molar". That's a pretty fair concentration and explains why it is economically viable to collect salt by evaporating seawater. If we had 500 mL of seawater (½ L), it would obviously contain 0.3 moles of NaCl (and a bunch of other stuff as well, but NaCl is the major component). So if n is the moles in a volume V of a solution of molarity M, the equation n = MV does the trick.
But we want to be able to make up solutions of known concentration, so that same equation, n = MV, will tell us how many moles of solute to put into a (volumetric) flask of volume V to obtain a solution of concentration M molarity. Then when I go to use such a solution, probably titrating drops of it out of a buret, the volume necessary to deliver n moles of the solution is clearly V = n/M. And so on, ad nauseum.

VM is clearly the Holy Grail of solution chemistry since it represents moles, and moles represent molecules, and it's numbers of molecules coming and going that we measure with stoichiometry. So we can scale VM by stoichiometric ratios to learn how much A has reacted with B. The standard example is that of an acid-base titration where nacid is obviously MacidVacid, but that's the same as the moles of base if the base has been precisely neutralized. Since the volume of the base was known, its original concentration can be determined by Mbase = nacid / Vbase, if the acid and base react 1:1. When would they not? When the acid was monoprotic but the base was dibasic, like Ca(OH)2, for example. But stoichiometry would have corrected for that. So you still have to know the stoichiometry of solution reactions.

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Last modified 22 June 1999. Chris Parr