I teach Redox and Electrochemistry as the first topic in AP Chemistry1. On day one of the course—without wasting time outlining the syllabus, marking scheme, and late policy —students carry out a laboratory activity: Determination of the concentration of FeSO4 by redox microtitration with KMnO4 of known concentration in an acidic solution (see supplemental material). The reaction is represented by the unbalanced equation:
FeSO4(aq) + KMnO4(aq) → Fe2(SO4)3(aq) + MnSO4(aq) (in H2SO4 solution) (eqn 1)
For simplicity, students are given the unbalanced net ionic equation:
Fe2+(aq) + MnO41–(aq) → Fe3+(aq) + Mn2+(aq) (in H2SO4 solution) (eqn 2)
When you—the teacher—run a pre-lab demonstration of the microtitration, emphasize that no external indicator, such as phenolphthalein, is required. The reaction is its own indicator. The reactants and products, save permanganate, are clear and colorless. When the purple color of KMnO4 no longer remains, the reaction has reached its stoichiometric point.
Using data from the microtitration, students attempt to determine [FeSO4] using stoichiometry, only to realize that they are unable to balance (eqn 2).
To wit:
i. Fe2+ loses one electron to become Fe3+ ; this is obvious. But what—in terms of electron transfer—happens to Mn in MnO41- when it becomes Mn2+?
ii. There are no oxygen atoms on the right-hand side of the equation; the equation-balancing that they learned in Chemistry is of no help.
This lays the groundwork for the necessity of the Oxidation Number (ON) Concept3. Simply put, something more must be understood before this equation can be balanced.
The Oxidation Number (ON) Concept—Using Logic
What follows is a concise, understanding-based distillation of what high school students need.
I start by walking students through the Oxidation Number Rules from their textbook or from Wikipedia4.
- An element in its uncombined form, such as Fe or Na has ON = 0. The ON of an atom in a homonuclear diatomic = 0; S in S8 and P in P4 also have ON = 0. This is explained below.
- The ON of a monatomic ion equals the charge on the ion. Cations have a deficiency of electrons and so will have a positive ON; the opposite is true for anions. For example, Na1+ has ON = +1; Cl1– has ON = -1.
- In a neutral molecule, the sum of the ON of each element = 0.
- In a compound or ion, the sum of the oxidation numbers equals the total charge of the compound or ion. This will be elaborated upon later.
- Fluorine in compounds has ON = −1; this extends to chlorine and bromine only when not bonded to a lighter (i.e., less electronegative) halogen, or to oxygen or nitrogen.
- Group 1 and group 2 metals in compounds have ON = +1 and +2, respectively.
- Hydrogen has ON = +1 but adopts −1 when bonded as a hydride to metals or metalloids.
- Oxygen in compounds has ON = −2, except when bonded to another oxygen (e.g., in peroxides) or to fluorine. When bonded to fluorine, Oxygen has ON = +2; in a peroxide, its ON = -1.
Some of these rules are obvious; others require memorization. Here are the keys to understanding the Oxidation Number concept:
1. The Use of Lewis Structure and the Electronegativity Concept
This applies to any covalent molecule or polyatomic ion for which a Lewis Structure can be drawn. Draw the Lewis structure for the molecule, clearly indicating valence electrons.
- The bonding electrons—two for a single bond, four for a double bond, and so on—are “taken” by the more electronegative atom. (In the case of a homonuclear diatomic, the bonding electrons are shared fifty-fifty.)
- Now draw a loop around the newly apportioned valence electrons; compare the quantity of valence electrons with the neutral atom, as it is found in the periodic table. An excess of electrons results in a negative ON; a deficiency of electrons gives a positive ON.
- The following examples are helpful:
a) Fluorine is the most electronegative element; its ON always = 0.
OF2:
Fluorine is more electronegative than oxygen, and so it will “take” the bonding electrons.
b) Oxygen is the second most electronegative element after fluorine. That said, F is not typically encountered in a high school chemistry classroom. This makes O the de facto “King of the Hill"5 in terms of electronegativity6; it explains why oxygen almost always has ON = -2.
H2O:
c) In peroxides, where O is bonded to another O, ON = -1. Take a look:
H2O2:
d) Elements such as C and N—and many others—have ON according to their bonding environment:
CH4:
Here, C is more electronegative than H and assumes an ON of -4.
CO2:
Because C is bonded to two oxygen atoms, it “gives” its bonding electrons to O, resulting in ON = +4.
HCN:
Here, C has ON = +2; it “loses” three of its electrons to N, but “takes” H’s electron.
And so it goes. Simply draw the Lewis Structure of the molecule. Apportion bonding electrons according to electronegativity, and compare the quantity of valence electrons to that of the neutral atom.
2. The “Pot” Method,
Once the ON rules are understood and learned, a valuable shortcut can be used in many cases. It requires simple algebra and what I call “no funny business”; that is, no F or peroxides. For example, to determine the ON of Mn in MnO4–, put four Os and one Mn in a “pot” with a charge of –1. Each O has an ON of -2, for a total of –8. Mn must then have ON = +7 to arrive at a sum of –1, the charge on the “pot”.
This is essentially an algebraic approach.
The “Pot” Method can be used to determine the ON of Cr as +6 in the dichromate ion, Cr2O72–.
Another illustrative example is for the various oxidation numbers of chlorine, as shown in Table 1.
Table 1. The Oxidation Numbers (States) of Chlorine
The previous examples illustrate the relationship between the maximum positive and maximum negative oxidation numbers of an atom and its position in the periodic table. Chromium is in Group 6, hence its maximum ON of +6 in Cr2O72– and in CrO42–. Similarly, Manganese is in Group 5; the ON of Mn in MnO41– is +5; Cl can have ON from +7 to -1.
3. Average Oxidation Number
It is possible for an element to have more than one ON in a compound.
In propane, C3H8, for example, using the “pot” method gives carbon an average ON = -8/3, which is algebraically sound, but doesn’t make sense: how could C have a fractional ON?
This conundrum is solved by drawing the Lewis structure of propane and apportioning the bonding electrons according to the electronegativity of the bonding atoms. This results in the terminal carbons having ON = –3; the central C has ON = –2.
The average the ONs of carbons = (-3)+(-2)+(-3)/3 = -8/3.
Fortunately, the average oxidation number of a given element is usually sufficient when balancing redox equations.
The next article in this mini-series will demonstrate an understandable—and quick—method for balancing redox equations.
1. A Logical Order of Topics for Honors Chemistry and AP Chemistry | Chemical Education Xchange
2. The Value of Microtitrations | Chemical Education Xchange
3. Oxidation number is used synonymously with Oxidation State
4. Oxidation state - Wikipedia (A much more in-depth explanation—more than what students need—can be found here.)
5. Apologies to Hank …
6. Liken F to the biggest bully in school. But that bully is—fortunately—absent most of the time. This makes O, the second biggest bully, but who’s always around, the one to be feared.
