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How to determine oxidation numbers

created by izubachi

(idea) by izubachi (21.1 hr) (print)   ?   (I like it!) 2 C!s Sat Feb 15 2003 at 4:40:59

When working with redox (reduction-oxidation) reactions in chemistry, especially with inorganic compounds, it is important to know the oxidation numbers of all compounds involved in the reaction. An oxidation number historically referred to the potential of binding to oxygen, however this interpretation is no longer considered accurate today and oxidation number now refers simply to charge. The individual charges of each ion as it combines can be found by consulting tables (such as the one given in oxidation numbers), however for efficiency's sake it is better to know the fundamental principles behind finding oxidation numbers and to apply them on the fly.

The oxidation numbers of redox reactions are governed by five basic rules.

  1. All free elements (those not in a compound) have an oxidation number of zero, including polyatomics such as oxygen (O2) and sulphur (S8). Of special note, water (H2O) is considered a free element and always has an oxidation number of zero.
  2. Monatomic ions have an oxidation number exactly the same as their charge. Easy.
  3. Florine (Fl) is always -1. Always. This rule does not apply to other halogens, just flourine.
  4. Oxygen is usually -2 with the exception of peroxide (O22-) which, despite the superscript, has an oxidation number of -1.
  5. Hydrogen is very nearly always +1, with some rare exceptions.
For the full picture, these rules must be used in conjunction to find the oxidation numbers of unknown portions of a compound. E.G. MnO4-. Since we don't have a specific rule to work with concerning Manganese, we'll need to deduce its oxidation number from what we do know. Oxygen in this compound is not in peroxide form, so it has an oxidation number of -2. There are four oxygens, so the total charge to that portion of the compound is -8. To balance this off, the Manganese would need to have a charge of +8, however it is important to note that the compound as a whole has a -1 charge (that little superscripted - sign). Therefore, Manganese must have a charge of +7 to allow that final singal negative charge to remain, and thus Manganese has an oxidation number of +7.

Breaking down the compound into each of its elements' individual charges seems a little silly. After all, the overall charge can immediately be known just from looking at the compound's formula; so why go any farther? The issue comes into play for redox reactions because individual elements will change places with each other. When they match up with their new partner on the other side of the equation, oxidation numbers are going to need to match up. If this requires some fiddling with added hydrogen or hydroxide ions, changes of compound number, or changes of charge, one will need all elements' oxidation number to take roll call, so to speak. By using the rules above, this can be done quickly and easily without tedious chart reference.


printable version
chaos

redox reaction Oxidation Numbers roll call halogen
Oxygen florine Chemistry Shortcuts to check for divisibility
oxidation reduction Hydrogen compound
Manganese inorganic
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