Download O usually has oxidation number of -2, except in peroxides where it is

Good general guidelines for assigning oxidation numbers (ON)
1. The Elements: The oxidation number of a pure element is zero
e.g. Fluorine in F2 has oxidation number of 0
2. Simple Ion Charge = Oxidation Number
e.g. Na1+ has oxidation number of +1, Mg2+ has oxidation number of +2
3. Hydrogen: Hydrogen has oxidation number of +1 in all compounds except in metal hydrides where it is
-1.
e.g. Water: H is bonded to O by a polar covalent bond. The e- pair
to O -> H is partially positive (δ+) and O is partially negative (δ-). H
oxidation number of +1.
is closer
has
4. Oxygen
O usually has oxidation number of -2, except in peroxides where it is assigned -1, and in OF2 where it is
assigned a +2 due to the higher electro negativity of F.
-In calcium oxide, CaO, oxide ion has a 2- charge. Its oxidation number is -2
-In H2O2, Each O is assigned the oxidation number of -1
5. Covalent Compounds: In compounds that do not contain hydrogen or oxygen, the more electronegative
element is assigned the oxidation number it would have in an ionic compound. (Provide electro
negativity table)
e.g. In SCl2, chlorine has a higher electro negativity so it is assigned -1 first and sulfur is assigned +2.
6. Compounds: The sum of the oxidation numbers in a compound is zero.
e.g. The sum of oxidation numbers in NaCl is (+1) + (-1) = 0
7. Polyatomic Ions
The sum of the oxidation numbers of the elements in a polyatomic ion must equal the ion charge. Consider
these examples. If there are two poly atomic ions in a compound deal with them first.
e.g. Carbonate ion,
. O has oxidation number of -2. There are 3O’s in the formula so the total
negative charge is 6-. Since the carbonate ion has a charge of 2-, the oxidation number of C must be
+4.
Or to say it another way
(have your periodic table out and know the families, alkali, alkaline earth, halogens and noble gases)
Not great rules but get you the feel of assigning oxidation numbers:
0)
elements have zero charge (or they would be ions); just like compounds
1)
Poly atomics choose first (Know your poly-atomics if more than 2 elements)
(“The needs of the many outweigh the needs of the few”)
2)
Remember they are called oxidation numbers (oxygen chooses -2, usually (can be -1 in
peroxides); Fluorine is -1)
3)
Give a little, alkali metals choose +1 (by giving up one electron). Hydrogen is above it all
and is special (usually +1, but hydrogen can be -1 when in a metal hydride)
4)
Alkali earth metals choose +2
5)
With all this giving up of electrons there must be some taking, Halogens take one
electron to become -1
6)
Continue with elements picking their preferred charges (work from outside columns to
the inner “valley of confusion”) until there is only one left; if the element is last to choose it
must have the charge that makes everything else sum to zero.
If you don’t choose first you might not get your first choice
(you can’t always get what you wanted, but if you try real hard you (might just get) what
you (“We, the compound”) need)
Balancing a chemical equation gives you lots of information. Some equations are more difficult to balance. The technique of
half reactions will allow you to balance some equations that you would find difficult otherwise.
Oxidation numbers help us keep track of the movement of electrons (what chemistry is really all about)
We need to learn how to assign oxidation numbers
RedOx notes:
Remember electrons are Negative (losing electron makes you
positive) we use the symbol e- to help reming us of that
You MUST remember the basic definitions, which are often summarized as
Oil Rig Oxidation is Loss of electrons; Reduction is Gain (of electrons)
Or
LEO GR
Loss of Electrons is Oxidation, Gain is Reduction
 In Oxidation half reactions, the oxidation number (value) of the
element increases (like a big Ox)
Na  Na+ + 1eIn Reduction half reactions, the oxidation number (value) gets smaller
(this is what reduction means)
2 e- + Cu+2  Cu
Oxidizing Agents allow oxidation to proceed (by being reduced)
Reducing agents allow reduction to proceed (by being oxidized)
Steps to take in solving the Redox problems
(how partial credit will be determined on the test)
1)
Assign oxidation numbers (elements have just as many
protons as electrons, Ions have a net charge).
2)
Identify the two half reactions: what is being oxidized and
what is being reduced. (also identify the agents)
3)
Balance each half reaction (keep the subscripts in mind now
so you don’t get confused later).
4)
Balance the two half reactions so that just as many electrons
are produced as are needed. This is called the net ionic
reaction
5)
Add the two half reactions and transfer these values to the
overall reaction.
6)
Take into account the spectators (adjust the numbers in step
5 as necessary). Then balance the rest of the spectators.
The ratios of the net ionic reaction help do this.