Reduction and Oxidation (redox)
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Reduction and oxidation reactions (redox) involve the transfer of electron density from one atom to another.
For example, here is the reaction of hydrogen and oxygen to produce water:
2H2 + O2 ==> 2H2O
The H2 and O2 both carry a charge of zero because
they are nonpolar.
But the product, water, is polar and the hydrogens have a partial
positive charge and the oxygen has a partial negative charge.
So, during the reaction, the hydrogen atoms lose some electron density and the oxygen atoms gain.
Many chemical reactions involve a shift of electron density and are called redox reactions.
Reduction and oxidation always occur together.
If one thing is reduced, another thing is oxidized.
The reactant that is reduced is called the oxidizing agent and accepts electrons.
The reactant that is oxidized is called the reducing agent and supplies electrons.
In the equation above, the H is oxidized and the O is reduced.
oxidation: 2H2 ==> 2H+ + 2e-
reduction: O2 + 2e- ==> O2-
The two electrons cancel out, giving you the original equation.
This method is very useful for balancing equations also.
Oxidation numbers tell which atoms gain or lose their electron(s) in a reaction.
The rules for oxidation numbers are as follows:
- Each atom in free form (not in compound), has an oxidation number
of 0. For example, oxygen gas (O2), charcoal (C), and iodine (I2) all have an oxidation number of 0.
- In ionic compounds consisting of monoatomic ions, each at has an oxidation number equal to its ionic charge.
- Group (IA) metals always have an oxidation number of +1.
- Group (IIA) metals always have an oxidation number of +2.
- Fluorine always has an oxidation number of -1.
- The oxidation number of oxygen is -2 except in peroxides or in flouride compounds. When combined with flourine, the only element with greater electronegativity, it has a positive oxidation number of +2. In peroxide compounds such as hydrogen perox
ide (H2O2), the oxidation number is -1.
- Hydrogen has a +1 oxidation number in all compounds except metal hydrides (for example: lithium hydride LiH). When combined with metal hydrides the oxidation number is -1.
The steps for obtaining the oxidation number in compounds are as follows:
- Set the total oxidation number equal to zero (for neutral compounds; for charged ions, make it equal to its charge).
- Use the above rules to find (an) element(s) that do not fit.
- Move the other oxidation numbers to the zero side of the equation.
- Divide the subscript of the unknown by the number that was on the 0 side of the equation.
From this, you have the oxidation number of the unknown.
Oxidation numbers in ion compounds can but determined by setting the side of the equation that was 0 in the previous equation to the ionic charge.
