General Chemistry/Redox Reactions/Oxidation and Reduction equations

General

Oxidation and Reduction reactions involve the donation and acceptance of electrons.

Specifically, at the most basic level one element gets oxidized by losing, or donating, electrons to the oxidizing agent. In doing this the oxidizing agent gets reduced by accepting the electrons lost, or donated, by the reducing agent (ie. the element getting oxidized).

If it seems as though there are two separate things going on here, you are correct to a degree: Redox reactions can be split into two half-reactions, one dealing with oxidation, the other, reduction.

Example:

Complete reaction: $F{e}_{(s)}+C{u}^{2+}\to F{e}^{2+}+C{u}_{(s)}$

Oxidation ½ reaction: $F{e}_{(s)}\to F{e}^{2+}+2e^{-}$

Reduction ½ reaction: $C{u}^{2+}+2e^{-}\to C{u}_{(s)}$

When the 2 reactions are summed, the result is the original equation

$F{e}_{(s)}\to F{e}^{2+}+2e^{-}$

$\underset{\_}{C{u}^{2+}+2e^{-}\to C{u}_{(s)}}$

(All electrons must cancel out). If you are adding two ½ reactions with disparate electrons being lost and gained then the electrons must be multiplied to create a common denominator.

Example:

$F{e}^{2+}\to F{e}^{3+}+e^{-}$

$\underset{\_}{H_2{O}_2+2e^{-}\to 2O{H}^{-}}$

Since the electrons must cancel, both ½ reactions must be multiplied by something in order to create equal numbers of electrons in both equations.

Adding 2 separate ½ reactions

...to be continued Template:BookCat