Oxidation-reduction reactions take place whenever a species that can give up electrons (reducing agent) comes in contact with another species that can accept them (oxidizing agent). They don't "care" where the electrons come from, as long as they can give/receive them.
NOTE: a species is any kind of particle involved in a reaction. i.e. molecule, ion, element, etc.
Consider the following redox reaction:
2 Fe + 3 Cl
2

2 FeCl
3
In the above reaction, iron is oxidized by losing ______ to become an ______ ion. This is the oxidation half of the the redox reaction and can be written as:
At the same time, each chlorine atom in Cl
2
is reduced by accepting ______ to become a Cl
-
ion. This is the reduction half of the redox reaction and can be written as:
Equations such as these represent half-reactions.
A half-reaction is one of the two parts of a redox reaction – the oxidation half alone or the reduction half alone.
Balancing Redox Equations by Half-Reactions
Ni + ZnCO
3

Zn + Ni
2
(CO
3
)
3
Step 1 : Figure out the oxidation numbers of each species.
Step 2 : Write the balanced oxidation and reduction half-reactions leaving out spectator ions.
Step 3 : Adjust the coefficients so that the number of electrons lost in oxidation EQUALS the number of electrons gained in reduction.
Step 4 : Add the balanced half-reactions together and return spectator ions (if there were any).

Diatomics are Weird!!! Diatomics need to be balanced BEFORE you state electrons gained or lost!
F
2
+ NaI  NaF + I
2
Practice : Balance the following by the half-reaction method
1.
CrBr
3
+ Ca  CaBr
2
+ Cr
2.
Fe + HF

FeF
3
+ H
2
3.
Al + ZnSO
4

Zn + Al
2
(SO
4
)
3
4.
Fe + I
2

FeI
3

As we have previously seen, the half reaction method is used to balance redox equations by balancing the oxidation and reduction halfreactions. A half reaction is an equation showing either the reduction or the oxidation of a species in an oxidation-reduction reaction (and the electrons either lost or gained) .
The first step in the half reaction method is to divide the overall reaction into two parts (half-reactions). Then each half reaction is balanced by adding coefficients to have the number of electrons gained by the reduction half reaction equal to the number of electrons lost by the oxidation half reaction. Finally the two half reactions are added to give a balanced equation.
When using the half-reaction method for acid and base solutions, we use the above steps but we must also account for the water (from the solution) and either the H + or OH ions from the acid or base.
Example : Balance the oxidation of sulfur with nitric acid in aqueous solution. S (s) + NO
3
(aq)  SO
2
(g) + NO (g)
Step 1 : Figure out the oxidation numbers for each element involved in the equation.
S + NO
3
 SO
2
+ NO
Step 2 : Write out the half reactions for the oxidation and reduction processes. (KEEP oxygen in the half reactions!!!!)
 ____ is oxidized because its oxidation number increases from __ to ___.
 ____ is reduced because it’s oxidation number decreases from __ to ___.
Oxidation :
Reduction :
 Make sure to balance the main atoms, leaving H and O for step 3. (In this example, S and N are both already balanced).
 DO NOT add in the electrons UNTIL the atoms are balanced!!!
Step 3 : For each half reaction, balance the oxygen atoms (by adding H
2
O) and then the hydrogen atoms (by adding H + ). a) Balance the oxidation half reaction
____ molecules of H
2
O can be added to balance oxygen. ____ H + ’s must then be added for it to be completely balanced
S  SO
2
+ 4e -
Balance the reduction half reaction
____ molecules of H
2
O can be added to balance oxygen. ____ H + ’s must then be added for it to be completely balanced
NO
3
+ 3e  NO
Step 4 : Multiply each half reaction by an appropriate number/coefficient to make the electron changes equal. In any redox reaction, the electrons lost in oxidation must be equal to the electrons gained in reduction .
 The oxidation half-reaction needs to be multiplied by ____, and
 the reduction half reaction needs to be multiplied by ____.
 Therefore, the electrons lost in oxidation and gained in reduction both equal ____.
: H
2
O + S  SO
2
+ e- + H +
: H + + NO
3
+ e-  NO + H
2
O
Step 5 : Add the half reactions and subtract terms that appear on both sides of the equation. (OR, cancel out the terms on both sides and just write out the final reaction)

The half reaction method can also be done in basic solution . In this case you do everything as stated above. However, you would then have to add the same number of OH- ions as H+ ions to BOTH sides of the half-reaction (to keep it balanced). On one side of the equation, the H+ and OH- will combine to give water. The other side will be left with the
OH- ions. (Cancel out water as necessary.)
Example : Balance the following in basic solution.
3
1-
2

1-
4
1-
Practice balancing the following half reactions: a) Balance H
2
GeO
3

Ge (in acid solution) b) Balance IO
3
1
I
2
(in acid solution) c) ClO
3
1
ClO
4
1 + O
2
(in basic solution)