Redox reactions are somewhat more complicated that the reactions we have considered thus far, and are often specified in an incomplete form. Thus balancing such reactions is a little more involved. We will discuss two ways to balance such reactions. One is the standard, stepwise procedure that uses the concept of half reactions. The other is the algebraic method developed in chapter 2. The former leads directly to balanced half reactions and indirectly to the overall reaction. The former leads directly to the overall reaction and indirectly to the half reactions.
Consider a simple oxidation-reduction reaction between copper and aqueous silver ions:
Since we know that electrons are transferred in the reaction, it is useful to think of this reaction as the sum of two half reaction in which the transferring electrons appear explicitly:
The sum of the first and twice the second yields the correct overall reaction.
We will see later that half reactions sometimes have a physical significance, namely that they represent reactions that actually occur at an electrode in an electrochemical cell. Thus, knowing the half reactions that lead to an overall reaction is useful.
Half reactions can also be used to develop a step-wise procedure for balancing redox reactions. The procedure is the following:
O to the side deficient in oxygen.
(for acidic solutions) or
(for basic solutions) to the side deficient in hydrogen and an equal amount
of H O to the opposite side.
Example: Use the above procedure to balance the following reaction in acidic solution:
The two half reactions are:
Balancing all elements except oxygen, the above reactions would not change, as they are already balanced with respect to Cu, N and S.
Balancing oxygens by adding water gives:
Since the solution is acidic, hydrogen is balanced by adding
as needed and adding additional water to balance. This gives
Finally, add electrons to balance the charge:
Now add the reactions in such a way that the electrons cancel out. We can do this by multiplying the first half reaction by 3 and the second one by 8, which will give 24 electrons in each reaction:
The sum of the two is
The algebraic method will now be applied to the same reaction. Again, we must
complete the reaction by adding and 
. However, we will add them with
unknown coefficients. The way to add these in is to look at the unbalanced reaction.
Since is positively charged, we look to see which side of the unbalanced
reaction is negatively charged. In this case, it is the left side, so we add
to the left side and to the right side. Note that this is only a
guess, but if we guess incorrectly, all that will happen is that the coefficients
of and will end up being the negatives of what they should be
and we can simply switch which side of the equation they appear on. Thus,
we wish to balance the reaction:
Putting unknown coefficients in front of each species gives
The balance conditions are:
, or
to 1, as there is always the possibility that these can be 0. Thus, here
we arbitrarily set y=1. Then the solution is
Multiplying all coefficients by 8 gives the balanced reaction:
To obtain the half reactions, we start with the above balanced equation and just need to do a little more algebra. We write the copper half reaction with electrons on the right since it is oxidized and the nitric acid half reaction with electrons on the left since it is reduced. Hydronium must be added to the same side as the electrons and water added to the opposite side. These are all added with unknown coefficients, giving:
Then we balance oxygen, hydrogen and charge in both half reactions. This leads to the following set of equations for the first half reaction:
whose solution is easily found to be
and for the second half reaction, the equation are
The last equation is solved immediately, using the fact that z=24. Thus, we find
so that the half reactions are
As a second example, consider balancing the reaction in acidic solution:
Since the solution is acidic, we again add hydronium. since the right side has a net positive charge, our initial guess is to add hydronium to the right and water to the left. Doing this and putting in unknown coefficients gives:
The balance conditions are:
Setting x=1 gives the solution:
Thus, we have an example where, in the overall reaction, no water or
appear. This does not mean that they do not appear in the half reactions.
Multiplying all coefficients by 2 gives the balanced reaction:
The half reactions are:
Now, let us guess as to which is the oxidation and which is the reduction. We will show that even if the guess is wrong, the coefficients will be correct but have negative signs. At first sight, it ``looks'' as if the first half reaction is an oxidation and the second is a reduction. Thus, consider writing:
For the first half reaction the oxygen, hydrogen and charge balance equations are, respectively:
which has the solution
so we know that we added the electrons, water and hydronium each to the wrong side and this is actually the reduction reaction. The coefficients are still numerically correct, and so we write the first half reaction as
Similarly, the balance equations for the second half reaction are
whose solution is
and so this half reaction is an oxidation:
Clearly the sum of these two gives the above overall reaction with electrons, water and hydronium all cancelling out in the sum.
A disproportionation reaction occurs when a species is both oxidized and reduced in the same reaction. An example is hydrogen peroxide:
In hydrogen peroxide, the oxygen has an oxidation number of -1, while in
water it is -2, and in O it is 0. Thus, one of the oxygens increases its
oxidation state by 1 and the other decreases it by 1.
Balancing such reactions is no different in the algebraic method. In the half reaction method, one needs to recognize disproportionation and write the same species in the same half reaction.
As an example, suppose we wish to balance in basic solution.
To use the half reaction method, one starts with the two half reactions
with Cl appearing in both and follows the usual steps.
In the algebraic method, we recognize a net negative charge on the right side.
Since we wish to balance this in basic solution, we guess that OH 
should
be added to the left side and water to the right. Adding unknown coefficients
gives
The chlorine, oxygen, hydrogen and charge balance conditions are, respectively:
Setting z=1 arbitrarily, the solution becomes:
and the balanced reaction is
