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Oxidation-Reduction Reactions

Oxidation-reduction (redox) reactions involve the transfer of electrons from one compound or species to another. In this section, we will discuss oxidation-reduction reactions and learn how to balance them.
Oxidation is the process by which an atom or species loses its electrons. In reduction, an atom or species gains electrons. Let's first consider oxidation and reduction separately.
Consider the conversion of iron from its neutral elemental state to its ionic form.


Notice that in the process iron lost electrons. The process is oxidation.
Consider another example. An example involving the conversion of bromine to its ionic form.


Notice that in the process bromine gained electrons. The process is reduction.
Now let's go one step forward. Consider the next reaction.


This reaction is a typical example of an oxidation-reduction reaction. The oxidation number of iron on the reactant side is 0. The oxidation number of bromine on the reactant side is also 0. What are the oxidation numbers of iron and bromine in FeBr3? Well, we know that bromine has an oxidation number of –1. So the oxidation number of iron in FeBr3 is +3. Thus iron is oxidized and bromine is reduced. The species that gets oxidized is called the reducing agent. The species that gets reduced is called the oxidizing agent. In this reaction, evidently iron acts as the reducing agent, and bromine acts as the oxidizing agent.

Oxidation results in an increase in the oxidation number. In the process of oxidation, electrons are lost.


Reduction results in a decrease in the oxidation number. In the process, electrons are gained.


Balancing Oxidation-Reduction Reactions

Balancing of an oxidation-reduction reaction is a little more complex than balancing a simple reaction. The main rule that you have to follow when balancing oxidation-reduction reactions is that the absolute value of the increase in oxidation number of all the atoms that are oxidized should equal the absolute value of the decrease in oxidation number of all the atoms that are reduced. Balancing oxidation-reduction reaction is sometimes time-consuming and quite often frustrating. We will look at two methods of balancing oxidation-reduction reactions.


Method A
  1. Write the unbalanced equation.
  2. Find the oxidation numbers of the atoms that undergo change in oxidation states and write on top of each atom the corresponding oxidation number.
  3. By this process, we will be able to see which atoms are getting oxidized and reduced.
  4. Compare and indicate the change in oxidation numbers from the reactant side and the product side, and write down the change in the oxidation numbers.
  5. Make the necessary changes by writing coefficients that will equalize the changes in the oxidation numbers. In other words, the net decrease in the oxidation numbers should equal the net increase in the oxidation numbers. Add water if necessary.
  6. Do a final check on whether all the atoms and charges balance out.

Method B
  1. Write the unbalanced equation.
  2. Separate the two half-reactions, write them out, and balance any of the atoms. From this point onward, we balance the reactions separately. Do the obvious or the simple balancing by inspection if possible.
  3. Balance the oxygen atoms by adding water on the appropriate side of the half-reaction.
  4. Balance the hydrogen atoms by adding H+ on the appropriate side of the half-reaction.
  5. Add sufficient number of electrons so that the charges are balanced.
  6. Once the half-reactions are balanced, combine the half-reactions and cancel out any common terms that appear on both sides of the equation to get the refined and balanced oxidation-reduction equation.

Example 2-3

 Balance the following oxidation-reduction reaction.


Method A
image\Ch 2 Example 2-3a.png
image\Ch 2 Example 2-3b.png
image\Ch 2 Example 2-3c.png

Method B
image\Ch 2 Example 2-3a1.png

Half-reaction I
image\Ch 2 Example 2-3a2.png

Half-reaction II
image\Ch 2 Example 2-3e.png

Before we combine the equations, the electrons need to be balanced out. So multiplying the balanced half-reaction # (I) by 4, we get
image\Ch 2 Example 2-3f.png

Combining the two half reactions
image\Ch 2 Example 2-3g.png

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