Redox Reactions

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Introduction to Redox Reactions

Oxidation-reduction reactions (often abbreviated as redox) involve the
transfer of electrons. Many important chemical reactions involve oxidation and
reduction. Photosynthesis, which stores energy from the sun in plants by
converting carbon dioxide and water into sugar, is a very important
oxidation-reduction reaction. Combustion reactions, which provide most of the
energy to power our civilization, also involve oxidation and reduction. An
example of this is the combustion of methane with oxygen:

CH4(g) + 2O2(g) CO2(g) + 2H2O(g) + energy

Even though none of the reactants or products is ionic, the reaction is nonetheless assumed
to involve the transfer of electrons from carbon to oxygen.

 

Balancing
Oxidation-Reduction Reactions

 

Suppose you are to balance the following reaction occurring in an acidic solution:

MnO4-(aq) + Fe2+Fe3+(aq) + Mn2+(aq)          

How would you balance such an equation? You might think to yourself that this is impossible, since the oxygen atoms “disappear” and that violates the Law of Conservation of Mass. But this reaction indeed occurs. Thus we need to develop a new approach to balance oxidation-reduction reactions.

 

What is Oxidation and Reduction?

For oxidation-reduction reactions that occur in aqueous solution, we break the reaction up into two parts, or two half-reactions:one involving oxidation and the other involving reduction. Oxidation involves an increase in oxidation state. Consider the following half reaction:

Sn2+(aq) Sn4+(aq)

Oxidation

The Sn2+ ion increases in oxidation state becoming “more positive”: it is oxidized. This means that oxidation involves losing electrons. The complete opposite is true for the process of reduction. Reduction involves a decrease in oxidation state. Consider the following half reaction:

Al3+(aq) Al(s)

Reduction

The Al3+ ion decreases in oxidation state-it becomes “more negative”: 
it is reduced. Thus, reduction involves gaining electrons.

A useful reminder is OIL RIG: Oxidation Is Loss,Reduction Is Gain

Now that you have a basic understanding of the concept of oxidation and reduction, now you must learn how to balance redox equations. It is not hard, but practice is critical. The reactions we’re concerned with will occur in either acidic or basic solutions. Therefore, there will be two ways to balance redox reactions depending whether or not it occurs in acidic or basic solution. (You will be told whether the solution is basic or acidic).

 

The Half-Reaction Method for
Balancing Redox Reactions in Acidic Solutions

Returning to the equation earlier:  MnO4-(aq) + Fe2+Fe3+(aq)+ Mn2+(aq) 

Let’s develop a simple method for balancing this equation in acidic solution. The following method applies to any redox reaction occurring in acidic solutions. I will break it down into steps to reduce confusion.

MnO4-(aq) + Fe2+Fe3+(aq) + Mn2+(aq)

1)     Identify which substance is reduced and which one is oxidized

               Mn7+(aq)  + 4O2-(aq)  + Fe2+(aq)
Fe3+(aq)  + Mn2+(aq)

You can see that Mn goes from  +7 charge to +2 charge, which means that Mn is reduced. You also notice that Fe goes from +2 to +3, thus Fe is oxidized.

2)     Write out the oxidation and reduction half reactions

MnO4- Mn2+             Fe2+Fe3+

Reduction                     Oxidation
The permanganate ion appears to be oxidized, but it is really reduced.

 

3)     Balance all elements except hydrogen and oxygen

All other elements are already balanced in this case

 

4)     Balance oxygen atoms using water

       MnO4- Mn2+ + 4H2O                Fe2+Fe3+  (no oxygen to balance)

5)     Balance hydrogen using H+ ions

 

8H+ + MnO4- Mn2+ + 4H2O                Fe2+Fe3+  (no hydrogen to balance)

 

6)     Balance the charges using e (electrons)

5e+ 8H+ + MnO4- Mn2+ + 4H2O                Fe2+Fe3+ + e

            

7)     Be sure both equations have the same number of electrons. If they don’t, then multiply one of them by some factor so they have the same number of electrons.

5e+ 8H++ MnO4- Mn2+ + 4H2O                Fe2+Fe3+ + e

                             5 electrons                                      1 electron

The two equations do not have the same number of electrons! But if we multiply the oxidation reaction by a factor of 5, then the equation will have 5 electrons:

5 X (Fe2+Fe3+ e) = 5Fe2+5Fe3+ +5e

 

8)     Add the two reactions together. Cancel out common species.

5e+ 8H+ + MnO4- Mn2++ 4H2O

   +   5Fe2+5Fe3+ +5e         (electrons should ALWAYS cancel out)      

Balanced Equation: 8H+ + MnO4- + 5Fe2+ Mn2+ + 4H2O + 5Fe3+

This may seem like a lot of work. But if you practice, you’ll be able to skip some of
the steps and balance these reactions fairly quickly.