Oxidation-Reduction Reactions - there are many examples

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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: it is oxidized. It becomes more positive because it loses electrons. Thus, oxidation = loss of electrons. The opposite is true for reduction. Reduction is a decrease in oxidation state (and thus a gain in of electrons). Consider the following half reaction:

Al3+(aq) Al(s)

Reduction

The Al3+ ion decreases in oxidation state:  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 by breaking up compounds into ions

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

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

2)     Write out the oxidation and reduction half reactions

MnO4- Mn2+             Fe2+Fe3+

Reduction                     oxidation

 

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  common species.

5e   + 8H+ + MnO4- Mn2++ 4H2O

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

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

With practice you practice, you’ll be able to skip some of the steps and balance these reactions fairly quickly.