CR_Oxidation - Reduction Reactions Lesson

Oxidation - Reduction Reactions

image of Antoine LavoisierThere is another way that chemists classify reactions. They are called oxidation-reduction reactions, or redox (reduction-oxidation) for short. Redox reactions involve the transfer of electrons between chemicals. Synthesis, decomposition, single displacement, and combustion reactions can all include reactions that are redox. So, don't think of redox reactions as a totally different family of reactions. Rather, you can classify reactions within the main types as either redox or non-redox.

In the late 18th century, Antoine Lavoisier  proposed the idea of combustion, reactions with oxygen. His theory was eventually accepted and chemists began to describe any reaction between an element or compound and oxygen as  oxidation.

The reaction between magnesium metal and oxygen, for example, involves the oxidation of magnesium.

LaTeX: 2Mg_(s)+O_{2(g)} \rightarrow 2MgO_(s)2Mg(s)+O2(g)→2MgO(s)

By the 20th century, scientists concluded that the oxidation process always involved the loss of electrons. Now, we refer to any reaction where electrons are lost as oxidation. It is no longer limited to just reactions involving oxygen.

It goes to reason that if oxidation is a process where electrons are lost that those electrons must then be gained by something. The gaining of electrons is known as reduction.  

An easy way to remember this is with this mnemonic device:

LEO says GER LEO = Loss of Electrons is Oxidation GER= Gain of Electrons is Reduction

That might sound odd to you. Why would the gaining of something be known as reduction? As the theory of oxidation and reduction developed, scientists started looking at these reactions in a different way. Instead of looking at the loss and gain of electrons, the concept of oxidation numbers was developed.

The concept of oxidation numbers is just a bookkeeping system used to keep track of electrons in chemical reactions. All chemicals have an oxidation number. It is the charge that element would have if it were composed of ions. Based on oxidation numbers, the definitions of oxidation and reduction were expanded to the following:

Oxidation is the increase in the oxidation number. Reduction is the decrease in oxidation number.

Oxidation and reduction always occur together. You can't have one without the other. In order to see both of these happening within the same reaction, you need to look within the reaction, splitting it into what are called half-reactions. 

Identifying Oxidation Numbers

In order to determine if a particular reaction is a redox reaction and be able to write the half reactions, you must look at the oxidation numbers of each element. If an element undergoes a change in oxidation number, a redox reaction has occurred.  

Let's start by learning how to assign oxidation numbers. The following set of rules may seem overwhelming at first, but you will find this process progressively easier with practice!

Rules for Assigning Oxidation Numbers 1. The oxidation number of a free element is zero. (This is true no matter if the element is diatomic... Oz, O3, P4, S8, Al, Ca, etc. all have an oxidation number of 0.) 2. The oxidation number of monatomic ions is equal to the charge on the ion. (For example, the oxidation number of sodium in the Na+ ion is +1, and the oxidation number of chlorine in the Clion is -1.) 3. In a polyatomic ion, molecule, or compound, the sum of all the oxidation numbers must equal the overall charge. (This will be zero for a molecule or compound.) 4. In compounds, fluorine has an oxidation number of -1. 5. In compounds, hydrogen has an oxidation number of +1. 6. In compounds, oxygen has an oxidation number of -2.

Sometimes there is a conflict between rules. When this happens, apply the rule with the lower number and ignore the conflict.

Assigning Oxidation Numbers - Example Determine the oxidation number of each element in sulfate, SO4-2. Using the rules, we know that the sum of all the oxidation numbers here will be -2. If we were to write it in a mathematical formula, it would look like this: oxidation #'s of S + oxidation #'s of O = -2 There is 1 sulfur and 4 oxygens. So, let's rewrite this formula. I will abbreviate oxidation number to O#. 1(0# S) + 4(0# 0) = -2 According to the rules, in compounds, oxygen has an oxidation number of -2. Let's plug this in and solve for the O# for S. 1(0# S)+4(-2) = -2 1(O# S) +-8=-2 1(O# S) = -2+8 O# S = 6

Use this simulation Links to an external site. to explore Oxidation Numbers.

Now you will use the skill of assigning oxidation numbers to identify the oxidation and reduction processes within a chemical reaction. Watch this video to learn how.

Examples of Redox Reactions

Any time there is a change in oxidation number within a reaction, it can be classified as a redox reaction.   These can be synthesis, decomposition, single displacement, or combustion reactions. Here are some examples of redox reactions in their broader categories.

Synthesis:

The tarnishing of silver or the rusting of iron are both synthesis, redox reactions.

These specific reactions are generally referred to as corrosion. When iron rusts, it does so by reacting with oxygen. It forms a mixture of iron(II) and iron(III) oxides.

LaTeX: 2Fe_(s)+O_{2(g)} \rightarrow 2FeO_(s) \\ 2Fe_(s)+3O_{2(g)} \rightarrow 2FeO_{3(s)} \\ 2Fe(s)+O2(g)→2FeO(s)2Fe(s)+3O2(g)→2FeO3(s)

Decomposition: 

When hydrogen peroxide is placed in light, it decomposes to water and oxygen:LaTeX: 2H_2O_2 \rightarrow 2H_2O+O_22H2O2→2H2O+O2

Combustion:

Recall combustion reactions could be complete or incomplete.   You were told the specific products that were formed from each of these, but another way to determine those products is based on oxidation number. 

  • Complete combustion  means the higher oxidation number is formed.
  • Incomplete combustion  means the lower oxidation number is formed.

Single Displacement:

Occasionally, a redox reaction occurs where one element state undergoes both oxidation and reduction. Such a redox reaction is known as a   disproportionation reaction. The element undergoing disproportionation must have at least three different oxidation states.   This would be the original oxidation number (of that element as a reactant), and the two products (one with a higher oxidation number and one with a lower oxidation number).   

Here is an example of a disproportionation reaction:

0 Cl2 +1-2 H2O +1-1 HCl +1 +1-2 HCLO Cl2 + H2O → HCl + HCLO reduction between Cl2 and HCl oxidation between Cl2 and HClO

You can see the chlorine starts out as oxidation # zero. Part of this chlorine is oxidized to +1 and part of it is reduced to -1.

Remember to work on the module practice problems as you complete each section of content.  

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