Oxidation + Reduction Half-Reactions

In summary, the conversation discusses the procedure and guidelines for carrying out oxidation, reduction, and net reactions. The process involves one element becoming a reducing agent and another becoming an oxidizing agent, with the overall goal of balancing the reaction equation. The example of Li and O2 is used to illustrate this process, with the final balanced equation being 4Li + O2 ---> 2Li2O. The conversation also touches on the importance of balancing the charges and the role of intermediates in reaction kinetics. The expert summarizer provides clear explanations and helps the original poster understand the concept.
  • #1
JDK
27
0
Hello,

I was curious if anyone could explain to me the simple procedure/guidleines to carrying out oxidation (half-reactions), reduction (half-reactions) and overall net reactions. I've read my Chem book over and over, but I don't quite understand it. Even a pointer to a good site with a decent explanation would be great. Any help is awesome. Thanks so much.

Here's a simple example I need some help with...

Li + O2
 
Chemistry news on Phys.org
  • #2
Oxidation reduction reactions involve an certain element obtaining an electron(s) from a second element. In this case one element becomes the reducing agent, gives its electrons away and the element receiving this electron becomes the oxidizing agent (since it oxidizes the other element).

You need not to be concerned about the exact mechanism of the reaction. First, try writing out the balanced equation.

4Li + O2 --->2Li2O

So which element was oxidized? Li does not have a charge while LiO is an ionic compound in which electrons are "shared" according to electronegativity. In this case oxygen is much more electronegative and it essentially obtains two electrons from Li through bonding.

Now write out the half reactions. Oxygen is reduced.

O2 + 4e- ---> 2O (with a 2- charge)

Li is oxidized.

4Li--->4e- + 4Li (with a 1+ charge)

*So overall

O2 + 4Li + 4e- ---> 2O (2-) + 4Li (1+) + 4e-

A third reaction is also important. The products of the latter two steps combine to form 2Li2O.

2O (2-) + 4Li (1+) ---> 2Li2O

Combine this with the above equation and you will return to the original balanced equation.

Hope this answers your question.
 
  • #3
Hmmm... *reads over information*

It somewhat does. I have a question... can you explain this line and how you got to it...

4Li + O2 --->2Li2O

Normally I would see Li and 02 occurring together like so... Li2O (lithium oxide). In this case I'm stumped. I have a hunch it has to do with the fact that oxygen is diatomic. If you can explain that line for me I have a feeling I might get the rest of it. Excuse my inability to catch onto this stuff quickly. Chem 20 was relatively easy up until this. Thanks again.
 
  • #4
You don't need to worry about the specifics in how the reactions occur. A chemical equation simply pertains to the initial and final states (at least for now; you will learn about intermediates later in reaction kinetics). In this case we know we have a reaction between Li and Oxygen gas.

If we wrote out the reaction as

Li + O2 --->Li2O, the equation would not balance indicating that such reactions simply do not.

It is logical that since the reactants makeup the products the number of moles would have to be equivalent on both sides.

Thus we know the initial and final states of the reaction as

4Li(g) + O2 (g)--->2Li2O (g)

Just tell me if you are still confused.
 
  • #5
Ahhh, I get it. Alright, I understand everything up until this line...

O2 + 4Li + 4e- ---> 2O (2-) + 4Li (1+) + 4e-

Why does the 4e- still remain in the products and then disappear afterwards? Thank-you.
 
  • #6
O2 + 4Li + 4e- ---> 2O (2-) + 4Li (1+) + 4e- is essentially

O2 + 4Li ---> 2O (2-) + 4Li (1+)

The products of this reaction combine to form 2 molar equivalents of Li2O.

2O (2-) + 4Li (1+)--->2Li2O

We are just summarizing the important steps (states) of the reaction.
 
  • #7
Thanks so much for your wonderful responses. I think I get it all completely now. Your help has been appreciated. :smile:
 

Related to Oxidation + Reduction Half-Reactions

1. What is oxidation and reduction?

Oxidation is the process of losing electrons, while reduction is the process of gaining electrons.

2. What are oxidation and reduction half-reactions?

Oxidation and reduction half-reactions are the individual reactions that make up a larger redox reaction. Oxidation half-reactions involve the loss of electrons, while reduction half-reactions involve the gain of electrons.

3. How do I balance oxidation and reduction half-reactions?

To balance oxidation and reduction half-reactions, you need to ensure that the number of electrons lost in the oxidation half-reaction is equal to the number of electrons gained in the reduction half-reaction. You can do this by adding coefficients to each species in the reaction.

4. What is the purpose of balancing oxidation and reduction half-reactions?

The purpose of balancing oxidation and reduction half-reactions is to ensure that the total charge of the reactants and products is equal. This is necessary for the overall redox reaction to be balanced.

5. How do I identify oxidation and reduction half-reactions?

Oxidation half-reactions involve an increase in oxidation state, while reduction half-reactions involve a decrease in oxidation state. Additionally, oxidation reactions typically involve the loss of hydrogen or gain of oxygen, while reduction reactions typically involve the gain of hydrogen or loss of oxygen.

Similar threads

Replies
15
Views
3K
Replies
16
Views
2K
  • Biology and Chemistry Homework Help
Replies
3
Views
2K
Replies
18
Views
23K
Replies
13
Views
13K
  • Biology and Chemistry Homework Help
Replies
1
Views
3K
  • Biology and Chemistry Homework Help
Replies
4
Views
6K
  • Biology and Chemistry Homework Help
Replies
10
Views
22K
  • Biology and Chemistry Homework Help
Replies
5
Views
4K
Back
Top