[John_Kimble]

Hello!

I can't quite understand how these redox reactions go.

For example:

4 Fe   +   3 O₂       2 Fe₂ O₃

So the oxidation states are 0, 0, +3 and -2 respectively.

Therefore, from my understanding the half reactions should look like:

4 Fe (0)     2 Fe (3+)   +   3e^-

O₂ (0)   +   2e^-       2 O₃ (2-)

Obviously this is wrong, since there's different # of electrons gained and lost.

The book says the correct answers are:

Fe       Fe (3+)   +   3e^-

O₂   +   4e^-       2 O^2-


Why do we leave out the 4 and 3 before Fe and O in reagents, but then write a 2 before O, but nothing for Fe from the product???


I can do the simple ones where the equations don't have to be balanced (all singular molecules/atoms), but once they have to be balanced things get too messy for me, and I can't for the life of me figure out how to do these. I've watched a bunch of videos and read plenty about it, but it's always the simple examples or just no explanaiton.

So if someone would be kind enough to explain how to do these redox half reactions when different numbers of moles are involved it would really help.

Thanks!!!

[AWK]

The book says the correct answers are:

Fe       Fe (3+)   +   3e-

O2   +   4e-       2 O2-


Why do we leave out the 4 and 3 before Fe and O in reagents, but then write a 2 for O, but nothing for Fe from the product???

This nothing means 1

[John_Kimble]

Great, thanks! This helped quite a lot, but I ran out of examples with answers, so I could test it if I'm understanding it right further

Does anyone maybe have any redox examples with answers? I'm googling with no success so far.

[John_Kimble]

So if I'm getting this right, we take the total number of atoms (e.g. 2 + 2 for Fluorine in the product), and then in the half reaction we can divide both by the same whole number.

e.g.

instead of:      4 Fe (0)       2 Fe₂ (3+)   +   3e^-

we divide the whole thing by 4, to get the:     Fe (0)       Fe (3+)   +   3e^-.


But then the oxygen should be:

6 O (0)   +   2e^-    6 O (2-)      / :6

to get:    O (0)   +   2e-       O (2-)

but it seems that it is:   3 O₂ (0)       6 O (2-)      ; divide by 3, to get the book answer     O₂ (0)  +   4e-     3 O (2-)

Why don't we then write the 3 O₂ as 6 O , so we can then divide both by 6, to get just O for both?


Oh lord, this is doing my head in, I haven't been so frustrated in a long long while. Thanks for any help I may get 

[Borek]

First, balance atoms. This will give you

Fe(0)  Fe³⁺
and
O(0)₂  2O^2-

Then, add electrons to balance charge:

Fe(0)  Fe³⁺ + 3e^-
and
O(0)₂ + 4e^- 2O^2-

Then, multiply these half reactions by such coefficients, that after adding them electrons will cancel out:

4Fe(0)  4Fe³⁺ + 12e^-
and
3O(0)₂ + 12e^- 6O^2-

add:

4Fe(0) + 3O(0)₂ + 12e^- 4Fe³⁺ + 6O^2- + 12e^-

or

4Fe(0) + 3O(0)₂ 2Fe(+3)₂O(-2)₃

[John_Kimble]

Thanks a lot, Borek!! I'm getting the correct results now!

Just one thing. I'm solving them, so that I assume that I don't know the number of atoms in the compounds in the product. Is this the correct method?

example:  this is given, and I have to write the half reactions:  Si   +   F₂       SiF₄


Now, when I start to balance the atoms I do it like this: ?

Si (0)    Si (4+)

F₂ (0)    2 F (-)

Is this correct? And if it is, what is the intuition behind it?

Thanks again.

[Borek]

This is not always that easy, as you can't always list separate ions (actually what I did listing them separately was not exactly correct). Truth is, for simple reactions like the ones you listed so far, there is no need for treating them as redox to balance them, it is much easier to balance by inspection. The more complicated ones in most cases will be easy to separate into reduction and oxidation, and the final product will be not combination of both reduced and oxidized forms. Sometimes it will be this way, but it is hard to explain without an example, so let's wait till you will hit this type of problem.

[John_Kimble]

The net ionic equation for the Breathalyzer test used to indicate ethyl alcohol concentration in the body is:
 
Assign an oxidation number to each element (atom) in the dichromate ion, ethanol, acetaldehyde and water. Write the partial redox reactions and define oxidizing agent in that process!

equation= http://postimage.org/image/8cibe3sbb/

So the changes in oxidation numbers I found are: 2nd C; from -1 to +1      and     Cr; from +6 to 3+

The half reactions I get are:

Cr₂ (+6)   +   3 e^-       2 Cr (3+)

C (-1)       C (+1)   +   2 e^-


Would that be correct? It seems that we'll only have exercises where we have to write half reactions.

Thanks for the help

[Borek]

3CH₃CH₂OH + Cr₂O₇^2- + 8H⁺  3CH₃CHO + 2Cr³⁺ + 7H₂O

Oxidation numbers you listed look OK to me, but these

Cr₂ (+6)   +   3 e^-       2 Cr (3+)

C (-1)       C (+1)   +   2 e^-

are not half reactions. You should have something like

Cr₂O₇^2- + H⁺ + e^-  Cr³⁺ + H₂O

(just balanced).

[John_Kimble]

Thanks again, Borek.