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Topic: Oxidation and reduction  (Read 4033 times)

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Offline fcb

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Oxidation and reduction
« on: March 27, 2010, 07:12:37 AM »
CH4 + 2O2 ---> CO2 + 2H2O

can someone help me and show me how they did it

thanks

Offline Schrödinger

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Re: Oxidation and reduction
« Reply #1 on: March 27, 2010, 12:20:35 PM »
How they did what - frame the equation or balance it?
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Offline fcb

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Re: Oxidation and reduction
« Reply #2 on: March 28, 2010, 08:36:17 AM »
How they did what - frame the equation or balance it?

I am sorry, I should have explained it better. (I was tired). I dont understand how you determine what is oxidized and what is reduced. I just dont understand the process in this particular case. I understand basic ones, Just not this one.

Offline Matias Ekstrand

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Re: Oxidation and reduction
« Reply #3 on: March 28, 2010, 08:52:15 AM »
Do you know how to determine oxidation states?

If so, begin by determining the oxidation states of the atoms on the left and right of the reaction.
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Offline Like_A_Whisper

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Re: Oxidation and reduction
« Reply #4 on: March 30, 2010, 11:30:53 AM »
Hmm, basically you need to study out the material for redox reactions. I still don't understand what it is you want but I think this is:

C-4H+14 + O02 ---> C+4O-22 + H+12O-2

So now we see what components changed their oxidation number on the left and the right side...
-We see that the C(carbon) has changed its oxidation number from -4 to +4...
-We see that the O(oxygen) has changed its oxidation number from 0 to -2...
-We see that the H(hydrogen) hasn't changed its oxidation number so we don't use it to solve this chemical equation.

Now we present the half reactions:
C-4 -8 electrons :rarrow: C+4      | 1 - Half reaction of oxidation(Because the component(carbon) gives away electrons)
                                    | 4 - This is the smallest number that contains both numbers( In our case 1 and 4 )
O20  +2 electrons :rarrow: 2O-2   | 4 - Half reaction of reduction(Because the component(oxygen) assumes electrons)
Now we use these numbers in the reaction...

CH4 + 2O2 = CO2 + 2H2O
So the oxygen is an oxidation device and during the reaction it is reducing itself(Assuming Electrons).
The Carbon is a reduction device and during this reaction it is oxidazing itself.(Giving away electrons)
I'm sorry, I'm not from USA or UK so I don't know the local term(phrase) for some of the words above. Anyways i think you understood me.
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