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 + O
02 ---> C
+4O
-22 + H
+12O
-2So 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
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 ) O
20 +2 electrons
2O
-2 | 4 -
Half reaction of reduction(Because the component(oxygen) assumes electrons)Now we use these numbers in the reaction...
CH4 + 2O
2 = CO
2 + 2H
2O
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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