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Chemical reaction:

$$\ce{CH4(g) + 2O2(g) -> CO2(g) + 2H2O(g)} $$

The example in our textbook for this type of problem involves the following reaction:

$$\ce{C + 2S -> CS2} $$

Here, I can clearly see that $\ce{C}$ starts off with an oxidation state of zero and ends with an oxidation state of +4, meaning that it underwent oxidation.

In my given reaction, I see how the $\ce{C}$ in $\ce{CO2}$ has an oxidation state of +4, but I don't quite understand why it doesn't also have an oxidation state of +4 in $\ce{CH4}$. My assumption here is that, in the case of $\ce{CO_2}$, $\ce{C}$ has an oxidation state of +4 and $\ce{O2}$ has an oxidation state of -4; and in the case of $\ce{CH4}$, $\ce{C}$ has an oxidation state of +4 and $\ce{H4}$ has an oxidation state of -4. If that is incorrect, could someone explain why?

Edit: I just realized I overlooked a table that says that hydrogen has an oxidation state of +1, which would mean that in $\ce{CH_4}$, $\ce{C}$ would have an oxidation state of -4. I gather that this means that $\ce{C}$ loses 8 electrons in the reaction, which is why it is considered to undergo oxidation.

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  • $\begingroup$ How do you know that C is plus and O is minus in $\ce{CO2}$? $\endgroup$ – Ivan Neretin Oct 20 '15 at 18:08
  • $\begingroup$ A table in my textbook says that O has an oxidation state of -2, and that the sum of oxidation states in a neutral molecule is zero, which suggests that, in $CO_2$, C is +4 and each O is -2. $\endgroup$ – dt5591 Oct 20 '15 at 18:12
  • $\begingroup$ Does the table say that in respect to a specific compound, or in general? $\endgroup$ – Ivan Neretin Oct 20 '15 at 18:13
  • $\begingroup$ In general - it is a table showing oxidation states of non-metals (F: -1, H: +1, O: -2, Group 7a: -1, Group 6a: -2, Group 5a: -3 $\endgroup$ – dt5591 Oct 20 '15 at 18:17
  • $\begingroup$ Please use the \ce{...} syntax when referring to chemicals. Plain MathJax, i.e. $ without anything else results in italic type for compounds which is wrong and looks ugly. Also, MathJax does cool things such as automatically turning $\ce{CH4}$ or $\ce{H2O}$ into $\ce{CH4}$ and $\ce{H2O}$, respectively. For more information, check out this meta post. $\endgroup$ – Jan Oct 20 '15 at 20:21
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I believe I have answered my own question - I assumed that H had an oxidation state of -1, when it appears to have an oxidation state of +1.

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    $\begingroup$ See, it is not like H is always +1 in any compound. Normally people rely on electronegativities to decide which element is in positive oxidation state and which is negative. $\endgroup$ – Ivan Neretin Oct 20 '15 at 18:19

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