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I have the following reaction:

$$\ce{CH_4 (g) + 2O_2 (g) \rightarrow CO_2 (g) + 2H_2 O(l)}$$

How do I determine how many electrons are transferred in this reaction? More specifically, I need to know the value of $v_e$ which I am supposed to plug into the Nernst equation, which my textbook explains as the coefficient of the number of electrons transferred in each half-reaction. How does one go about splitting this into two half-reactions?

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  • $\begingroup$ You can post your "I figured it out" part as an answer to this question, you know :) $\endgroup$ – ManishEarth Apr 22 '13 at 8:07
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    $\begingroup$ Agreed. You should definitely post your answer. $\endgroup$ – Ben Norris Apr 22 '13 at 11:08
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The carbon atom must have an oxidation state (O.S.) of -4 on the left hand side since each hydrogen bonded to it must have an O.S. of +1. On the right hand side, the carbon atom has an O.S. of +4 since each oxygen bonded to it has an O.S. of -2. The carbon's oxidation state increased by 8. This is a hint that 8 electrons were transferred. To verify, examine the oxygen atoms. On the left, we only have free oxygen, therefore the O.S. is 0. On the right, we have 4 total oxygen atoms, each with an O.S. of -2. Thus our net change in oxidation state is 0 - 8 = -8. We can almost conclude that all these electrons were transferred from carbon, but we must check the oxidation states of the hydrogen atoms. On both sides, each H atom has an O.S. of +1, so there is no net O.S. change. Thus we can confirm that 8 electrons were indeed transferred in this reaction.

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