# Why does potential chemical energy decrease with increasing oxidation of carbon?

I saw an interesting spectrum in my biology class the other day. On the left end was methane - the least oxidized form of carbon - and the most energetic form of carbon (at least relative to what else was on the slide). On the right end was carbon dioxide - the most oxidized form of carbon - and the form with the lowest energy.

Now, I do agree with the slide - we can combust methane but we can't combust carbon dioxide - at least I don't think we can.

However, is there an intuitive explanation of why increasing oxidation state correlates with lower potential energy, at least for carbon?

Does increasing oxidation state correlate with lower potential energy for all elements?

Why does it seem that oxygen forms rather strong bonds with other elements in many cases?

From a very basic explanation from general chemistry:

• Increasing your oxidation state increases the charge. So that means that you move to the left on the periodic table increasing your atom radius and decreasing your ionization energy and electronegativity.
• If your IE decreases and you become more electronegative, your atom is more attractive for chemical reactions so it has a lower potential energy.

There is a less handwaving explanation from inorganic chemistry, but that is a little harder to understand and I don't want to formulate a MO diagram.

is there an intuitive explanation of why increasing oxidation state correlates with lower potential energy, at least for carbon?

The heat of formation of methane, carbon dioxide and carbon tetrafluoride are -74.9, -393.1 and -927.2 kJ/mol respectively (the trend remains unchanged even if you divide by MW and compare per unit mass). As we shift electron density from electropositive elements to electronegative elements we stabilize electrons, their energy is lowered.

Does increasing oxidation state correlate with lower potential energy for all elements?

I haven't checked extensively, but the trend seems to hold up. I'd expect it to hold up.

$\ce{BeH_2}$, $\ce{BeF_2}$ 127.2, -796.7 kJ/mol

$\ce{CaO}$, $\ce{CaF_2~~}$ 39.1, -789.1 kJ/mol

$\ce{NaH}$, $\ce{NaF~~}$ 126.2, -288.0 kJ/mol

Why does it seem that oxygen forms rather strong bonds with other elements in many cases?

I'm not sure what you mean by "strong bonds". The C-O bond in carbon dioxide is much more reactive than the C-H bond in methane. Perhaps you're alluding to molecular stability which takes us back to the top, e.g. compounds with polar bonds typically have stabilized electrons, electrons in lower energy orbitals, as discussed above.

• "Perhaps you're alluding to molecular stability" - I was - and I'll avoid the use of fluffy terms such as "strong bonds" or "stability" in the future! Insightful answer as usual! @ron Sep 22 '14 at 13:25

This scenario has a very simple explanation. The reduction of any electron to a more electronegative atom will release energy. To understand this, imagine a highly reduced methane molecule; the C-H bonds are unstable and the electrons in the H atoms is itching to escape the methane molecule to pair with an oxygen. In this example the reduction of oxygen by the hydrogen releases energy thereby lowering the reducing agent's( methane) free energy. Oxidation usually will result in a lower free energy state, due to the energy being released in the reaction.

TL DR; High energy reducing agent ---- donates high energy electron to oxidizing agent. Reaction releases energy which the reducing agent supplied.

oxidizing substance becomes more stable b/c it lost a high energy electron. Reducing substance becomes more stable b/c it attained an electron for an electronegative atom and the surplus energy is released.