Why are metal oxides generally more stable than their corresponding elemental forms?

Badges: 7
Report Thread starter 2 years ago
Why is the oxidation of certain metals such as iron favorable from a thermodynamic standpoint? What is it about these metal oxides that makes them inherently more thermodynamically stable than their starting, elemental forms? The only reason I can imagine is that iron, being a d-block element, has a low electronegativity and cannot stabilize valence electron density as well as oxygen, the second most electronegative element on the table.

Also, is oxidation always thermodynamically favorable?
Kian Stevens
Badges: 16
Report 2 years ago
Generally speaking, any exothermic process is thermodynamically favourable.
For many compounds - including the d-block oxides - their standard formation enthalpies are negative. These are of course exothermic processes.
Again, this does not have to be the case for just metal oxides. Many, many metal based compounds, i.e. metal halides, metal carbonates, have exothermic formation values.

Why are exothermic processes favourable? Taking individual elements, which may unstable by themselves, and forming a bond between them is favourable. This is because the elements' valence electrons reduce in energy to lower energy levels during bonding, and energy is released in the form of electromagnetic radiation. Hence, the more exothermic the formation of a compound is, the more stable the formed compound is compared to the starting elements.
The energy levels I mentioned are dependent on the distance between the valence electrons and the nucleus; electrons in larger radial orbits have larger potential energies, which in this context isn't very stable.

As for your last question: no, oxidation isn't always favourable.
The problem with redox reactions is that they have two processes: the reduction of one species and the oxidation of another, or vice versa.
These two processes carry half equations with them, and you can't just measure the thermodynamics of one half equation. It has to be of both, where they are compared to a reference value: the standard hydrogen electrode. This comparison will give an E_{cell} value, in which only positive values are favourable. It goes that E_{cell} = E_{cathode} - E_{anode}, where E_{cathode} is the reduction half equation and E_{anode} is the oxidation one. The reason why only positive values are favourable is because electrode potential and \Delta G can be linked: \Delta G = -zFE_{cell}. \Delta G must be negative for a reaction to be feasible, and so only positive values for E_{cell} will satisfy this.
Using the above equation for E_{cell}, it is obvious that oxidation is not always a favourable process. It will only be favourable if the E_{anode} value is more negative than the E_{cathode} value, as this will result in a positive value for E_{cell}. But... what if E_{anode} is less negative? This would actually result in a negative E_{cell} value, which would imply oxidation isn't a favourable process in that case.
Last edited by Kian Stevens; 2 years ago

Quick Reply

Attached files
Write a reply...
new posts
to top
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.


What support do you need with your UCAS application?

I need help researching unis (25)
I need help researching courses (12)
I need help with filling out the application form (9)
I need help with my personal statement (79)
I need help with understanding how to make my application stand out (44)
I need help with something else (let us know in the thread!) (3)
I'm feeling confident about my application and don't need any help at the moment (15)

Watched Threads

View All