Heated Copper and Oxidation
Date: Summer 2013
I have been trying to understand the process of copper + heat = oxidation (the black/brown debris left on the metal - not the verdigris type). It is my understanding that copper contains a free radical. When soldering/brazing copper, brass or bronze, high heat is employed, which always results in oxidation. What is this oxidation composed of? What I, also, do not understand is what is happening when the copper is heated? Is the copper's free radical joining with the O2 in the air to create the oxidation? Or is the O2 depleting the copper of its free radicals? Or am I completely wrong about O2 and radicals? Thanks for your help. I want to explain what is happening, to my students, and myself.
Thanks for the questions. The black/brown debris on the copper surface (when heated) is most likely copper(I) oxide which has a formula of Cu2O. There may also be copper nitride, Cu3N, but it would be minor, if at all. I would not think of copper as being a free radical. I would simply think of copper as giving one electron to oxygen to form the Cu+ ion; the oxygen molecule O2 accepts 4 electrons to become two O2- ions. Yes, when copper is heated in soldering or brazing, oxidation occurs. For this reason, one uses a wire brush to remove the slag from brazed joint.
I hope this helps.
The copper oxidation you are referring to is Cupric Oxide, or CuO, and
is formed when oxygen in the air combines with copper atoms on the
surface of metallic copper. Each copper atom on the surface donates
two valence electrons to an oxygen atom, thus causing the oxygen
atom to bind to the copper atom. The resulting oxide layer is relatively
thin at normal temperature, and serves to protect the underlying
copper atoms from further corrosion. At normal temperatures, this
oxide layer looks like a slight darkening (or tarnishing) of the copper
surface. This protective oxide layer is called a "passivation layer"
because it makes the copper surface "passive", or non-reactive. There
are no "free radicals" present or involved.
As the copper is heated, oxidation becomes more aggressive, and an
increasingly thicker oxide layer forms that is non-adherent and easily
flakes off. In a nutshell, the formation of an oxide layer on copper (and
on most other metals) is a simple oxidation reaction.
As you already know, verdigris is different from copper(II) oxide. Verdigris is a mixture of compounds such as copper(II) acetate, copper(II) carbonate, and copper(II) chloride. The combination of substances give it the familiar green color. Copper(II) oxide, on the other hand, is black/brown. However, both result from oxidation-reduction reaction with the Cu being oxidized to Cu(2+) and something being reduced. In the case of copper(II) oxide the reaction can be written, in the simplest form, as: 2Cu + O2 --> 2CuO. Thus, the O2 is reduced to 2O(2-). Heat, then, is something that speeds up the reaction. Left alone, pristine, shiny, copper will eventually turn dull brown as the O2 in air reacts with it. Heating the copper just supplies more than enough energy to make the reaction proceed faster.
Since this is considered an oxidation-reduction reaction, with the copper being oxidized, then the mechanism must involve a transfer of 2 electrons from each copper (Cu --> Cu[2+]), or a total of 4 electrons from the 2 copper atoms that get oxidized with each reaction. These electrons then convert each oxygen atom to O(2-). I don't think that the mechanism for the reaction involves free radicals. This is simply a transfer of electrons the way oxidation-reductions go.
Greg (Roberto Gregorius)
Before getting into your question, there were several web sites (including previous NEWTON web site) below;
As shown in those sites, copper and oxygen produce copper (I) oxide (4 Cu + O2 â†’ 2 Cu2O) and copper (II) oxide (2 Cu + O2 â†’ 2 CuO) which are semiconductors. As these are oxidation processes, copper loses electrons rather than radicals.
Weonkyu Koh from Los Alamos
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