Corossion of cell metals (electrochemistry)

Question

Write the equations of corrosion in a neutral medium and in a medium of HI.

Answer 1

Since you specified in the comments that this would be the corrosion of an electrode that is a mixture of cobalt and aluminum, we can go through the corrosion of the electrode under the conditions you have specified.

Corrosion of electrodes via a rusting process occurs in two fundamental steps: oxidation of the metal (removal of electrons), and dissolution of the resulting metal oxide (which usually happens very slowly in the absence of an acid).

To determine the stoichiometry of the metal oxides, we would look at their positions in the periodic table of elements and determine the easiest way for them to lose electrons to reach one of the stable states, which are listed in order of most to least stable: no electrons aside from core electrons; any non-core electrons are completely filling the d orbitals and no-where else; any non-core electrons are filling the d orbitals with exactly 1 electron in each orbital; any non-core orbitals are filling the d-orbital with exactly 1 electron per orbital and a full s orbital.

In the case of aluminum, there are only two stable electronic states, the Al metal and Al³⁺, which has only core electrons. In the case of cobalt, the first stable state that one reaches when oxidizing it is Co²⁺, where the s orbital is filled and there is 1 electron per orbital for every d orbital. In standard conditions on Earth, the major oxidizing agent is oxygen gas, O₂, which means that the equation for the oxidation step of the corrosion process is:

Rxn 1. 2Co(s) + O₂(g) --> 2CoO(s)

Rxn 2. 4Al(s) + 3O₂(g) --> 2Al₂O₃(s)

In a neutral medium, such as pure water, these oxides will take a very long time to dissolve--so much so that they are listed as insoluble in water when you look them up in a solubility table. For this reason, they form a pretty effective barrier between the pristine metal underneath and the solution touching it and therefore protect the metal underneath. However, in the presence of a strong acid, the oxide ions (O2-) that are bound to the metals are protonated to form water, which will then join the rest of the water and force the metal ions into the solution. The metal underneath is then exposed to the acid itself, which may be strong enough to directly oxidize the metal without the need to wait for more oxygen to dissolve and arrive at the metal surface. At that point, which depends upon the strength of the acid as well as its concentration, the H+ in the acid would use the electrons from the metal to become hydrogen gas. Additionally, if the conjugate base of the acid is a good ligand, then it will bind to the metal and likely help to keep the metal dissolved. Iodide (I-) is a good ligand for transition metal ions with small charges. For simplicity sake, I am writing the iodide as a spectator ion here. For dissolution of the surface metal oxide we have:

Rxn 3. CoO(s) + 2HI(aq) --> Co²⁺(aq) + H₂O(l) + 2I^-(aq)

Rxn 4. Al₂O₃(s) + 6HI(aq) --> 2Al³⁺(aq) + 3H₂O(l) + 6I^-(aq)

For the oxidation of the pristine metal directly by the acid, we have:

Rxn 5. Co(s) + 2HI(aq) --> Co²⁺(aq) + H₂(g) + 2I^-(aq)

Rxn 6. 2Al(s) + 6HI(aq) --> 2Al³⁺(aq) + 3H₂(g) + 6I^-(aq)

We can obtain the full reactions for the corrosion of the surface of this electrode that starts out dry and is then placed into the aqueous hydroiodic acid by combining the first step (reactions 1 and 2) with the second step--that is, by combining reaction 1 with reaction 3, and reaction 2 with reaction 4. This gives us the reaction equations:

Rxn 7. 2Co(s) + O₂(g) + 4HI(aq) --> 2Co²⁺(aq) + 2H₂O(l) + 4I^-(aq)

Rxn 8. 4Al(s) + 3O₂(g) + 12HI(aq) --> 4Al³⁺(aq) + 6H₂O(l) + 12I^-(aq)

Once reactions 7 and 8 are complete, reactions 5 and 6 become dominant. This is why, when placing metals made in strong acids appear to do nothing at first before eventually generating bubbles.