I believe that Cr on its own (as in chrome coating) oxidizes as Cr(III) when exposed to air, and similarly I think that it is the same for Cr as a component of stainless steel.

Is it always the case that an alloy containing chromium will oxidize as Cr(III) when exposed to air at room temperature, or are there alloys which might form Cr(VI) instead?

The situation which brings this question to mind is the case of a computer hard disk platter. In the case of a "head crash" of the hard disk, layers of the platter are ground to dust. The layers as I understand are composed of various things: Some kind of carbon polymer as an outer covering, magnetic layers of things like Co-Pt-Cr or Co-Ni-Cr (I found a source which suggests "traces of boron or tantalum" are also added to this layer), and also layers composed of things like Cr-Ti or Cr-Ni. I'm wondering whether any of these layers would cause hexavalent chromium to form.

  • $\begingroup$ Can Cr (VI) be formed? Certainly! But a few Cr(VI) ions probably would not be significant, or even measurable. $\endgroup$ – DrMoishe Pippik Jan 3 '19 at 22:55
  • $\begingroup$ Ah, so even though generally Cr(III) is formed in these cases, sometimes Cr(VI) ions still formed at room temperature in Cr and its alloys (this is the case even with solid Cr or stainless steel)? Does this mean that other oxidation states of Cr could be formed as well, and it is just that Cr(III) is only formed almost exclusively? $\endgroup$ – chemistry_is_difficult Jan 3 '19 at 23:35
  • $\begingroup$ There are ~6*10^23 Cr atoms in a mole, ~52 g. Thermodynamics is based on statistics, which gives the most likely outcome. However, individual Cr atoms and O2 molecules may have kinetic energy far beyond the average and might form unusual oxidation states. Your question asks "Is it always the case that an alloy containing chromium will oxidize as Cr(III)...", so of course there might be a few exceptional cases. $\endgroup$ – DrMoishe Pippik Jan 4 '19 at 0:50
  • $\begingroup$ Thanks for the clarification! I suppose that since the presence of Cr in an alloy does not change its kinetic energy, Cr will tend to oxidize in air as Cr(III) regardless of the alloy? (I hope "tend to" is the correct terminology in light of the information you've given me). I guess I am hampered here in my question because I'm not sure though why Cr tends to oxidize as Cr(III) rather than as Cr(VI) in air to begin with, I suppose because for Cr to reach the +6 oxidation state using only air requires that the atoms be unusually charged? $\endgroup$ – chemistry_is_difficult Jan 4 '19 at 1:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.