Decomposition energy in Pourbaix diagram

#1

Hi all,

I noticed that the unit of the decomposition energy in Pourbaix diagram have been updated to eV/atom. This actually changed the stability of entries compared with the results derived from eV/(normalized formula unit). However, the energy of the PourbaixEntry is normalized by number of non H or O atoms. I wonder if the decomposition energies deduced from this way are consistent within entries have different H/O compositions.

Also, I have another question about the corrected aqueous compound energy of H2O, which is -4.972 eV in MP. Is this value derived from E_H2O + T*S_H2O ? where E_H2O is the total energy of H2O calculated by DFT, T = 298 K and S_H2O is the entropy of H2O(l) at 298 K. (The data from Persson et al., DOI: 10.1103/PhysRevB.85.235438 seems to be incorrect).

Thanks.

#2

Hi xig,

Youve noticed correctly, and this change was made to have pourbaix stability be more in line with what’s typically reported from the compositional phase diagram. The reason it works is because the decomposition energy is normalized consistently to the number or atoms associated with the decomposing entry, which does not change over the pourbaix domain. The number of atoms associated with the decomposition product does change, but implicit in the decomposition energy is any residual H2O, H+, and e- required to form the decomposition product.

The value of the free energy of water is equivalent to what you’ve mentioned. It can also be derived from the equilibrium potential of water oxidation (1.23 V * 4e-). Please follow up if I can clarify further.

#3

Thanks for your reply.