Theoretical Study Of Oxygen Evolution Reaction On Ni-Fe/Co Binary LDH

In this work,First principles periodic Density functional theory(DFT)calculations were used to investigate the electrochemical oxygen evolution reaction(OER)on NiFe/Co layered double hydroxide.The Gibbs free energy of each elementary reaction was calculated,afterwards the Potential Determination Step(PDS)of different metals was deduced and their catalytic activity were compared.And further explores the performance comparison of NiFe-LDH at different ratios in the presence of defects.The following conclusions were obtained:(1)For the NiFe-LDH with a ratio of 2:1,the catalytic activity of(001)surface and(010)surface are discussed.(001)surface takes O atoms to do the reaction site.(010)surface has two kinds of cutting methods,we can choose Fe atoms or Ni atoms to do the reaction site.The overpotential of Potential Determination Step(PDS)is sorted by superior to inferior,Fe atoms(0.378eV)<Ni atoms(0.469eV)<O atoms(0.946eV),indicating that metal atoms ofNiFe-LDH(010)surface as catalytic sites,the reaction is more likely to achieve,and the Fe sites of the catalytic activity better.(2)In this paper,the overpotential of 2:1 ratio Ni-Fe/Co LDH were studied,and combined with the catalytic performance of CoFe-LDH,find that the order of OER catalytic activity in the(010)surface is NiFe-LDH>CoFe-LDH>NiCo-LDH.Fe sites(0.378 eV)has a better catalytic activity on NiFe-LDH(010)surface,at the same time Fe sites(0.580eV)has a better catalytic activity on CoFe-LDH(010),and Ni site(0.670 eV)has a better catalytic activity on NiCo-LDH(010).(3)In the case of NiFe-LDH(001)surface with a ratio of 2:1,the catalytic activity of the metal as the reaction sites are better than that the O sites when the OH defect was present.When the ratio of Ni to Fe is 2:1,for the NiFe-LDH(010)surface with a metal exposure,for the perfect surface and the surface with H or OH defects,the catalytic activity sorted by superior to inferior,(010)surface with OH defects>no defective(010)surface>(010)surface with H defects.Compared with NiFe-LDH in different ratios,we know that when the presence of OH defects in the surface,the surface with metal exposed is better than the(001)surface with OH exposed.On the metal exposed surface,the best performance is the ratio of Ni to Fe 2:1,followed by 3:1,and last is 1:1.But on the OH exposed(001)surface,the best performance is the ratio of Ni to Fe 2:1,followed by 1:1,the last is 3:1.This study can be combined with the experiment to help understand the oxygen evolution reaction,and has a certain reference meaning for the discussion of OER reaction mechanism.At the same time the calculation method can continue to be used in OER calculation on other LDH or metal oxide,metal hydroxide.