Dear colleagues:
When metal oxides and graphene oxides are reduced electrochemically, the more negative the reduction potential, the more black they get. In addition, what is the relationship between the black color and the increasing electron density, conductivity and increased sp3/sp2 ration provoked by the reduction process?
Best
Marco
Glad to answer your question. Oxygen vacancies can narrow the bandgap of the semiconducting oxides. see Nanoscale, 2020,12, 4030-4039. For the electron density and conductivity, the increased oxygen vacancies level increases the carrier density and improves the conductivity for an N-type semiconductor, and the effect is the opposite for a P-type semiconductor. Foundmentally, electrical conductivity and optical color are both determined by the complex dielectric function which can be tuned by the oxygen vacancy level.
Marco Antonio Uscanga Olea
You have linked many parameters to each other. I will try to simplify your question. When oxides are reduced, metals are formed and they are not black at all. Therefore, we will remove them.
Graphene oxide is reduced to graphene. The sp3 / sp2 ratio does not increase, but decreases. The electrical conductivity increases. If experimentally confirmed, i.e. there is a transition to the region of high frequencies, then from Planck's formula we obtain a transition to Wien's formula. We mean the law of distribution of energy in the spectrum of absolutely black body. In this area, atomic oscillators study energy not continuously, but in certain portions - quanta, and the quantum energy is proportional to the oscillation frequency. From here you can explain the behavior of the other parameters.
The black color is due to the interaction of light with the free electrons of that are released upon oxygen removal. These free electron might also increse the electrical and thermal conductivity.
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