Band-edge positions of a semiconductor determine how \"energetic\" charge carrie

Question

Band-edge positions of a semiconductor determine how "energetic" charge carriers it offers for driving redox reactions upon photonic excitation. Therefore, different band-edge positions will have different potentials as to carrying out redox reactions. The following schematic shows arbitrarily chosen band-edge positions relative to normal hydrogen electrode (NHE) for arbitrarily chosen materials (A, B, C). Realizing that minimum potential (theoretical) required to split water is 1.23 eV, answer the following: a) Why do you think they chose normal hydrogen electrode as a reference? b) Name the materials which absorb in UV-range only. Give reasons. Name the materials which absorb in visible region only. Justify your arguments. Which material is most likely to reduce water? Why? Which material is most likely to oxidize water? Why? Is there any material in the given list that can split water to give H_2 and O_2 simultaneously (Note that this is called overall water splitting)? b) Discuss how you can modify the material that absorbs in UV range only so that it also absorbs in visible range too. People have engineered materials that meet the band-edge position requirements for overall water-splitting. However, they are not yet able to split water in commercial scales. What do you think the possible limiting factors are for such incompetent materials? Any idea to resolve those issues?

Explanation / Answer

a) A reference electrode is an electrode which has a stable and well-known electrode potential.So hydrogen has a well-known electrode potential which is also equals to zero and this makes comparison easy.

b) E = hc/

= wavelength (m)

h = Planck's constant = 4.135667516 x 10-15 eV*s

c = speed of light = 299792458 m/s

Part 1

Material B will absorb in UV reason as the energy is around 3.3 eV (~380 nm) which corresponds to UV region

Part 2

Material C will absorb in UV reason as the energy is around 1.6 eV (~780 nm) which corresponds to visible region.

Part 3

Material A and B can reduce water as the threshold potential for reduction of water is 1.23 eV and A and C both have potential equals to or more than this.

Part 4

Material C will oxidize water as the threshold potential for reduction of water is 1.23 eV and C Has potential less than this.

C) by adding more and more alkyle groups.

E = energy (eV)

= wavelength (m)

h = Planck's constant = 4.135667516 x 10-15 eV*s

c = speed of light = 299792458 m/s

Part 1

Material B will absorb in UV reason as the energy is around 3.3 eV (~380 nm) which corresponds to UV region

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