Prelaboratory Questions And Problems

*1. The possibility of using a semiconductor to cause a photocatalytic reaction is governed by the location of the valence and conduction bands (VB and CB, respectively) relative to the location of the desired reactions. For example, if one desires to split water into hydrogen plus oxygen, one needs to pay attention to the corresponding standard potential for the reduction of water into hydrogen (0.0 V vs the Standard Hydrogen Electrode, SHE) and that for the oxidation of water into oxygen (1.23 V). The electrons to be produced at the semiconductor must have a more negative standard potential than that required for the reduction of water, and the holes must be more positive so as to oxidize water. Assuming that the location of the bands in the selected semiconductor remains unchanged upon illumination, select the option that contains a combination of values suitable for water splitting. The locations of the VB and CB in a potential axis are given in volts, respectively.

2. Is photocatalysis en electrochemical phenomenon? Explain.

*3. In a semiconductor-based photosynthetic system, light drives a non-spontaneous reaction. This is called a thermodynamically up-hill reaction. On the other hand, if the reaction is thermodynamically favored (i.e., thermodynamically down-hill), light simply serves to overcome the activation energy and the reaction is said to be photocatalytic.

There are cases where the reaction that is complementary to the target reaction defines the energetics of the overall process (i.e., makes it photosynthetic or photocatalytic). For example, in the recovery of metal ions M"+ from a spent solution by irradiation of a semiconductor (e.g., Ti02), as in the present experiment, there are two possible cases: Case a:

Case b:

These two cases are exemplified below for M = Cu (see Rajeshwar, 1995 and Reiche, 1979):

+ C2H6 + 2C02 Calculate AG0 for each process (per mole of Cu) and label them as photosynthetic or photocatalytic. Note: Take

Ecu^/cu = -°-34 v' ^(h/H2o = 1.23 V, and C02/CH3C00- = —0.40 V.

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