In this section, we consider a complete photoelectrochemical (PEC) cell such as that shown in Figure 15.21. The cell consists of a photoelectrode and a metal counter electrode that are immersed in a common electrolyte. For example purposes, we assume that the photoelectrode is an n-type semiconductor, which acts as the anode in the cell. The electrolyte contains both the oxidized and reduced forms of a redox couple whose equilibrium potential lies in between the conduction and valence band edges of the semiconductor (i.e., in the band gap).
Figure 15.21 Schematic diagram of a photoelectrochemical cell.
During operation of the cell, electron–hole pairs are generated by illumination of the photoelectrode. The electron–hole pairs are separated due to the electric field in the space-charge region. The holes that are generated migrate to the surface of the semiconductor and participate in oxidation of the reduced form of the redox couple. The excited electrons move through the external circuit, including the load, to the counter electrode, where reduction of the oxidant species takes place. The doping level of the semiconductor influences the width of and voltage gradient in the space-charge layer. It also determines the Fermi level of the majority carriers (electrons for an n-type material), which is close to the conduction band for an n-type semiconductor. In this example with a single redox couple, there is no net reaction and no change in the electrolyte composition. The product of the PEC cell is therefore electrical power due to the work performed by the excited electrons as they move through the external circuit.
Illustration 15.7 provides a quantitative example of how such a PEC cell with an n-type photoanode works. Included in the calculations are surface overpotentials, ohmic losses in solution, and losses in the external circuit containing the load. The calculations are performed for a specified photocurrent. Alternatively, the methods described above can be used to estimate the photocurrent if sufficient information is available. Ohmic losses due to current flow in the semiconductor itself and mass transport losses in solution were not considered in these calculations.
ILLUSTRATION 15.7
In this example, we examine the power that is available from a photoelectrochemical cell consisting of an n-type semiconductor electrode and an inert counter electrode. In the presence of light, the semiconductor electrode serves as the anode and the counter electrode as the cathode. Energetic electrons from the anode pass through the load and then drive the cathodic reaction. We will consider ohmic losses in solution and through the load, as well as the surface overpotentials associated with the two reactions. The surface overpotentials and ohmic losses in solution all reduce the power that can be applied to the load. The current–voltage behavior of an n-type semiconductor electrode operating in the dark is described by the following expression:
is 1 × 10−6 A·cm−2. The photocurrent can be assumed to be constant and is equal to 25 mA·cm−2. The aqueous electrolyte contains 0.1 M Fe2+, 0.1 M Fe3+ in H2SO4. The conductivity of the electrolyte is approximately 0.06 S·cm−1. The counter electrode is Pt, and the reversible iron reduction reaction on this electrode is characterized by an exchange current density of 0.005 A·cm−2 and a transfer coefficient of 0.5. The redox potential for the Fe2+/Fe3+ electrode is 0.526 V with respect to a SCE reference electrode in this solution, and the equilibrium potential of the semiconductor electrode in the dark (versus SCE) can be assumed to have the same value. Mass-transfer effects may be neglected. Assume a load of 1 Ω. The distance between the two electrodes is 2 mm. The area of both the anode and cathode is 10 cm2. Assume that only one side of each electrode is exposed to the electrolyte and that the current density is uniform. Please determine the following: (i) the voltage drop across the load, (ii) the cell current, (iii) the power available from the cell, (iv) the ohmic drop in solution, and (v) the surface overpotential associated with the counter electrode.
SOLUTION:
Strategy: Write the appropriate expressions for the current in terms of the potential and set them equal to each other in order to solve for the values of potential in the cell. Once the potentials are known, the current is readily calculated. We will write expressions for each of the two electrodes, the ohmic drop in solution, and the voltage drop through the load.
Semiconductor electrode: As shown in Equation 15.21, the expression for an illuminated n-type semiconductor is
where , the photocurrent, has been added to the expression for the current in the dark. For a PEC cell with an n-type electrode, the photocurrent is positive or anodic. Hence, iph is positive and equal to the oxidation that takes place as a result of holes created by light absorption. Neglecting resistive losses in the solid semiconductor electrode itself, ηanode = (ϕsemi − ϕs,a) −(ϕsemi − ϕs,a)eq where the solution potentials are referred to SCE and (ϕsemi − ϕs,a)eq = U = 0.526 V (the equilibrium potential of the semiconductor in solution in the dark). We arbitrarily set the absolute value of the potential of the semiconductor electrode ϕsemi = 0 as the reference point for the entire calculation (arbitrary), similar to what we did previously for a metal electrode. With this specification, everything in the above expression is known except the current density and the value of the potential in the solution at the anode.
Unknowns: i and ϕs,a
Ohmic drop in solution: Assuming a 1D system,
Note that current flow is positive from the anode to the cathode, consistent with what we have done previously.
Unknowns: i and ϕs,a, ϕs,c.
Load: Electrons flow from the anode to the cathode through the load, with current flow in the opposite direction. Therefore,
where iA I = the total current that flows in the external circuit between the electrodes, and RΩ is the resistance between the electrodes (i.e., the load) in ohms (specified as 1 Ω for this problem). is the potential of the metal at the cathode. Note that , as specified above. The equation is just Ohm’s law through the external circuit (V = IRΩ).
Unknowns: i and
Cathode:
where ηcath (ϕm − ϕs,c) − (ϕm − ϕs,c)eq and (ϕm − ϕs,c)eq = Veq,c = 0.526 V. This is just a BV expression for the reduction of ferric ion to ferrous ion at the Pt cathode. Note that since cathodic current is defined as negative,
Unknowns: i, ϕm and ϕs,c.
Numerical Solution: The above expressions yield four equations and four unknowns: i, ϕs,a, ϕs,c, and ϕm.
We can eliminate i by setting the relationships equal to each other. This process gives us three equations for the three unknowns: ϕs,a, ϕs,c, and ϕm. This is similar to what we did in Section 3.4 for a Zn-Ni cell.
Simultaneous solution of these equations yields
We can now answer the above questions:
The voltage drop across the load, which is the voltage of the PEC cell, = ϕm − ϕsemi = 0.137 V, since ϕsemi = 0.
The cell current can be calculated from any of the equations for the current. The simplest of these is probably the one for the potential drop across the load:
The power (IV) available from the cell is simply iAϕm = 0.0188 W.
The ohmic drop in solution is .
Maximum Power
The power available from a PEC varies with the potential of the cell (terminal to terminal) and the current. At one extreme, open circuit, the voltage difference is at its maximum value. However, the current and, therefore, the power output both are zero at open circuit. At the other extreme, the current is at its maximum when the cell is shorted, but the voltage difference between the terminals is zero, which again yields a power output of zero. The point on the i–V curve for the PEC cell where the product of i times V has the greatest value is the maximum power available from the cell. The i–V curve and the corresponding power curve for Illustration 15.7 both are shown in Figure 15.22. For the conditions given, the maximum power corresponds to a cell voltage of 0.13 V and a cell current density of 14.4 mA·cm−2. This value depends on the potential losses in the cell (e.g., ohmic and kinetic losses), as well as on the magnitude of the external load.
Figure 15.22 i–V curve, and on right ordinate, power versus cell potential for the cell from Illustration 15.7.
Cell Efficiency
The energy conversion efficiency is expressed by
(15.26)
Another measure of performance used in the solar community is called the fill factor, which is defined as follows:
(15.27)
The denominator is the product of the open-circuit voltage, which is the maximum voltage, and the short-circuit current, which is the maximum current; it represents a hypothetical power that is used as a convenient reference point.
How can we optimize efficiency? Part of this process is by now very familiar. We know that by reducing ohmic losses in the cell, for example, the maximum power (iV)max is increased, and this operating point occurs at a higher current density. For a photochemical cell, there are additional important aspects. A detailed analysis of efficiency is beyond our scope; but let’s briefly consider a few points. First, it is critical to carefully align the load requirements with the conditions under which the power is maximized. As seen in Figure 15.21, the power can drop off quickly as conditions move away from the optimum. Depending on the application, this alignment will likely be achieved through a combination of cells in series and parallel. Also, the need to maximize light capture is frequently at odds with the reduction of electrochemical losses in the cell. For example, a clear pathway to light may require a circuitous path for current between the electrodes, increasing ohmic losses. Hence, the compact, bipolar stacks optimal from an electrochemical perspective stand in contrast to a structure that allows a clear path for radiation to enter and be absorbed. Additionally, the catalysts needed to reduce kinetic overpotential or the metal connections needed to reduce resistance losses may partially block access to light. Some incident radiation may be reflected or absorbed in the electrolyte, representing additional losses. The semiconductor must be thick enough to capture the light. On the other hand, a thicker electrode is more resistive and will increase ohmic losses. The magnitude of the saturation current is influenced by the difference between the potential of the redox reaction and that of the relevant valence or conduction band, as well as by the doping level of the semiconductor. Clearly, the design of PEC cells extends well beyond the factors typically considered in electrochemical cell design.
Types of PEC Cells
The type of PEC cell shown in Figure 15.21 is a regenerative cell, the product of which is electrical power. The cell is called a regenerative cell because the redox couple that reacts at the photoelectrode is regenerated at the counter electrode as illustrated in Figure 15.23a. No net chemical change takes place in this type of cell. In contrast, a PEC can also be used to generate chemical products (e.g., solar fuels) in a photoelectrolytic cell such as that shown in Figure 15.23b, which involves two separate redox reactions at different energies (potentials). Because they convert light into chemical energy, or fuels, photoelectrolytic cells are often referred to as “artificial photosynthesis.” Note that a membrane is used to keep the desired chemical products from reacting at the other electrode.
Figure 15.23 Energy diagrams for (a) a regenerative and (b) a photoelectrolytic cell. Source: Adapted from Grätzel 2001.
The design requirements for an efficient photoelectrolytic cell are more stringent than those required for a regenerative cell. In both cases, the band gap must be suitable for light absorption. In addition, photoelectrolytic cells require alignment of both desired reactions with the appropriate band-edge energies. For example, for water to be oxidized at the photoanode, the valence band energy must be less than the energy that corresponds to the oxygen evolution reaction. On the SHE scale, this would require a potential greater than 1.229 V. The electron energy at the counter electrode must be greater than that of the hydrogen evolution reaction in order for hydrogen to be produced in the configuration shown in Figure 15.23. This means that the potential of the counter electrode would need to be more negative than the equilibrium potential of the hydrogen reaction. To handle both reactions, the band gap would need to be greater than 1.229 V (the difference between the reversible potentials for the O2 and H2 evolution reactions) and the band edges would need to be appropriately aligned to accommodate both reactions. The pH is also a consideration since the equilibrium potentials of both the O2 and H2 reactions, as well as the band-edge positions of the photoelectrodes, relative to a reference electrode in solution all change with pH.
Satisfaction of the band gap and alignment constraints is, unfortunately, not sufficient. For example, CdS meets the band gap and alignment requirements, but is not suitable for use because it is not stable and undergoes photocorrosion. In addition, the hydrogen evolution reaction is very slow on CdS. Other factors that affect photoelectrode performance include the diffusion length for majority and minority carriers, carrier mobility, and rates of recombination. The use of cocatalysts to promote a desired electrochemical reaction (e.g., hydrogen or oxygen evolution) is an important strategy used to enhance performance. Protective layers are also being explored to enhance stability. Nanostructured materials show promise for improved performance and for the creation of new structures and composite materials with superior properties.
An alternative approach to the use of a single photoelectrode is to place two photoelectrodes in tandem as illustrated in Figure 15.23. This strategy allows the alignment and band gap constraints to be more easily met and provides greater flexibility for system optimization. For example, in a cell designed to generate oxygen and hydrogen, the valence band of the photoanode must be aligned so that oxygen evolution can take place, which requires a low electron energy as illustrated. This enables electrons from the water move to the semiconductor, creating oxygen gas as the product at the n-type photoanode. The low electron energy corresponds to a relatively high potential on the standard SHE scale. In the tandem configuration, the size of the band gap for the photoanode material is not constrained by the need to also accommodate the hydrogen reaction, and can be optimized for the system. On the other side of the cell, the electron energy of the conduction band of the photocathode must be high enough to reduce protons and form hydrogen gas (relatively low potential versus SHE). The band gap of this material is also not constrained. However, the energy of the conduction band of the photoanode must be greater than that of the valence band of the photocathode in order to transfer electrons between those two bands as shown. These concepts are shown in Illustration 15.8, which gives an example of a two photoelectrode system. As already noted, the strategy shown in Figure 15.24 involves the transfer of electrons from the photoanode to the photocathode. This electron transfer can be eliminated through the use of a redox mediator that can be reduced at the photoanode and oxidized at the photocathode, permitting even greater flexibility for the design of practical systems. The interested reader is directed to the references at the end of this chapter for more information.
ILLUSTRATION 15.8
Cu2O and BiVO4 can be coupled in a tandem PEC cell with two semiconductor–liquid junctions to produce hydrogen and oxygen, where the two semiconductors are wired to complete the connection as illustrated in Figure 15.23. Given the data in Figure 15.9, which of the two semiconductors should be n-type and which should be p-type? A ruthenium catalyst is used for the hydrogen evolution reaction (HER) and a cobalt catalyst for the oxygen evolution reaction (OER). With which electrode is each catalyst used? (Bornoz et al., J. Phys. Chem. C, 118, 16959 (2014) provide further details.)
SOLUTION:
From Figure 15.9, we approximate the following (hydrogen scale numbers calculated from the others):
Semiconductor Eg Ec Ev Ec H Ev H
Cu2O 2.2 −3.5 −5.7 −0.94 1.26
BiVO4 2.4 −4.5 −6.9 0.06 2.46
From these energy data, Cu2O better aligns with the hydrogen evolution reaction (cathodic), and BiVO4 better aligns with the oxygen evolution reaction (anodic). This can be seen by considering each reaction separately. With BiVO4, the valence band energy is quite low, permitting oxygen evolution. The corresponding voltage on the SHE scale is 2.46 V, well above the equilibrium potential for oxygen. The highest electron energy corresponds to the conduction band of Cu2O. This corresponds to a potential on the SHE scale that is lower than the equilibrium potential for hydrogen and would therefore result in hydrogen evolution, the desired cathodic reaction. In both cases, the equilibrium potential (energy) for the preferred reaction lies near the middle of the band gap. Cu2O is the photocathode and should be p-type. Similarly, BiVO4 is the photoanode and should be n-type. The hydrogen catalyst should be integrated with the photocathode and the oxygen catalyst with the photoanode. Finally, the energy of the conduction band for BiVO4 is higher than that of the valence band of Cu2O, enabling the electron transfer needed to complete the circuit.
Figure 15.24 Photoelectrolytic cell with two photoelectrodes.
Closure
We introduced the basic physics of semiconductors and described the interface that develops when they are put into contact with an electrolyte. Current flow under dark conditions has also been described. The most important feature of the semiconductor electrolyte system is that it is photoactive. Thus, light energy can be converted to electrical energy or to drive electrochemical synthesis. A key aspect of this chapter is the combination of light and electrochemistry to explore photoelectrochemistry.
Further Reading
Archer, M. D. and Nozik, A. J., eds. (2008) Nanostructured and Photoelectrochemical Systems for Solar Photon Conversion, Imperial College Press.
Chen, X., Shen, S., Guo, L., and Mao, S.S. (2010) Semiconductor-based photocatalytic hydrogen generation. Chem. Rev., 110, 6503–6570.
Memming, R. Semiconductor electrochemistry. Available at http://onlinelibrary.wiley.com/book/10.1002/9783527613069
Smith, W.A. Sharp, I.D. Strandwitz, N.C. and Bisquert, J. (2015) Interfacial band-edge energetics for solar fuels production. Energy Environ. Sci. 8, 2851.
Tan, M. Laibinis, P.E. Nguyen, S.T. Kesselman, J.M. Stanton, C.E., and Lewis, N. (1994) Principles & Applications of Semiconductor Photoelectrochemistry, Prog. Inorg. Chem., 41, 20–144.
Problems
15.1. Sketch the interface between a metal and an electrolyte. Assume that the metal electrode has a small positive charge. Identify the inner and outer Helmholtz plane and the diffuse layer. What determines the thickness of the diffuse layer? Make a similar sketch for the semiconductor–electrolyte interface for an n-type semiconductor and identify the depletion region and the Helmholtz plane. What determines the thickness of the depletion region?
15.2. Thermal energy is represented by kT. Calculate the value of kT at room temperature and express the energy in terms of eV. Compare the value that you calculated with the band gap for Si. Explain why the intrinsic conductivity of semiconductors is low under normal conditions.
15.3. Determine the doping (in ppb) of phosphorous or arsenic that must be added to silicon to achieve a concentration of 1015 cm−3. The density of silicon is 2329 kg·m−3. Would this doping create an n- or a p-type semiconductor? What would be its resistivity?
15.4. Analogous to Figure 15.10, sketch the charge distribution and band bending for a p-type semiconductor brought into contact with an electrolyte. Assume that before equilibration the Fermi level of the redox couple is in the middle of the conduction and valence bands.
15.5. Data for the capacitance versus potential for an n-type Si semiconductor are provided in the table. The potential is measured with respect to a saturated calomel electrode, and the electrolyte is ammonium hydroxide. Create a Mott–Schottky plot (C−2 versus potential). From these data, determine the flat-band potential and the doping level in the semiconductor. Use a dielectric constant of 11.9. (Data are adapted from J. Electrochem. Soc., 142, 1705 (1995).)
Potential [V] C [F·cm−2]
−0.5 1.459E-08
0 1.101E-08
0.5 8.771E-09
1 7.715E-09
1.5 7.036E-09
15.6. Repeat Problem 15.5, but use data that are provided for a p-type semiconductor. For a p-type semiconductor, Equation 15.8 is replaced with the following equation:
Potential [V] C [F·cm−2]
−1 2.108E-08
−1.5 1.571E-08
−2 1.265E-08
−2.5 1.125E-08
−3 1.031E-08
15.7. An intrinsic semiconductor has a 2 eV band gap. If the Fermi level of the electrode corresponds to 0.4 V relative to a SCE reference, sketch the energy levels and show the corresponding energies in eV of the band edges using a vacuum scale.
15.8. Calculate the strength of the electric field at the semiconductor–electrolyte interface. Use a dielectric constant of 20, and assume the thickness of the depletion layer is 100 nm, with a doping level of 1016 cm−3.
15.9. For a 0.1 M 1 : 1 salt, calculate the concentration of charge carriers in solution. Compare this with typical values for doped semiconductors: 1015–1018 cm−3.
15.10. Using the semiconductor data from Illustration 15.3, determine at what electrolyte concentration is the Debye length of the same order of magnitude as the thickness of the depletion region for an n-type semiconductor doped to 1016 cm−3. Assume a 1 : 1 electrolyte.
15.11. An n-type semiconductor is brought into contact with an electrolyte with a redox couple. Sketch the interface before and after contact with two different electrolyte solutions. The first where the redox potential is a little below the Fermi level of the semiconductor, and the second where the redox potential is a little above the valance band edge. Thus, four sketches are required. Be sure to show the excess charges in the semiconductor and the solution as well as the thickness of the depletion region and the Fermi levels.
15.12. Starting with Equation 15.17 (Beer–Lambert law), show that the penetration depth (distance at which the intensity of light is reduced by a factor of 1/e) is inversely proportional to the absorption coefficient. Using Equation 15.20, determine the penetration depth in crystalline silicon for light with wavelengths of 400, 600, and 1000 nm.
15.13. Describe three ways of (i) increasing the number of electrons in the conduction band of a semiconductor, and (ii) increasing the number of holes in the valence band of a semiconductor. In both cases, we are generating mobile charge carriers.
15.14. To effectively create a photocurrent, the semiconductor must absorb essentially all of the incident light. What’s more, only light that is absorbed in or just outside the depletion layer results in charge separation. Discuss the implications of these restrictions when designing photoelectrodes for energy conversion. Specifically, compare the thicknesses and doping of Si and GaAs electrodes. Assume that the energies of the incident photons are 0.05 eV above their respective band gaps: Si 1.12 eV (indirect band gap) and GaAs 1.4 eV (direct band gap); and use the absorption coefficient from Equation 15.20 and Figure 15.17.
15.15. There are a number of parameters that arise is semiconductor–electrolyte systems: band gap, flat-band potential, doping levels, Fermi level, absorption coefficient, and width of the depletion layer. Describe each of these parameters. Specifically address whether these are intrinsic properties of the material, design variables that can be tuned, or one that will change depending on the operating conditions, such as light intensity or applied potential.
15.16. Given the spectral content of solar radiation, discuss the potential of the following semiconductors in (i) absorbing incident radiation and (ii) in efficient solar energy conversion: Si, GaAs, CdTe, and TiO2. What about their use to electrolyze water?
15.17. In Section 15.3, it is stated that “the potential drop across the Helmholtz double layer will be much less than that across the depletion layer of the semiconductor.” Given how capacitors in series behave, justify this claim. Remember that the capacitance of the double layer is much higher than that of the depletion region in a semiconductor.
15.18. In Section 15.4, a simplified version of the Mott–Schottky equation was developed for an n-type semiconductor (see Equation 15.8). What is the analogous expression for a p-type semiconductor? For what range of overpotentials does it apply? Why?
15.19. Using the data from Illustration 15.3, calculate quantum yield at an overpotential of −0.1 V assuming that the diffusion length is 200 μm. How small could the diffusion length be and still have a quantum yield of 0.9 or more?
15.20. An approximate value for the photon flux in direct sunlight is 2 × 1021 photons/m2·s. If the quantum yield were 1, what current density would this result from the photon flux? Name at least two reasons why this current density is not achieved in practice.
15.21. The open-circuit potential for the semiconductor electrode was determined in Section 15.7. Another important characteristic is the short-circuit current. Develop an expression for this current.
15.22. Using the data from Illustration 15.7, create a polarization curve (i versus Vcell) for photon currents of 50, 250, and 500 A·m−2.
15.23. One method to determine Vfb is to measure the capacitance as was outlined in Illustration 15.3. Another approach is to measure the onset of photocurrent. The key to this method is to use monochromatic light of energy just slightly above the energy of the band gap. Under these conditions, α is tiny and only a small amount of the incident light is absorbed in the depletion region. Starting with Equation 15.22, show that the photocurrent is proportional to the width of the depletion region, W. Substitute this result into Equation 15.6 to obtain the following relationship:
15.24. Using the data provided in the table for TiO2 and the result from the previous problem (Problem 15.23), determine the flat-band potential of the system.
η [V] iph [mA·cm−2]
0.704 0.909
0.557 0.845
0.441 0.778
0.321 0.706
0.200 0.612
0.129 0.553
15.25. For an n-type semiconductor with a flat-band potential of −0.4 V, what is the relative concentration of electrons at the surface of a semiconductor compared to the bulk: (i) when there is no applied potential (, and (ii) at an applied overpotential of 0.2 V. Comment on the implications of these results on the kinetics of the reaction and the current behavior.
15.26. In Section 15.3 the starting point for developing the description of the depletion layer was that the associated energy level of the redox couple was between the conduction and valence bands. What would change if the potential of the redox couple were either above the conduction band or below the valence band? Would it change with the doping (n or p)?
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