One of the most challenging issues for future technologies is to maximize the energy gain from renew — Physical Chemistry — Kinetics Chemistry Question
Designing a photoelectrochemical cell
One of the most challenging issues for future technologies is to maximize the energy gain from renewable sources: solar, wind, hydropower, geothermal and biomass. Although they represent clean and highly abundant sources of energy with a tremendous physical potential, they are intermittent, which applies mainly to solar energy. This means that they are not available when and where needed, or at least not all the time: the sun sets, wind does not blow, etc. One of the possible solutions to this problem is to store energy in a medium that is long-lasting and dispatchable. Chemical bonds represent such a medium. In general, this is the concept of solar fuels.
Such a system is already readily available in nature: photosynthesis. Plants use sunlight to make fuel (carbohydrates) out of water and carbon dioxide. To be able to do that, however, plants need fertile soil, water and favourable climate. On the other hand, artificial photosynthetic systems are not limited by such constraints and are capable of producing a fuel with a higher energy density, like hydrogen. Photoelectrochemical (PEC) water splitting is a powerful, yet complex process. By completing the upcoming tasks, you will get a basic insight into photoelectrochemistry.
Once the substrate for testing electrode is covered with a highly conducting and relatively non-reacting coating (e.g. gold), a photocatalyst can be added. Thanks to their chemical stability in aqueous environment, metal oxide-based semiconductors are suitable materials for photo-electrochemical applications. Recently, titanium dioxide has emerged as an excellent photocatalyst. It is an n-type semiconductor and can be used as a material for photoanodes. The whole complex mechanism of photoreactions occurring on irradiated n-type semiconductors can be simplified as follows. A photon with sufficient energy (wavelength) strikes the surface of a semiconductor and an electron from the highest occupied molecular orbital (HOMO) is excited to the lowest unoccupied molecular orbital (LUMO), leaving a positively charged hole (h+) behind. When applying an external electric field, the excited electrons are driven through the system to the counter-electrode where they participate in reduction reactions while photogenerated holes participate in oxidation reactions. The observed flow of electrons is called net photocurrent.
Let us call Eg the difference between the energy levels of HOMO and LUMO. It denotes the minimum excitation energy (maximum wavelength) of irradiation. To choose the optimal photocatalyst for a redox reaction, two main presumptions exist: 1) the Eg of a semiconductor has to be “accurately” wider than the potential of a redox reaction; 2) the energy level of HOMO has to be below the energy level of the oxidation half-reaction while the energy level of LUMO has to be above the energy level of the reduction half-reaction.
Which of the following half-reactions have reduction potentials dependent on pH?
a) Br2 + 2 e− → 2 Br−
b) NO3 - + 3 H+ + 2 e− → HNO2 + H2O
c) ClO3 − + 6 H+ + 6 e− → C l - + 3 H2O
d) Cr2O7 – + 14 H+ + 6 e− → 2 Cr3+ + 7 H2O
e) 2 CO2 + 2 e- → (COO)2 -
f) 2 IO3 - + 12 H+ + 10 e- → I2 + 6 H2O
g) S2O8 2- + 2 e- → 2 SO4 2-
h) TiO+ + 2 H+ + e- → Ti3+ + H2O
Model Answer
Reduction potentials for reactions b), c), d), f) and h) are dependent on pH.
Using the Nernst-Peterson equation and considering [Aox] = [Ared], derive a formula for the dependence of the reduction potential of the following reaction on pH:
Aoz + n H+ + z e- → Ared + n/2 H2O
What is the nature of this indepence (logrithmic, exponenncial, quadratic, etc) ?
Model Answer
The potential dependence on pH is a linear function with intercept equal to E° and slope equal to:
E = E° - RT/(zF) * ln([Ared] / ([Aox][H+]^n)) = E° - RT/(zF) * (ln([Ared]/[Aox]) - n * ln([H+]))
Considering [Aox] = [Ared]:
E = E° + nRT/(zF log(e)) * log([H+]) = E° - nRT/(zF log(e)) * (-log([H+])) = E° - nRT/(zF log(e)) * pH
This dependence is linear.
Let us consider two possible reactions occurring in an electrolyte:
Box + 3 e− → Bred Eo = + 0.536 V
Cox + 2 e− → Cred Eo = + 0.824 V
a) Which of the two possible reactions will occur under the conditions of p = 1 atm, T = 298.15 K? Will substance B oxidize C or will substance C oxidize B? Write a balanced chemical equation for the reaction between substances B and C.
b) Determine the standard potential for such a reaction.
c) Calculate the equilibrium constant for this reaction.
Model Answer
a) Standard potential E°(C) is more positive than E°(B), hence substance C is a stronger oxidizer and will therefore oxidize substance B. The balanced equation is:
3 Cox + 2 Bred → 3 Cred + 2 Box
b) Standard reaction potential Er° is 0.288 V:
Cox + 2 e− → Cred EC° = +0.824 V
Bred → Box + 3 e− EB° = −0.536 V
Er° = EC° - EB° = 0.824 − 0.536 V = +0.288 V
c) K = exp(zFEr° / RT) = exp(6 * 96485 * 0.288 / (8.3145 * 298.15)) ≈ 1.62 × 10^29
Now, let us consider the electrochemical system of two reactions that can occur in a cooled experimental cell, where one of them is pH-dependent and the other is not:
Dox + e− → Dred ED = +0.55 V
Eox + e− + H+ → Ered EE = +0.95 V
a) Calculate the change in potential (in millivolts) as a function of pH for the pH-dependent reaction. The given potentials correspond to pH = 0 and T = 262 K. The only parameter that can be changed during the experiment is the pH value of electrolyte.
b) Draw a straight line plot of the dependence of the reduction potentials (for both D and E) on pH in the range from 0 to 13.
c) Find the value of pH at which the equilibrium constant for the oxidation of substance D is K = 2.56 × 105.
d) Show on the drawn plot the region of pH where D will oxidize E.
Model Answer
a) Reaction E is pH-dependent and its potential drop is 52 mV per pH unit (z = 1, n = 1, T = 262 K).
b) Plot E [V] vs pH from 0 to 13. Horizontal line for ED at 0.55 V, sloping line for EE starting at 0.95 V.
c) The reaction potential Er = EE – ED is calculated from the equilibrium constant:
Er = RT ln(K) / zF = (8.3145 * 262 * ln(2.56 * 10^5)) / (1 * 96485) ≈ 0.28 V
Thus, EE = ED + Er = 0.55 V + 0.28 V = 0.83 V. Since EE = 0.95 - 0.052 * pH, this potential is achieved at pH = 2.31.
d) The two lines cross at pH = 7.7 (roughly); D will oxidize E in the pH range from 7.7 to 13.
Calculate the time needed to electrolytically cover a 5 × 10 × 0.5 mm metallic plate fully immersed into 10 cm3 of the solution of a gold precursor with c(Au3+) = 5 mmol dm-3 with 5 mg of a gold protective film. Consider that only gold (MAu = 197 g mol−1) is being deposited on the surface of the metallic plate, no side reactions occur, the surface area of the contacts with the electrode is negligible, and there is a constant current of 25 mA during the process.
Model Answer
Using the formula for electrolysis:
t = (m * z * F) / (M * I) = (0.005 g * 3 * 96485 C mol-1) / (197 g mol-1 * 0.025 A) ≈ 294 s
In the following picture you can see a schematic view of an energy diagram comparing four materials (F–I) by positions of their HOMOs and LUMOs towards the investigated redox reaction.
[VISUAL]
a) Which materials could be used as photocatalysts for the reaction outlined in the diagram?
e) Calculate the maximum wavelengths (in nm) of irradiation sources needed to excite your chosen materials. Based on your results, decide whether you can or cannot use UV and/or VIS light for the irradiation.
Model Answer
a) Only materials G and I can be used to catalyze the given reaction, because their HOMOs lie below Eox and their LUMOs are higher than Ered.
e) For material G (Eg = 3.2 eV), it can be irradiated only by UV light with a wavelength lower than:
λ(G) = hc / Eg ≈ 388 nm
For material I (Eg = 2.0 eV), it can be irradiated by either visible or UV light, because the maximal wavelength is:
λ(I) = hc / Eg ≈ 620 nm