Chapter 2 Echem Thermodynamics

3 • The way a fuel cell works is basically the opposite of the principle of an electrolysis cell. • In a fuel cell, chemical energy from hydrogen and oxygen (or air) is directly converted to electrical energy, i.e. without a combustion process. All fuel cells essentially consist of two electrodes (cathode and anode) and an electrolyte (medium for transporting the ions), which separates the two electrodes from each other. • Fuel cells are usually classified by the type of electrolyte used. Other traits they may differ in include Service Temperature, Efficiency, and Field of application. Fuel cells

6 Renewable energy Solar hydrogen technology makes it possible. Using electrolysis, available solar or wind energy is used to produce hydrogen. This is temporarily stored and used for generating power via the fuel cell anytime is needed. A clean solution indeed!

9 PEM fuel cell • PEM fuel cells use a thin, proton-conducting polymer membrane as an electrolyte. Both sides of the membrane are coated with a layer of catalyst material. • The catalytic effects of the electrode (e.g. platinum) cause the hydrogen gas at the anode to break down into protons and electrons even at room temperature. • The H + ions (protons) traverse the proton-conducting membrane to get to the cathode side and the electrons travel to the cathode, thereby doing electrical work. As a result, water is produced at the cathode side. • Individual fuel cells, summarized into a unit and connected with each other in series, result in a "stack" of cells. The stacks' output can be varied as desired by adjusting the number of individual cells. CHG8191- Dr. Elena Baranova

12 Thermodynamic efficiency ( h th ) in the conversion of chemical energy to electric energy H 2 (g) 2H + + 2e - 1/2O 2 (g) + 2e - + 2 H + H 2 O(l) (Anode) (Cathode) D r G o = –237.13 kJ/mol D r H o = –285.83 kJ/mol D r S o = –163.4 J/K·mol Standard state of reactants and products (1 atm ; 298 K) H 2 (g) + 1/2O 2 (g) H 2 O(l) Oxygen Energy (+) Hydrogen Energy (–) Fuel Cell Electricity Energy = V·I·t Heat Water

15 Reverse process : Electrolysis of water (298 K ; 1 atm) 1.23 V 1.23 V o o o r r r o r o 1 r G H T S G 285.83 298(0.1634) G 237.13 kJ mol - D = D - D D = - D = = (101.3x10 3 Pa)(1.5 mol)(22.4x10 -3 m 3 /mol)(298 K/273 K) o o r r o r o 1 r U H P V U 285.83 3.72 U 282.11 kJ mol - D = D - D D = - D = H 2 O(l) H 2 (g) + ½O 2 (g) Change in internal energy

18 Thermodynamic efficiency ( h th ) in the conversion of chemical energy to electric energy o o o o r r r r th o o o r r r Δ G Δ H TΔ S Δ S η = 1 T Δ H Δ H Δ H - = = H 2 (g) + 1/2O 2 (g) H 2 O(l) D S for this reaction is negative then T D S is the amount of heat dissipated into the environment (even in a reversible operating fuel cell). We can define a characteristic temperature at which the fuel cell delivers only heat (entropic heat). It is the temperature where D r G=0 (T D r G=0 ). r r r th G=0 r r G=0 r Δ S T η 1 T 1 T Δ H Δ H with T Δ S D D = - = - = G 0 r T 5000 K D = = For the overall reaction of hydrogen oxidation

21 Heat produced by the combustion reaction (Q 1 ) at T 1 : H 2 (g) + 1/2O 2 (g) H 2 O (l) Waste heat (Q 2 ) at T 2 Efficiency of heat engine (Carnot’s theorem) 1 2 2 2 1 1 1 1 Q Q Q T W CE = = 1 1 Q Q Q T - = - = -

22 Carnot efficiency (CE) and thermodynamic fuel cell efficiency ( h th ) 2 T Carnot efficiency CE 1 T Þ = The Carnot efficiency (CE) is a monotone increasing function approaching unity (100% efficiency) at very high temperature The thermodynamic fuel cell efficiency ( h th ) is a linear decreasing function approaching unity (100% efficiency) at low temperature There is a break-event point at the temperature (T BE ) where CE = h th o BE BE G 0 r BE o G 0 r T T 1 1 T T T T T D = D = - = - = × with T o =298 K and T BE = 1200 K G 0 r T 5000 K D = = T BE CE h th r th G=0 T Thermodynamic fuel cell efficiency η 1 T D Þ = -

23 The theoretical (reversible) cell potential is a function of operating temperature and pressure: where, a stands for activity or the ratio between the partial pressures of reactants (H 2 and O 2 ) or product (H 2 O) and atmospheric pressure (for liquid water product a H2O = 1). E o is a standard equilibrium potential E o eq = 1.229 V H 2 (g) + 1/2O 2 (g) H 2 O(l) Nernst equation for PEM fuel cell

24 Theoretical cell potential at different temperatures and pressures Actual cell potentials are always smaller than the theoretical ones due to irreversible losses. Voltage losses in an operational fuel cell are caused by several factors such as: • kinetics of the electrochemical reactions (activation polarization), • internal electrical and ionic resistance, • difficulties in getting the reactants to reaction sites (mass transport limitations), • internal (stray) currents, crossover of reactants T(K) Atm. 200 kPa 300 kPa 298.15 1.230 1.243 1.251 333.15 1.200 1.215 1.223 353.15 1.184 1.200 1.209

25 Electrochemical Thermodynamics Application to fuel cells Exercices https://www.youtube.com/watch?v=8rofx6Gaz40 https://www.youtube.com/watch?v=LDwS31OE7ak

26 General If no other indication is given, consider the following numerical data when solving problems in Chapter 1 : Standard (1 atm, 298 K) equilibrium potential : =1.229V o eq E D H f o S o C p o kJ / mol J / mol K J / mol K C 0 5.73 8.52 H 2 0 130.5 28.8 O 2 0 204.8 29.3 CH 4 -74.6 186.3 35.3 CO -110.5 197.7 29.1 H 2 O (l) -285.8 69.9 75.2 H 2 O (g) -241.8 188.8 33.5 Water vapor pressure (at 40°C £ T £ 60°C) : p H 2 O (mbar)=1.286×10 -5 e 0.04966T(K) Consider : 1 atm ≈ 1 bar valid for H 2 (g)+½O 2 (g) H 2 O(l)

27 Problem 1.1 A PEM fuel cell is working with pure H 2 as fuel at the cathode and with (a) pure O 2 or (b) air at the anode. 1) Calculate for both cases the open circuit potential (E eq ) at T=40 and 60°C and at different operating pressures of 1, 3 and 5 bar. For simplicity, neglect the temperature dependence of the standard potential. 2) Explain the opposite temperature dependence of cell voltage at atmospheric and high gas pressures.

28 Problem 1.2 In a solid electrolyte fuel cell utilizing YSZ as the solid electrolyte, H 2 and O 2 are consumed at the anode and cathode, respectively, according to the reactions: H 2 (g) + O 2- H 2 O(g) + 2e - O 2 (g) + 4e - 2 O 2- The cell is working at 800 ° C with p H2 =80 kPa and p H2O =20 kPa at the anode and p O2 =100 kPa at the cathode. Calculate 1) the open-circuit potential of the cell, 2) the thermodynamic efficiency of the cell.

29 Problem 1.3 Consider the following overall reaction taking place in a solid oxide fuel cell (SOFC) at 800 ° C : CH 4 (g) + 1/2O 2 (g) → CO(g) + 2H 2 (g) 1) Estimate the thermodynamic efficiency of this fuel cell. 2)Analyze this fuel cell from the point of view of the first and second law of thermodynamics. 3) Does the cell produce or absorb heat?

30 Problem 1.4 A PEM fuel cell working at 25°C produces a power of 10 W at 0.75 V from a single cell. 1) Calculate the amount of hydrogen and air (Nm 3 /h) needed to operate the stack consisted of 20 cells. 2) What is the electrochemical and the actual efficiency of the cell? Assume 100% utilization of reactants (100% conversion efficiency).