Download Tutorial 1 / SS 2013

Fuel Cells in Energy Technology
Tutorial 1 / SS 2013 - solutions
Prof. W. Schindler, Jassen Brumbarov /
Celine Rüdiger
08.05.2013
1. Thermodynamics
c) To understand the principles of energy conversion of heat engines and electrochemical
energy converters, some important definitions and formulas from the field of thermodynamics
should be discussed. Explain the meaning of the following thermodynamic relations!

First Law of Thermodynamics:
dU = δQ + δW
The energy of the universe is constant. The First Law of Thermodynamics states that energy is
conserved. The internal energy U denotes the total energy of a system (kinetic/thermal energy,
potential energy, electric energy…); dU is the infinitesimal change of the internal energy. By doing
work on/by a system (δW  directed motion) or by transferring heat to/from the system (δQ 
disordered motion), the internal energy of the system is changed. If energy is added to the system, the
internal energy will rise. Thus, these terms are positive. If energy is released from the system (for
example by releasing heat or if the system does work), the terms are negative and the internal energy
decreases. U is a state function: a property of a function that is not dependent on the way in which the
system gets to the state in which it exhibits that property.

Second Law of Thermodynamics:
dS  0 for a closed system
In any spontaneous process there is always an increase in entropy of the universe. S is the entropy. It
is a measure of order/disorder of a system. The Second Law of thermodynamics states that the
entropy change in an isolated system can either be zero (reversible processes) or it increases
(irreversible, spontaneous processes). The Clausius statement of the Second Law is: “It is impossible
for any system to operate in such a way that the sole result would be an energy transfer by heat from
a cooler to a hotter body.” (Third law: the entropy of a perfect single crystal at 0 K is zero)

Entropy:
dS  δQ/T
Here again, “ > “ is used for irreversible processes and “ = ” is used for reversible processes. In a
process a  b, S = Sb – Sa.
In a Carnot cycle, the entropy change is reversible. For the adiabatic processes dS = 0, for the
isothermal processes dS = δQ/T.

Enthalpy:
H=U+p.V
Enthalpies are often listed in literature for chemical reactions, because this is the actual energy that is
set free. If you simply burn a mole of hydrogen at constant pressure, then 286 kJ (= H for the
formation of liquid water) is the amount of heat you get out. Nearly all of this energy comes from the
thermal and chemical energy of the molecules themselves, but a small amount comes from work done
by the atmosphere as it collapses to fill the space left behind by the consumed gas. A chemical
reaction is exothermal (releases heat), if the enthalpy decreases. For a closed system and an
infinitesimal process, where the only work done is the volume work (dU = δQ – pdV), the change in
enthalpy is dH = TdS + Vdp. For constant pressure (open chemical reactions), the enthalpy is the heat
evolved or consumed (TdS = δQ).

Gibbs Free Energy: G = H – T . S
According to the definition of the Gibbs free energy and assuming constant pressure for the reaction,
G is the enthalpy (heat content of reaction) minus the thermal energy (“disordered energy”  cannot
be used to do work) of the reaction. A chemical reaction is spontaneous, if the free enthalpy
decreases (for constant temperature the free enthalpy change is ΔG = ΔH - TΔS).

Work:
dW = -p . dV (for volume work)
In physics, work can be expressed in many ways (forced motion of a body along a path dW = Fds;
motion of charge q in an electric field of strength E along the path s dW = qE(s)ds, conservative
electric field: W = -qU). Dealing with gases (mostly ideal gases), the definition above is most suitable
to describe a number of thermodynamic problems. The sign of work is defined in that way, that the
internal energy of the gas is decreased if the gas is doing work (expansion  dV > 0). Especially in
pistons, where the gas is expanded or compressed, the work done can be easily calculated by
integrating over the volume of the gas in the piston. Remember that the pressure changes in such
processes, so that p has to be substituted according to the ideal gas law. (W = ∫V i Vf –pdV, ideal gas:
pV = nRT  W = -nRT ln (Vf / Vi) )

Efficiency:
 = benefit / cost
The definition  = benefit / cost is a general definition. In a heat engine cycle, the energy is added by
heat (Qin), which is converted to a net work output (W cycle). Consequently, the efficiency of a heat
engine is  = W cycle / Qin. Heat pump cycles have the efficiency  = Qout / W cycle, as work is done on the
system to transfer heat from a cold to a warm reservoir.
Carnot engine: Qin = Th(Sb – Sa), heat pump: Qout = Th(Sb – Sa).

Ideal Gas in an isolated system:
a) Ideal gas law: p . V = n . R . T
p is the pressure in kg/(m . s²), V the volume in m 3, n the number of moles, R is the gas constant with
R = 8,314 J . mol-1 . K-1 and T the temperature in K (0 °C = 273.15 K). An ideal gas is approximated by
a gas with a very low density.
b) Work done by an ideal gas (expansion):
W = ∫V i Vf –pdV, ideal gas: pV = nRT  W = -nRT ln (Vf / Vi)
c) Adiabatic changes of an ideal gas: T . Vк-1 = const1, p · Vκ = const2,
κ = Cp/Cv
In adiabatic changes of an ideal gas that undergoes a process from step 1 to step 2, no heat is
exchanged (dQ = 0). The change of the internal energy is determined only by the involved amount of
work dU = dW. With the definition of the specific heat (= per mole) at constant volume: C V = (∂U/∂T)V ,
the first law of thermodynamics becomes dU = CVdT = -pdV. Using the ideal gas law for n = 1 mole,
one obtains CVlnT = -RlnV + const. With R = CP - CV and к = CP/CV one obtains the equation T 1 . V1к-1
= T2 . V2к-1 for the temperatures and volumes before and after the adiabatic change of the state.
d) Isothermal changes of an ideal gas: pV = nRT = const., dS = δQ/T
e) Carnot Cycle of an ideal gas:
Qin = Th(Sb – Sa), W cycle = (Th-Tc)(Sb – Sa)
f)
Efficiency of a heat engine and of a heat pump:
 = W cycle / Qin   = 1 – Tc/Th,  = Qout / W cycle   = Th/(Th – Tc)
Apply thermodynamics for a chemical reaction: The synthesis of ammonia at standard conditions
(pressure of 1 atm) and at 25°C is accompanied by a change of standard entropy and of standard
enthalpy:
N2 (g) + 3H2 (g)  2NH3 (g),
ΔS0 = - 198.7 J/K, ΔH0 = - 46.11 kJ/mol.





Calculate the change in standard Gibbs free energy!
Explain why the second law of thermodynamics is fulfilled!
Is the reaction exothermal?
Does the reaction run spontaneously?
Calculate the heat of the reaction in J!
The entropy of the educts is lower than the entropy of the products
(order increases: 4 small molecules produce 2 bigger molecules)
ΔH0 = -46.11 kJ/mol, ΔS0 = -198.7 J/K
ΔG0 (1mol) = ΔH0 - TΔS0 = [2x(-46.11·103) – 298x(-198.7)] kJ = -33 kJ (at 298 K)
Exothermal (ΔH < 0) and spontaneous (ΔG < 0)
Heat of reaction: this reaction is reversible, so we can use the change in entropy to calculate the
released heat during the synthesis:
δQ = TΔS = 298 K x (-198.7 J/K) = -59.2 kJ
To determine the entropy change of an open system like a chemical reaction in open air, the
surrounding of the reaction system has to be taken into account, too! Even if the entropy of the
products is lower than the entropy of the educts, the entropy of the surrounding will increase by a
greater amount due to heat transfer from the reaction to the surrounding (disordered thermal motion).
Those reactions are exothermal. The overall entropy will increase.
2. Oxidation, reduction and oxidation numbers
a) What are redox reactions? Explain the term oxidation and reduction as well as oxidising
and reducing agent. What is the oxidation number?
Redox reaction: chemical reaction in which an electron transfer from one reactant to another occurs.
A + B → A+ + B-
A redox reaction can be separated into an oxidation reaction and a reduction reaction. In the oxidation
reaction a species A releases electrons. A is called the reducing agent (donor). In the reduction
reaction a species B gains electrons. B is called the oxidising agent (acceptor).
A → A+ + e-
B + e- → B-
The oxidation number is a measure of the relative densitiy of electrons around an atom. In a chemical
compound it is the ionic charge that the atom would have if all other atoms were removed (the electron
pairs stay at the more electronegative species). For one-atomic ions the oxidation number
corresponds to the charge.
b) Write down the half-reactions for the oxidation of aluminium, lithium, magnesium and the
half-reactions for the reduction of fluorine, chlorine and sulphur. Write down the formulas
for magnesium oxide, aluminium fluoride, calcium fluoride and magnesium nitride! You
may need the periodic table below.
Oxidation half-reactions:
Al
Li
Mg
→
→
→
Al3+ + 3eLi+ + eMg2+ + 2e-
Reduction half-reactions:
F2 + 2eS + 2eCl2+ 2e-
→
→
→
2F1S22Cl-
Magnesium oxide: MgO
Mg is a Group II representative metal; it loses 2 electrons, forming a cation with "+2" charge. Oxygen
is a representative nonmetal from Group VI, hence it gains two electrons forming an ion with a "-2"
charge.
Aluminium fluoride: AlF3
Al is a Group III representative metal; it loses 3 electrons, forming a cation with a "+3" charge. Fluorine
is a representative nonmetal from Group VII, hence it gains only 1 electron forming an ion with "-1"
charge.
Calcium fluoride: CaF2
Ca is a Group II representative metal; it loses 2 electrons forming a cation with a "+2" charge. Fluorine
is a representative nonmetal from Group VII, hence it gains only 1 electron forming an ion with "-1"
charge.
Magnesium nitride: Mg3N2
Mg is a Group II Representative metal; it loses 2 electrons, forming a cation with "+2" charge. Nitrogen
is a Group V representative nonmetal, hence it gains 3 electrons to form an anion with a "-3" charge.
To obtain the simplest ratio of the species in a compound, multiply the charges by the smallest whole
number so that the total positive charge equals the total negative charge. This simplest ratio of the
ions is then written in the conventional chemical formula as CaF 2, Mg3N2 etc.
Remember that many atoms can have different charges as ions or in compounds; there is for example
Fe2+ and Fe3+ and the oxygen atom has different oxidation numbers in O 2 (0), H2O2 (-I) and H2O (-II).
Hence, the results of question 1 are only guidelines. To formulate chemical reactions in the right way,
one has to know all reaction conditions and a profound knowledge of the chemistry involved.
3.
Fuel cells in general
What are the constituent parts of a fuel cell (for instance PEM FC)? Make a sketch!
What is the function of each part? How does the FC convert chemical energy into electric work?
How is the chemical energy defined? Which amount of the chemical energy can be converted into
electric work by an ideal electro-chemical energy conversion device at standard conditions
(pressure 1 atm, temperature 25 °C)?
The main constituents of every kind of FC are: Fuel input (e.g. gas of defined pressure) to maintain
constant performance (current), metal electrodes to collect the current, a supported catalyst to drive
the redox reactions (the reactions take place at the interface of the supported catalyst and the
electrolyte), an ion conducting electrolyte which is gas-tight and not electron conducting to close the
electric circuit, an output for the exhaust (e.g. water: is produced and collected on the cathode side). If
a load is connected to the outer circuit, a current flows.
Electrochemical energy conversion: separation of the oxidation and the reduction reaction  the
produced electrons flow from the anode to the cathode through the external electric circuit.
If the two reactions weren’t separated, a Knallgas reaction would occur (oxyhydrogen explodes and
oxygen is reduced to water)  only heat is produced, no electric work!
The total chemical energy is the reaction enthalpy ∆H of the redox reaction of the specific FC setup.
For the H2 | O2 FC the chemical energy for standard conditions (pressure of 1atm) is given by the
standard reaction enthalpy of the total reaction:
H2 (g) + ½ O2 (g)  H2O (l)
ΔH0 = -286 kJ/mol (at 298 K)
For constant pressure and temperature, the amount of energy an ideal FC can convert into electric
work is given by the Gibbs free energy change during the reaction:
W = ΔG = ΔH – TΔS,
for a H2 | O2 FC at standard conditions: ΔG0 = -237.3 kJ/mol (at 298 K)