Fuel Cells
5. Fuel Cells
Combustion is a process during which irreversible exothermic oxidation reactions convert the chemical energy of fuels into thermal energy and then, oftentimes, into electrical energy through an elaborate conversion process that results in low conversion efficiencies. Typically, only about 35% of the chemical energy of a fuel is eventually converted into electri- cal energy in a steam power plant. Fuel cells offer a more efficient alternative. In a fuel cell chemical energy is converted into electricity through the release of electrons by chemical reac- tion. The process is almost reversible and hence the irreversibility is greatly reduced in these cells in comparison with combustion. Prior to description of the fuel cell, let us illustrate the basics of oxidation states and the electron transfer during chemical reaction.
a. Oxidation States and electrons Consider the reaction of butane C 5 H 12 with O 2 producing the products CO 2 and H 2 O. The oxidation states for this reaction are provided by the number of electrons gained or lost. The mass of a C atom is twelve times larger than of H atom so that the electrons surrounding H atoms are pulled more strongly towards the carbon atoms (Chapter 1). The five C atoms in the fuel gain y=12 electrons. The oxidation state of each C atom in the fuel is -12/5 (= y/x, where x denotes the number of carbon atoms in the fuel) while that of each H atom is +1. The oxygen
molecule is symmetric and its oxidation state is zero. Consider each of the five CO 2 molecules in the products. Atomic O atom is heavier than C atoms. The four outer electrons of each of the carbon atoms are pulled towards the O atom. In this context, the oxidation state of each C atom is +4 while the oxidation state of each O atom is -2. There is a net loss of 5 ×4 = 20 electrons
from the 5 carbon atoms in 5 CO 2 . Thus as the carbon in C 5 H 12 is burned, the oxidation state of
C changes from -12/5 to +4 in CO 2 . There is a net loss of (12/5) + 4 electrons (=y/x + 4) per C atom in the fuel. For this example, the 5 C atoms in the fuel lose 32 (= x × ((y/x) +4)= y+4 x)
electrons. The oxidation state of H atom is unchanged. The oxidation states of O atoms in CO 2 and H 2 O are -2 and -2. Hence, the O atoms in the five CO 2 molecules and six H 2 O molecules gain 5 ×2×(-2) +6×1×(-2) = 32 electrons, which are transferred from the five C atoms in C 5 H 12
to the oxygen atoms in the products during the chemical reaction. The fuel cell achieves such a transfer through an external load. Generalizing for any fuel denotes by C x H y , the number of electrons transferred is 4x + y. In the case of pure H 2 (i.e., x=0), the analogous number of electrons is two.
b. H 2 -O 2 Fuel Cell The operational principle of a H 2 -O 2 fuel cell is illustrated in Figure 4 . At the anode
H 2 → - 2e + 2H + , and at the cathode
2H + 2e + - 12 / O
2 H O (1) .
electrical work
porous anode
Figure 4: Principle of operation of a fuel cell. In a fuel cell chemical reactions occur at the ambient temperature T o . In a hydrogen–powered
fuel cell, separate streams of H 2 and O 2 are converted into liquid water at T o . Consequently, the absolute availability for the H 2 at the inlet is g H 2 . The maximum possible work under the
steady state steady flow characteristic of a fuel cell can be determined using the familiar rela- tion:
d(E cv -T 0 S cv )/dt = Σ Q ˙ Rj , ( 1 −
) +( Σ N ˙ k ψ k ) i - ( Σ N ˙ k ψ k ) e - W ˙ cv - ˙I (31)
T Rj ,
W ˙ opt = ( Σ N ˆg ˙ k ) i –( Σ N ˆg ˙ k ) e (32) The fuel cell efficiency is
(33) The maximum possible efficiency is
η fc = w/ ∆h c .
( η fc ) opt =w opt / ∆h c = (– ∆g)/∆h c = –( ∆h R –T ∆s R )/ ∆h c .
Assuming the higher heating value applies, ∆h c = HHV = – ∆h R , then
(35) If the value of ∆s R (= s prod -s reac ) >0 at the fuel cell operating temperature, (e.g., by adding heat
( η fc ) opt =1+T ∆s R / ∆h c .
to maintain isothermal conditions for an endothermic reaction), then ( η fc ) opt >1. The availability or exegetic efficiency for a hydrogen–oxygen fuel cell can be defined as
η avail = W/(exergy of H 2 ).
sider also the scenario for the reaction of a stoichiometric amount of H 2 with O 2 . What is the maximum possible fuel cell efficiency. Assume that ∆h c = 285830 kJ kmole –1 . Solution The fuel cell temperature T = T o = 25ºC, and the two reactant streams enter sepa- rately. Consequently,
W ˙ opt =( Σ k Ng ˙ k ) i –( Σ Ng ˙ k ) e .
Assume that N ˙
H 2 = 1 kmole s . In that case, N O 2 = 1/2 kmole s , and N HO 2 = 1 kmole s –1 , and
( Σ N ˙ k g k ) i = g H 2 + (1/2) g O 2 , where
g –1
H 2 = 0 – 298 × 130.574 = –38911 kJ kmole , and
g –1
H 2 = 0 – 298 × 205.033 = –61100 kJ kmole , i.e.,
( Σ N ˙ k g k ) i = –69461 kJ per kmole of H 2 .
Likewise,
( Σ N ˙ k g k ) e = –306675 kJ per kmole of H 2 , so that
w opt =G reactants –G products = –69461 + 306675 = 237214 kJ per kmole of H 2 .
At the cathode of a fuel cell positive potential is applied; the reaction is given as 2H + + 2e – + 1/2O 2 →H 2 O(l),
and at the anode the pertinent reaction is
2 → 2e + 2H .
Overall, a molecule of H
2 generates 2 electrons, i.e., a kmole of H 2 or 6.023 × 10 molecules generates 2
×6.023×10 26 electrons. An electron carries a charge of 1.602
2 generates a charge of 1.602 ×10 × 2 × 6.023 × 10 26 Coulomb.
×10 –19 Coulomb so that a kmole of H
The electrical work W elec = Voltage in volts × charge in coulombs =
237214 kJ per kmole of H
2 × 1000 J kJ ÷ (1.602×10 –19 × 2 × 6.023 × 10 26 Coulomb) = 1.229 V.
The optimum fuel cell efficiency
( η fc ) opt = w opt / ∆h c = 237214/285830 = 0.83.
Remarks
A short formula for determining the voltage of a fuel cell is Volts = ( ∆G per kmole of fuel in kJ)
× 1.036×10 –5 ÷(Number of electrons generated per molecule of the fuel). Fuel cells may be connected in series to obtain a higher voltage than an individual cell
provides. For fuel cells using hydrocarbon fuels, the anodic reaction is
x H y + 2x H 2 O → x CO 2 + (4x+y)H + (4x+y)e ,
and the cathodic reaction is (x+y/4)O + (4x+y)H +
2 + (4x+y)e → (2x+y/2)H 2 O.
The overall reaction can be represented as
Ambience, T 0