• Efficiency and reaction conditions
The proton exchange membrane is a solidstate electrolyte that functions at around 80 °C. Compared to the 1000 °C at which the solid oxide electrolytes become conductive, this is a low temperature.
What temperature is preferable?
One of the key banners under which fuel cells are marketed is their efficiency. We must consider how this efficiency comes about and what factors influence it. To do this we must look at the thermodynamics governing the fuel cell.
 Considering the energy of the system  the Gibbs free energy change.
Let's consider the energy of a hydrogen fuel cell system as follows:
INPUTS: 
PROCESS: 
OUTPUT: 
Hydrogen 
FUEL CELL 
Electrical Energy = VIt 
Heat 

Oxygen 
Water 
It’s easy to calculate the electrical power and energy output of the system:
Power = VI ; Energy = VIt
The “chemical energies” of inputs and outputs are a little more difficult to define. It is the change in Gibbs free energy that we must consider in this case (or more precisely G_{f} – the energy of formation  because we use the convention of comparison to pure elements in their standard states), which is the energy available to do external work. In the case of a fuel cell system this external work is pushing electrons through the external circuit, and past whatever impedances we put in their way. Work done by changes in volume and temperature between inputs and outputs is not harnessed by the fuel cell (although this is possible in turbine hybrid systems).
The G_{f} of both O_{2} and H_{2} is zero, a useful result when dealing with a hydrogen oxygen fuel cell. ΔG_{f} refers to the difference in Gibbs free energy of formation between the inputs and the outputs, and is therefore a specific measure of the energy released by the reaction.
ΔG_{f} = G_{f} of products – G_{f} of reactants
We usually consider this quantity to be per a mole of chemical. Let us find the chemical energy released during a nominal fuel cell reaction:
H_{2} + ½O_{2} → H_{2}O
The Gibbs free energy of a system is defined as:
G = H – TS
Which leads to the change in free energy being expressed as:
(1) \(\Delta {\overline g _f} = \Delta {\overline h _f}  T\Delta \overline s \)
Note that we’ve gone lowercase and added little lines above the letters. This is to signify that we are dealing with molar quantities, so that the units will be Joules per mole or something similar.
The value Δ
_{f} of is the difference between _{f} of the products and _{f} of the reactants. So for the reaction H_{2} + ½O_{2} → H_{2}O, we have:(2) \( \Delta {\overline h _f} = {({\overline h _f})_{{H_2}O}}  {({\overline h _f})_{{H_2}}}  {1 \over 2}{({\overline h _f})_{{O_2}}}\)
And Δ
is the difference between entropy of the products and reactants so that:(3) \( \Delta \overline s = {(\overline s )_{{H_2}O}}  {(\overline s )_{{H_2}}}  {1 \over 2}{(\overline s )_{{O_2}}}\)
These values of T” subscript to the enthalpy, , means the enthalpy at temperature T.
and _{f}vary with temperature and pressure according to the equations given below. A full derivation of these equations is beyond the scope of this TLP but can be found in thermodynamics textbooks. It should be noted that we use 298 K as the standard temperature, which is necessary as an integration limit. The “(4) $${\overline h _T} = {\overline h _{298}} + \int\limits_{298}^T {{{\overline C }_P}} dT$$
Similarly, the entropy, T is given by:
, at temperature(5) $${\overline s _T} = {\overline s _{298}} + \int\limits_{298}^T {{1 \over T}} {\overline C _P}dT$$
Values for standard enthalpies and entropies are obtainable from tables and some are given below:

_{f} (J mol^{1}) 
(J mol^{1} K^{1}) 
H_{2}O liquid 
285,838 
70.05 
H_{2}O Steam 
241,827 
188.83 
H_{2} 
Zero (element in standard state) 
130.59 
O_{2} 
Zero 
205.14 
We need to know the molar heat capacity at constant pressure,
_{p}. This is not constant with temperature but can be described by empirical equations. Examples below, (units are J g mol^{1} K^{1}):For H_{2}:
(6) \({({\overline C _P})_{{H_2}}} = 56.505  22,222.6{T^{  0.75}} + 116,500{T^{  1}}  560,700{T^{  1.5}}\)
And for steam:
(7) \({({\overline C _P})_{{H_2}{O_{(g)}}}} = 143.05  58.040{T^{0.25}} + 8.2751{T^{0.5}}  0.036,989T\)
And O_{2}:
(8) \({({\overline C _P})_{{O_2}}} = 37.432 + 2.010,2 \times {10^{  5}}{T^{1.5}}  178,570{T^{  1.5}} + 2,368,800{T^{  2}}\)
One may now proceed to substitute values of Δ , which can in turn be substituted into equation (1) to yield the allimportant Δ _{f} value.
_{p} from equations (6)(8) into equations (4) and (5) to yield integrals that solve to values of _{f} and for H_{2}O_{(g)}, H_{2} and O_{2} at any temperature. Substituting these values into equations (2) and (3) gives us values for Δ _{f} andThe table below was calculated using these equations:
Temperature (°C) 
Δ _{f} (kJ mol1) 
25 (liquid) 
237.2 
80 (liquid) 
228.2 
80 (steam) 
226.1 
100 
225.2 
200 
220.4 
400 
210.3 
600 
199.6 
800 
188.6 
1000 
177.4 
which can be handily represented in the following graph:
The graph shows us, somewhat obviously, that the energy obtainable from a fuel cell reaction decreases with the temperature at which the reaction is carried out. This however is not the only reason for lack of efficiency at higher temperatures, as we shall see.
Irreversibilities
This Δ
_{f} value represents the energy released by the reaction:H_{2} + ½O_{2} → H_{2}O
If there were no losses, if the fuel cell process operated completely reversibly, the fuel cell should be able to convert 100% of this Gibbs free energy into useful electrical energy. However losses, or irreversibilities, creep into every system. For example, some of the energy released in the reaction will inevitably leave the system as heat. Once it has floated away in warmer air, that energy cannot be recovered and turned back into useful electrical energy; so we call it irreversibility rather than a loss.
Open circuit voltage
Back to efficiency
Calculating the efficiency of a fuel cell system is not easy, partly because it is hard to define what the term refers to in this situation. It would help to look at a situation where efficiency is a little more obvious. The efficiency limit of a heat engine (such as a gas or steam turbine or an internal combustion engine) is defined as:
$${\rm{Carnot\;\; Limit}} = {{{T_1}  {T_2}} \over {{T_1}}}$$
T_{1} stands for the maximum temperature of the engine and T_{2} represents the temperature of the exhaust, all in Kelvin. The limit stems from the fact that there is always going to be some energy, proportional to the T_{2}, which is “lost”.
We know that fuel cell systems are not seconded to this limit; so how do we give a measure of their efficiency? We have already seen that it is the Gibbs free energy that is converted into electrical energy. If not for irreversibilities, all of this could be converted to electrical energy giving a 100% maximum efficiency. We could therefore define the efficiency of a fuel cell system as:
$${{{\rm{Electrical\; energy \;output}}} \over {{\rm{Gibbs\; free\; energy\; change}}}}$$
When this is the case however, no matter what the reaction conditions are, the efficiency limit is 100% and therefore the figure isn’t really much use to anybody.
A common measure of the energy contained in a fuel is its calorific value. This is a measure of the heat that would be produced by burning the fuel. A more precise measurement is the change in enthalpy of formation, Δ
_{f} and as with the Gibbs energy, the convention is that a negative value translates to more energy released during reaction. Using this value we can obtain a more useful efficiency value for a fuel cell:$${{{\rm{Electrical\; energy \;output \;per \;mole \;of \;fuel}}} \over {\Delta {{\bar h}_f}}}$$
Recall however that the Δ
_{f} of steam is different to that of liquid water, hence there are two possible efficiency values for a given process depending on the state of the output. See table below listing the Δ _{f} values of the reaction H_{2} + ½O_{2} → H_{2}O, depending on the form of the product.H_{2}O liquid 
285,838 
H_{2}O Steam 
241,827 
Papers quoting an efficiency value will usually say whether it refers to the higher heating value, the HHV (in which steam is the product); or the lower heating value, the LHV.
We can express maximum possible efficiency as:
$${{\Delta {{\overline g }_f}} \over {\Delta \overline {{h_f}} }} \times 100\% $$
Which is often termed the “thermodynamic efficiency” of the fuel cell. We can use this equation along with equation 9 to come up with various efficiency limits and maximum reversible EMF values. These are listed in the table below, which is an extension of the earlier table:
Temperature (°C) 
Δ _{f}(kJ mol1) 
Max EMF (V) 
Efficiency Limit 
25 (liquid) 
237.2 
1.23 
83% 
80 (liquid) 
228.2 
1.18 
80% 
100 
225.2 
1.17 
79% 
200 
220.4 
1.14 
77% 
400 
210.3 
1.09 
74% 
600 
199.6 
1.04 
70% 
800 
188.6 
0.98 
66% 
1000 
177.4 
0.92 
62% 
Are higher temperatures better?
Points to note about the graph above:
 The graph above is quite specific to the hydrogen fuel cell. If we were to look at the cell fuelled by CO:
CO + ½O_{2} → CO_{2}
We’d see the Δ
_{f} becomes less negative even faster with increasing temperature so that the efficiency limit of 82% at 100 °C falls to 52% at 1000 °C. On the other hand, the Δ _{f} of the methane fuelled reaction:CH_{4} + 2O_{2} → CO_{2}+ 2H_{2}O
Hardly changes with temperature so the efficiency limit stays consistent. We must therefore remember that the temperature dependency of efficiency is characteristic to a reaction.
 It can be seen that fuel cells do not always have a higher efficiency limit than fuel cells.
Since the efficiency limit of the H_{2} fuelled cell reduces with operational temperature, one might be tempted to conclude that hydrogen fuel cells should be run at the lowest possible temperature. There are several reasons why this is not the case:
 In higher temperature systems, the heat produced can be more useful. Turbine hybrid systems can be used to utilise the energy of the exhaust gases. This isn’t as easy if the fuel cell is run at low temperatures.
 The voltage losses (discussed later in this section) of the cell, the irreversibilities which are an inevitable part of the process, are generally more significant at higher temperature.
Fuel utilisation and reaction optimisation
In conclusion, the factors determining the efficiency of the system are manifold and various. We can estimate the efficiency obtained by simply measuring the open circuit voltage of the fuel cell V_{c} and we can make sense of this by looking at the reaction and how the Gibbs free energy is transferred between reactant and product species.