Let us consider a classical hydrogen cell with porous electrodes. Such electrodes are permeable to g — Physical Chemistry — Kinetics Chemistry Question
Fuel cells
Let us consider a classical hydrogen cell with porous electrodes. Such electrodes are permeable to gases and the cathode also to liquid water. The cathode is fed with oxygen and the anode with hydrogen. The produced water is led out from the cathode compartment. The space between the electrodes is separated by a membrane which is permeable only to H+ ions, so that they can conduct the electric current. Such a fuel cell is quite efficient as there is no other way for hydrogen and oxygen to react except for electron-transfer via electrodes and H+ exchange through the membrane. Assume all gases behave as ideal. In this task, assume standard temperature 298 K and standard pressure 1 bar.
It is possible to construct a fuel cell very similar to the one described above, but working with butane and oxygen.
A modified construction of the butane fuel cell uses an oxide-conducting electrolyte, in which the following electrode half-reactions occur:
O2 + 4 e− → 2 O2−
4 CO2 + 5 H2O + 26 e− → C4H10 + 13 O2−
Another fuel cell works with the formal combustion of methanol. The EMF of such a cell at the standard temperature of 298 K is 1.21 V, and at 373 K it drops by 10 mV.
Useful data:
ΔfH°(H2O(l)) = −286 kJ mol−1 S°(H2(g)) = 131 J K−1 mol−1
ΔfH°(H2O(g)) = −242 kJ mol−1 S°(O2(g)) = 205 J K−1 mol−1
ΔfH°(CO2(g)) = −393 kJ mol−1 S°(C(s)) = 6 J K−1 m
ΔfH°(C4H10(g)) = −126 kJ mol−1 S°(CO2(g)) = 214 J K−1 mol−1
S°(H2O(l)) = 70 J K−1 mol−1 ΔfG°(C4H10(g)) = −17 kJ mol−1
S°(H2O(g)) = 189 J K−1 mol−1
Determine the standard electromotive force (EMF) of the above described fuel cell working at 298 K with 1 bar hydrogen and 1 bar oxygen. Assume that water is produced in the liquid state.
Model Answer
First, find the driving force, i.e., the Gibbs energy of the reaction H2 + ½ O2 → H2O under standard conditions (298 K and 1 bar). Then, convert it to the EMF (voltage).
The standard reaction enthalpy and entropy are
ΔrH° = ΔfH°(H2O(l)) = −286 kJ mol−1
ΔrS° = S°(H2O(l)) − (S°(H2(g)) + ½S°(O2(g))) = 70 − (131 + 205/2) = −163.5 J K−1 mol−1
The standard change of Gibbs energy is
ΔrG° = ΔrH° − TΔrS° = −286 − 298 × (−163.5 × 10−3) = −237.3 kJ mol−1
The standard EMF is then
Determine the standard EMF of the above described fuel cell working at 298 K with 1 bar hydrogen and 1 bar oxygen. Assume that water is produced in the gas state.
Model Answer
The solution is similar to the previous one with the difference of water state.
ΔrH° = ΔfH°(H2O(g)) = −242 kJ mol−1
ΔrS° = S°(H2O(g)) − (S°(H2(g)) + ½S°(O2(g))) = 189 − (131 + 205/2) = −44.5 J K−1 mol−1
The standard change of Gibbs energy is
ΔrG° = ΔrH° − TΔrS° = −242 − 298 × (−44.5 × 10−3) = −228.7 kJ mol−1
The standard EMF is then
Calculate the ideal thermodynamic efficiency (thermodynamic or maximum or ideal efficiency is the ratio between the maximum extractible work and the heating value) of the fuel cells described in previous questions at (a) the standard temperature of 298 K and (b) 373 K. Neglect the enthalpy and entropy temperature dependence in all the calculations.
Model Answer
The ideal thermodynamic efficiency is:
For both cells and for various temperatures, we get:
Write the balanced chemical equations for the cathode and anode half-reaction.
Model Answer
Cathode: O2 + 4 e− → 2 H2O
Anode: C4H10 + 8 H2O → 4 CO2 + 26 H+ + 26 e−
Calculate the EMF of the butane–oxygen fuel cell. Assume that butane is fed to the electrodes at the standard temperature and 1 bar and that it reacts with oxygen at 1 bar. Assume that water is produced in the liquid state.
Model Answer
The overall reaction is:
2 C4H10 + 13 O2 → 8 CO2 + 10 H2O
The reaction as accompanied by the transfer of 52 electrons. Hence, at standard temperature:
ΔfG°(H2O(l)) = −237.3 kJ mol−1
ΔfG°(CO2(g)) = −393 − 298 × ((214 − (6 + 205)) × 10−3) = −393.9 kJ mol−1
ΔfG°(C4H10(g)) = −17 kJ mol−1
ΔfG°(O2(g)) = 0
ΔrG° = (8 ΔfG°(CO2(g)) + 10 ΔfG°(H2O(l))) − (2 ΔfG°(C4H10(g)) + 13 ΔfG°(O2(g))) =
= (8 × (−393.9) + 10 × (−237.3)) − (2 × (−17) + 13 × 0) = −5 490 kJ mol−1
Calculate the ideal thermodynamic efficiency of the butane fuel cell.
Model Answer
The ideal thermodynamic efficiency is determined as:
Determine the standard EMF of this modified butane fuel cell with an oxide-conducting electrolyte.
Model Answer
It is the same as in the previous answer. The overall reaction is the same.
Write balanced chemical equations for the cathode and anode half-reaction. Write also the overall reaction that takes place in the methanol fuel cell.
Model Answer
Anode: CH3OH + H2O → 6 H+ + 6 e− + CO2
Cathode: O2 + 4 H+ + 4 e− → 2 H2O
Overall: 2 CH3OH + 3 O2 → 2 CO2 + 4 H2O
Write down the Nernst equation for the EMF of this cell. Choose the appropriate standard states for the reactants and products.
Model Answer
Any answer with correctly expressed activities (e.g. using molar fractions) is assumed to be correct.
Calculate the standard reaction enthalpy and entropy of the reaction taking place in the methanol fuel cell related to lowest integer coefficients in the chemical reaction involved in the fuel cell.
Hint: Use the van ’t Hoff equation.
Model Answer
We use van ’t Hoff equation, in which we substitute EMFs for equilibrium constants. We obtain reaction enthalpy and Gibbs free energy changes, which we use to calculate the entropy change: