Electrolysis of Water

By providing energy from a battery, water (H2O) can be dissociated into the diatomic molecules of hydrogen (H2) and oxygen (O2). This process is a good example of the the application of the four thermodynamic potentials.

The electrolysis of one mole of water produces a mole of hydrogen gas and a half-mole of oxygen gas in their normal diatomic forms. A detailed analysis of the process makes use of the thermodyamic potentials and the first law of thermodynamics. This process is presumed to be at 298K and one atmosphere pressure, and the relevant values are taken from a table of thermodynamic properties.

Quantity
H2O
H2
0.5 O2
Change
Enthalpy
-285.83 kJ
0
0
DH = 285.83 kJ
Entropy
69.91 J/K
130.68 J/K
0.5 x 205.14 J/K
TDS = 48.7 kJ

The process must provide the energy for the dissociation plus the energy to expand the produced gases. Both of those are included in the change in enthalpy included in the table above. At temperature 298K and one atmosphere pressure, the system work is

W = PDV = (101.3 x 103 Pa)(1.5 moles)(22.4 x 10-3 m3/mol)(298K/273K) = 3715 J

Since the enthalpy H= U+PV, the change in internal energy U is then

DU = DH - PDV = 285.83 kJ - 3.72 kJ = 282.1 kJ

This change in internal energy must be accompanied by the expansion of the gases produced, so the change in enthalpy represents the necessary energy to accomplish the electrolysis. However, it is not necessary to put in this whole amount in the form of electrical energy. Since the entropy increases in the process of dissociation, the amount TDS can be provided from the environment at temperature T. The amount which must be supplied by the battery is actually the change in the Gibbs free energy:

DG = DH - TDS = 285.83 kJ - 48.7 kJ = 237.1 kJ

Since the electrolysis process results in an increase in entropy, the environment "helps" the process by contributing the amount TDS. The utility of the Gibbs free energy is that it tells you what amount of energy in other forms must be supplied to get the process to proceed.

Reverse process: Hydrogen fuel cell
Index

Internal energy concepts

Reference
Schroeder
Ch 5
 
HyperPhysics***** Thermodynamics R Nave
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Hydrogen Fuel Cell

Hydrogen and oxygen can be combined in a fuel cell to produce electrical energy. A fuel cell uses a chemical reaction to provide an external voltage, as does a battery, but differs from a battery in that the fuel is continually supplied in the form of hydrogen and oxygen gas. It can produce electrical energy at a higher efficiency than just burning the hydrogen to produce heat to drive a generator because it is not subject to the thermal bottleneck from the second law of thermodynamics. It's only product is water, so it is pollution free. All these features have led to periodic great excitement about its potential, but we are still in the process of developing that potential as a pollution-free, efficient energy source (see Kartha and Grimes).

Combining a mole of hydrogen gas and a half-mole of oxygen gas from their normal diatomic forms produces a mole of water. A detailed analysis of the process makes use of the thermodynamic potentials. This process is presumed to be at 298K and one atmosphere pressure, and the relevant values are taken from a table of thermodynamic properties.

Quantity
H2
0.5 O2
H2O
Change
Enthalpy
0
0
-285.83 kJ
DH = -285.83 kJ
Entropy
130.68 J/K
0.5 x 205.14 J/K
69.91 J/K
TDS = -48.7 kJ

Energy is provided by the combining of the atoms and from the decrease of the volume of the gases. Both of those are included in the change in enthalpy included in the table above. At temperature 298K and one atmosphere pressure, the system work is

W = PDV = (101.3 x 103 Pa)(1.5 moles)(-22.4 x 10-3 m3/mol)(298K/273K) = -3715 J

Since the enthalpy H= U+PV, the change in internal energy U is then

DU = DH - PDV = -285.83 kJ - 3.72 kJ = -282.1 kJ

Since the entropy decreases in the process of combination, the amount TDS must be provided to the environment as heat at temperature T to keep the entropy constant. The amount of energy per mole of hydrogen which can be provided as electrical energy is the change in the Gibbs free energy:

DG = DH - TDS = -285.83 kJ + 48.7 kJ = -237.1 kJ

For this ideal case, the fuel energy is converted to electrical energy at an efficiency of 237.1/285.8 x100% = 83%! This is far greater than the ideal efficiency of a generating facility which burned the hydrogen and used the heat to power a generator! Although real fuel cells do not approach that ideal efficiency, they are still much more efficient than any electric power plant which burns a fuel.

Index

Internal energy concepts

Reference
Schroeder
Ch 5

Reference
Kartha & Grimes
 
HyperPhysics***** Thermodynamics R Nave
Go Back