Process thermodynamic - Free energy

Gibbs Energy

To better explain the Gibbs energy concept applied to the alumina reduction, we will use first as an example the process of electrolysis of the water, depicted in the picture below, analyzed in condition of reversibility:

Electrolysis of water

The total energy necessary to electrolyze the water is equal to ΔH = 285.83 kJ/mol. This amount includes:

  • The change in the internal energy U linked with the change in the kinetic, potential and chemical bonds energy of the species involved in the electrochemical reaction
  • The energy required by the gas bubbles (O2 and H2) formed during electrolysis to expand under the pressure p:   work

being, as we know:

Change in enthalpy

Now arise a question: do we have, with the battery, to provide all the energy linked to the change in enthalpy? Let’s see…

The change in the internal energy U is equal to:

Change in internal energy

Where L is the work applied to the system while Q is the heat given to the system. In the case of a reversible reaction at constant temperature T the term Q is equal to TΔS. Furthermore, in our case the work L has to parts:

  • The electric work inputted by the battery: WEl
  • The work done by the gases expanding: -pΔV

Hence we have:

Change in internal energy

Substituting the (8) into the (6):

Change in enthalpy

Now, in our example, the only kind of external energy we are deliberately providing to the water is the electrical energy of the battery, and this energy is equal to:

Change in Gibbs Energy - Free Energy

Since the reaction is characterized by an increase in entropy (ΔS > 0), the energy WEl that the battery has to provide from the external is less than the total energy required by the electrolysis reaction. The “missing” energy is taken as heat from the environment.

The term ΔH-TΔS is equal to the Gibbs energy, or “free” energy. So we can write:

Change in Gibbs Energy - Free Energy

Let’s take again in consideration the (1), the basic alumina reduction reaction, in reversible conditions (hence at 100% current efficiency), occurring in an environment at temperature T. For this reaction remain valid all the arguments we have developed for the previous example. Let’s calculate now the amount of electrical energy needed by the (1) to occur.

If we take again in considerations the reaction (4), with the same assumptions (electrolyte saturated with α alumina, PCO2 = 1 atm, temperature of 977°C), inserting the proper figures from the thermodynamic tables we have:

Change in Gibbs Energy - Free Energy

At 100% current efficiency we have:


The difference:

Change in enthalpy

Is equal to the heat that is taken by the reaction (1) from the external environment. In the case of an alumina reduction cell the external environment is the bath itself, so the bath at its high temperatures (around 950°C) gives heat to the reaction (1). Obviously, the bath can not give indefinitely heat to the reaction, but this heat needs to be restored from the external. Practically, this means that we need to spend electrical energy also for the term TΔS.

Now we will calculate the minimum voltage to apply to a cell for the reaction (1) to occur in reversible conditions. In the (1) 12 moles of electrons are involved to produce 4 moles of aluminum.

In other terms, if F is the Faraday constant (the amount of electrical charge carried out by a mole of electrons) and E is the electrical potential we have:

Amount of electrical charge involved

Inserting the values we get:

Decomposition potential

Depending on the temperature at which occur the reaction (940°C ÷ 960°C normal operating ranges in the industry). This voltage is equivalent to the minimum energy, in reversible conditions, to give from the outside for the reaction (1) to occur. The voltage needed to compensate for the term TΔS is equal to:

TDeltaS energy requirement as voltage