Saturday, January 17, 2009

Hall-Heroult Process

Hall-Heroult Process 

The Hall-Heroult process is a continuous industrial process for the production of aluminum. The process is carried out in a molten cryolite salt bath. For more information on how the process works take a look at the wikipedia entry here. To calculate Al production, there are several reactions that one must look at. The first set of reactions is the dissolution of Al2O3 in the bath: 

Na3AlF6 -> 3Na+ + AlF63- 


4AlF63- + Al2O3 -> 3Al2OF62- + 6F- 

The production of Al is at the cathode through: 

Al2OF62- + 6e- -> 2Alo + 6F- + O2- 

And at the anode: 

C + 2O2- -> CO2 + 4e- 

For the ease of calculating Al production, the reactions at the cathode and anode can be simplified to: 

Al3- + 3e- -> Alo at the cathode 


C + 2O2- -> CO2 + 4e- at the anode 

Aluminum production: 

Textbooks on Electrochemistry may use different notation for the stoichiometry so be careful. The notation used for this case is:

aj = species on opposite side of e-

bj = species on the same side as e-

ne = electrons 

Keeping this in mind you will have aAl = 1, bAl3- = 1 and ne = 3. The Faraday`s law can be written in several different ways, but the one used here for the production of Al is:

I = -(ne/aAl)*F*(dnAl/dt) 

where:  I = current at the electrode (the cathode is this case)

            F = Faraday`s constant, 96,485.3 C mol-1

            dnAl/dt = the production of Al for a given time period 

Note: there is a negative sign, because electrons are used up. aAl in the denominator has to be substituted with bAl3- if calculating the required Al3- ions for the production of Alo

The equation can be changed to give the production of Al to:

dnAl/dt = (|- Icathode|*aAl)/(ne*F) in mol/s 

Why is it in mol/s? Take a look at the I/F term. The units for current are Amperes, which are C sec-1 and C mol-1 for the Faraday`s constant. So (C sec-1)/(C mol-1) = mol sec-1. Also, why is the current |-Icathode|? It is that way, because we know that I = I anode = |-Icathode|. 

At this point you can put all of the knowns in the above equation with I being the only unknown: 

dnAl/dt = (I*1)/(3*96,485.3) = I/289,455.9 mol/s 

Using the molecular weight of Al, we can convert the units to g/s: 

dnAl/dt = (I/289,455.9)*26.9815 g/mol = I/10,727.94 g/s 

For example, if we have current of 150,000 amps, the production rate will be: 

dnAl/dt = 150,000/10,727.94 = 13.98 g/s 

Don`t forget that there might be some side reactions taking place at the cathode, which will reduce the current at the cathode used in the production of Al. If the current efficiency is given, multiply the total current by it to obtain the cathodic one. 

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