it is very useful

1. Principles of Non Ferrous
Extraction MetallurgyExtraction Metallurgy
Date 13th November, 2014

2. Decomposition potential (E) of Al2O3
dissolved in cryolite
o Situation is Al2O3 is in cryolite and you apply voltage
to decompose Al2O3
o We don’t decompose pure Al2O3 but Al2O3 in
solution (cryolite)
• You can express equation under ideal condition• You can express equation under ideal condition
0.5 Al2O3 (in cryolite)= Al(l)+0.75 O2 (g), ∆G1
• Now how to assume 0.5Al2O3 (in cryolite)!!
• Say you take, Al (l) + 0.75O2(g) =0.5 Al2O3 (c), ∆G2
• 0.5 Al2O3 (c) = 0.5 Al2O3 (l), ∆G3
• 0.5 Al2O3 (l)+ Na3AlF6 (l)= ½ Al2O3 (in cryolite), ∆G4
• Thermodynamics says, ∆G2+ ∆G3+ ∆G4= ∆G1

3. Decomposition potential (E) of Al2O3
dissolved in cryolite..contd
• From Ellingham diagram at 1000 °C for reaction
4/3Al(l) + O2(g)= 2/3Al2O3 (c), ∆G2 =-206k cal or
for per mole of Al, ∆G2 =-154.5k cal /mol
• Similar ∆G3 can be found( for condense state to• Similar ∆G3 can be found( for condense state to
liquid state) from standard data = 3800 cal
• For ∆G4, ∆G4= RT ln a1/2
Al2O3 for reaction
0.5 Al2O3 (l)+ Na3AlF6 (l)= ½ Al2O3 (in cryolite),
• Taking T = 1000 deg C and R =1.987 cal/deg C

4. Decomposition potential (E) of Al2O3
dissolved in cryolite..contd
• We are going to use one assumption here
• The bath contains 8% alumina and rest cryolite,
than mole fraction of alumina comes out to 0.15.
Assuming solution is ideal (activity is equal to
mole fraction)….leads to ∆G4= RTln(0.15) 1/2=-mole fraction)….leads to ∆G4= RTln(0.15) 1/2=-
1884 cal
• ∆G1= -(∆G2+ ∆G3+ ∆G4)=152584 cal
• ∆G1=23,066 ZE (considering F=23.061 kcal per
volt gram equivalent)[for Al deposition Valency is
Z=3]
• E comes out to be 2.20V

5. Decomposition potential (E) of Al2O3
dissolved in cryolite..contd
• In laboratory if we decompose of Al2O3 using
Al and Pt electrode than values comes out to
be 2.1 to 2.15 V (this value is marginally less
than observed by theoretical calculation)
Now let us try to answer this question by the
assumption we have used: bath contains 8%
alumina rest is cryolite and mole fraction to be
0.15

6. Apply Temkins model to Al2O3 in
Cryolite
• Al2O3 in 3NaF.AlF3
• According to Temkins approach there are 2Al3+,
3O2-, 3Na+ and 1AlF6
3-
• Consider you assume for simplicity 10 mole• Consider you assume for simplicity 10 mole
percent of Al2O3 in cryolite than 90 mole
percent is 3NaF.AlF3
• Activity of Al2O3= [ionic fraction of Al3+]2[ionic
fraction of O2-]3
• =[nAl3+/Σn+][nO2-/Σn-]

7. Apply Temkins model to Al2O3 in
Cryolite..contd..
• [2/2+27]2[3/3+9]3 ∼ zero
• This means you have considered 0.15 as
activity of Al2O3 and you have got the value of
2.21 V and now if you put above values for ∆G42.21 V and now if you put above values for ∆G4
than you may be very near to observed
experimental value in laboratory which comes
out to 2.1 V
• There is cognizance between the theoretical
and experimental value which assume Al
cathode and platinum anode

8. In practice we use Consumable graphite
• Consumable graphite means it reacts with oxygen to
either form CO or CO2..other wise if you use Pt
anode than oxygen gas formation will take place but
no consumption will take place (for this we require
large tonnage of Pt)large tonnage of Pt)
• This means the type of reaction we assume taking
place in cell will be 0.75O2(g) + 0.75C (s)= 0.75CO2
(g), ∆G5=-71.2kcal
• 0.75O2(g) + 1.5C(s)= 1.5CO (g), ∆G6=-81 kcal
In practice what do you think which of the above
two reaction will take place at higher temperature??

9. In practice we use Consumable
graphite..contd..
• ½ Al2O3 (in cryolite) + 0.75C= Al(l)+0.75 CO2
(g) OR
• ½ Al2O3 (in cryolite) + 1.5C= Al(l)+1.5 CO (g)• ½ Al2O3 (in cryolite) + 1.5C= Al(l)+1.5 CO (g)
• For both the reaction E comes out to be less
than 2 V..around 1.19 V for former and 1.05V
for later

10. Actual Decomposition potential
• For actual plant operation voltage is maintained at
5V to 7V..here are break ups
• Voltage needed for electrolytic reduction= 1.7V
• Voltage drop across carbon lining= 0.6V• Voltage drop across carbon lining= 0.6V
• Voltage drop due to anode resistance= 0.5V
• Voltage drop due to resistance of electrolyte= 1.8V
• Voltage drop due to contact resistance, joints= 0.5V
Total= 5.1V

11. What are advantages and
disadvantages of consumable graphite
electrodes??
Less decomposition voltage andLess decomposition voltage and
formation of CO and CO2

12. Soderberg consumable anode
US 7384521 B2
• Here you continuously feed at anode to put
paste of carbon to take care of graphite
consume during Hall-Heroult processconsume during Hall-Heroult process

13. Can we take care of oxygen coming at
anode
• Yes by injecting Hydrogen..you can make
steam..this will save your graphite..calculation have
indicated that decomposition potential comes out
to 1.3V…
• Even calculation for injecting methane at anode is• Even calculation for injecting methane at anode is
carried out which give decomposition potential to
be 1.06V with less amount of graphite consumption
• Such injection are only possible if we have
technology (injection reaction control), and
availability of Hydrogen and methane gas at
cheaper rates.

14. 2% of over all
energy in Al
production
goes in
manufacturingmanufacturing
of graphite
electrodes
Replaceable
electrodes

15. Cryolite is consumed continuously
and in most of the places natural
cryolite is not available thereforecryolite is not available therefore
synthetic process is must to pr
oduce cryolite

16. Synthesis
of Cryolite