Principles of Non Ferrous

Extraction MetallurgyExtraction Metallurgy

Date 13th November, 2014

Decomposition potential (E) of Al2O3

dissolved in cryoliteo 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

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 a1/2Al2O3 for reaction

0.5 Al2O3 (l)+ Na3AlF6 (l)= ½ Al2O3 (in cryolite),

• Taking T = 1000 deg C and R =1.987 cal/deg C

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

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

Apply Temkins model to Al2O3 in

Cryolite

• Al2O3 in 3NaF.AlF3

• According to Temkins approach there are 2Al3+,

3O2-, 3Na+ and 1AlF63-

• 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-]

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 ∆G4 2.21 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

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??

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

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

What are advantages and

disadvantages of consumable graphite

electrodes??

Less decomposition voltage and Less decomposition voltage and

formation of CO and CO2

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

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.

2% of over all

energy in Al

production

goes in

manufacturing manufacturing

of graphite

electrodes

Replaceable

electrodes

Cryolite is consumed continuously

and in most of the places natural

cryolite is not available therefore cryolite is not available therefore

synthetic process is must to pr

oduce cryolite

Synthesis

of Cryolite