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123172 Laboratory Module 1 Ouestions AlzOg is dissolved in molten cryolite and aluminium. What effect does a change in a) The cell Potential is one of the reactants in the manufacture of alumina...

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123172 Laboratory Module 1 Ouestions AlzOg is dissolved in molten cryolite and aluminium. What effect does a change in a) The cell Potential is one of the reactants in the manufacture of alumina concentration have on production l. b) The energy requirements for aluminium (Hint: this relates to part B of the lab work) Z. If you didn,t use a carbon anode but instead had an "inert" anode (these have been developed) then the reaction would be: 2AlzOz --' 4AI+ 3Oz so oxygen is evolved instead of COz a) In what form is the energy supplied Ui Witt this require 1nor.
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Module 1 Ouestions 123172 Laboratory of in the manufacture of the reactants and is one in molten cryolite is dissolved l. AlzOg on have in alumina concentration does a change What effect aluminium. a) The cell Potential production for aluminium requirements b) The energy the lab work) to part B of (Hint: this relates (these have been "inert" anode had an but instead a carbon anode you didn,t use Z. If be: reaction would then the developed) --' 4AI+ 3Oz 2AlzOz COz instead of is evolved so oxygen supplied is the energy a) In what form process? traditional than the electrical energy 1nor. oi leiJ Witt this require Ui (whY?) a process have? might such other advantages c) What work) part A of the lab relates to (Hint: this in hot aqueous be dissolved (Na3AlF6)' It can also fluoride in molten 3. AlzOl dissolves from aqueous of Al3* tnui. by electrolysis aluminium not Why is NaOH solution. potentials. standard electrode in terms of solution? Discuss the lab work) to part C of (Hint: this part relates

Answered Same Day Dec 23, 2021

Solution

David answered on Dec 23 2021
121 Votes
1) The reactions involved in the aluminum extraction process is –
   
 
 
2 3+
2 3
3+
2
2
2 2
2Al O 6O 4Al
At cathode: 4Al 12 4Al
At anode: 6O 3O 12
C +O CO
aq aq
aq e
aq e


 
 
 
 


a) To calculate the cell potential –
2
3
O0
4
3+
P0.0592
log
Al
cell cellE E
n
 
  
  

So for every increasing 1 M concentration of alumina, there would be increased concentration
and pressure of the products. So the ration of the P(O2) to Al3+ is...
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