Describe the hall heroilt process for electrolytic refining of al
Answers
Answered by
0
The Hall–Héroult process is the major industrial process for smelting aluminium. It involves dissolving aluminium oxide (alumina)(obtained most often from bauxite, aluminium's chief ore, through the Bayer process) in molten cryolite, and electrolysingthe molten salt bath, typically in a purpose-built cell. The Hall–Héroult process applied at industrial scale happens at 940–980°C and produces 99.5–99.8% pure aluminium. Recycled aluminum requires no electrolysis, thus it does not end up in this process............!!!!!!!!!!!!!!!
pls mark BRILLIANTLEAST
pls mark BRILLIANTLEAST
Answered by
0
Hey !!
Here is your answer...
➡ HALL HEROULT METHOD
↪ Aluminum oxide is reduced to aluminum metal in a special electrolytic cell invented by Hall Heroult.
↪ In the year 1886, American Chemist Charles Martin Hall and French Chemist Paul Heroult developed electrolysis method for production of aluminium.
↪ Electric current cannot pass through solid form of alumina. It's melting point is also very high ( 2348 k temperature ). Electrolysis at such high temperature is very expensive.
↪ Cryolite ( Na3AIF6 ) is added so that electrolysis can be carried out easily. This mixture works as better electric conductor than molten alumina. The melting point can be brought still lower by adding feldspar ( CaF2 ).
↪ The mixture of alumina ( AL2O3 ) and Cryolite are electrolysed in an iron vessel having inner surface carbon layered.
↪ In the cell, the rods of carbon are joined by copper clamp as anode and carbon layered graphite as cathode.
↪ On passing electric current molten aluminum is deposited on cathode and dioxgen gas is produced at the anode. The molten aluminum collected at the bottom of the cell is taken out.
↪ During reaction, the positive ion AL3+ het reduced on cathode (-) where negative ion O2- get oxidised at anode (+) in presence of carbon to form CO2.
C + O2 ==> CO2
↪ Reaction of cathode and anode are as follows :-
Cathode :- 2AL3+ + 6e- ====> 2AL
Anode :- 6O2- ===> 3O2 + 12e-
Hope it helps you....
THANKS ^-^
Here is your answer...
➡ HALL HEROULT METHOD
↪ Aluminum oxide is reduced to aluminum metal in a special electrolytic cell invented by Hall Heroult.
↪ In the year 1886, American Chemist Charles Martin Hall and French Chemist Paul Heroult developed electrolysis method for production of aluminium.
↪ Electric current cannot pass through solid form of alumina. It's melting point is also very high ( 2348 k temperature ). Electrolysis at such high temperature is very expensive.
↪ Cryolite ( Na3AIF6 ) is added so that electrolysis can be carried out easily. This mixture works as better electric conductor than molten alumina. The melting point can be brought still lower by adding feldspar ( CaF2 ).
↪ The mixture of alumina ( AL2O3 ) and Cryolite are electrolysed in an iron vessel having inner surface carbon layered.
↪ In the cell, the rods of carbon are joined by copper clamp as anode and carbon layered graphite as cathode.
↪ On passing electric current molten aluminum is deposited on cathode and dioxgen gas is produced at the anode. The molten aluminum collected at the bottom of the cell is taken out.
↪ During reaction, the positive ion AL3+ het reduced on cathode (-) where negative ion O2- get oxidised at anode (+) in presence of carbon to form CO2.
C + O2 ==> CO2
↪ Reaction of cathode and anode are as follows :-
Cathode :- 2AL3+ + 6e- ====> 2AL
Anode :- 6O2- ===> 3O2 + 12e-
Hope it helps you....
THANKS ^-^
Attachments:
Similar questions