
Which of the following reactions is a part of the Hall’s process?
A.${\text{A}}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}} + 2{\text{NaOH}} \to {\text{2NaAl}}{{\text{O}}_2} + {{\text{H}}_2}{\text{O}}$
B.${\text{F}}{{\text{e}}_{\text{2}}}{{\text{O}}_{\text{3}}} + 2{\text{Al}} \to 2{\text{Fe}} + {\text{A}}{{\text{l}}_2}{{\text{O}}_3}$
C.${\text{AlN}} + 3{{\text{H}}_2}{\text{O}} \to {\text{Al}}{\left( {{\text{OH}}} \right)_3} + {\text{N}}{{\text{H}}_3}$
D.${\text{A}}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}}.2{{\text{H}}_{\text{2}}}{\text{O}} + 2{\text{N}}{{\text{a}}_2}{\text{C}}{{\text{O}}_3} \to {\text{C}}{{\text{O}}_2} + 2{\text{NaAl}}{{\text{O}}_2} + 2{{\text{H}}_2}{\text{O}}$
Answer
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Hint:The Hall-Heroult’s Process is the major industrial process for smelting of aluminium. It involves dissolving alumina in cryolite and electrolysing the resultant solution.
Complete step by step answer:
In the Hall-Heroult’s process, the alumina obtained by the refining of the ore, bauxite is mixed with cryolite and subjected to electrolysis. The reason for adding cryolite is that aluminum melts at very high temperatures of ${2300^0}{\text{C}}$ and the addition of cryolite lowers the temperature to around ${700^0}{\text{C}}$. As a result of the electrolysis, aluminium is discharged at the cathode while carbon dioxide is discharged at the anode because graphite electrodes are used for the purpose of electrolysis.
The overall cell reaction is as follows:
$2{\text{A}}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}} + 3{\text{C}} \to 4{\text{Al}} + 3{\text{C}}{{\text{O}}_2}$
Before the alumina obtained from bauxite is subjected to electrolysis it is purified by the use of sodium carbonate to obtain sodium aluminate, which is reduced by carbon dioxide and then heated to give pure alumina. The reaction is as follows:
${\text{A}}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}}.2{{\text{H}}_{\text{2}}}{\text{O}} + 2{\text{N}}{{\text{a}}_2}{\text{C}}{{\text{O}}_3} \to {\text{C}}{{\text{O}}_2} + 2{\text{NaAl}}{{\text{O}}_2} + 2{{\text{H}}_2}{\text{O}}$
Hence, the correct option is D.
Note:
The process of aluminium extraction from its ore bauxite, commonly called the Bayer’s Process, involves a number of steps. Initially, the metal ore is dissolved in a solution of caustic soda, from where sodium aluminate is obtained. Sodium aluminate is subjected to hydrolysis followed by heating to get pure alumina from which aluminium is extracted.
Complete step by step answer:
In the Hall-Heroult’s process, the alumina obtained by the refining of the ore, bauxite is mixed with cryolite and subjected to electrolysis. The reason for adding cryolite is that aluminum melts at very high temperatures of ${2300^0}{\text{C}}$ and the addition of cryolite lowers the temperature to around ${700^0}{\text{C}}$. As a result of the electrolysis, aluminium is discharged at the cathode while carbon dioxide is discharged at the anode because graphite electrodes are used for the purpose of electrolysis.
The overall cell reaction is as follows:
$2{\text{A}}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}} + 3{\text{C}} \to 4{\text{Al}} + 3{\text{C}}{{\text{O}}_2}$
Before the alumina obtained from bauxite is subjected to electrolysis it is purified by the use of sodium carbonate to obtain sodium aluminate, which is reduced by carbon dioxide and then heated to give pure alumina. The reaction is as follows:
${\text{A}}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}}.2{{\text{H}}_{\text{2}}}{\text{O}} + 2{\text{N}}{{\text{a}}_2}{\text{C}}{{\text{O}}_3} \to {\text{C}}{{\text{O}}_2} + 2{\text{NaAl}}{{\text{O}}_2} + 2{{\text{H}}_2}{\text{O}}$
Hence, the correct option is D.
Note:
The process of aluminium extraction from its ore bauxite, commonly called the Bayer’s Process, involves a number of steps. Initially, the metal ore is dissolved in a solution of caustic soda, from where sodium aluminate is obtained. Sodium aluminate is subjected to hydrolysis followed by heating to get pure alumina from which aluminium is extracted.
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