
Assertion: In Hall’s process for preparation of $Al$, graphite anode has a long life.
Reason: ${O_2}$ is liberated in the process at the graphite anode.
(A) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(B) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
(C) Assertion is true but Reason is false.
(D) Assertion is false but Reason is true.
(E) Both Assertion and Reason are false.
Answer
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Hint: As we are already aware that the Hall-Heroult process is an industrial technique for smelting Aluminium and involves a mixture of aluminium oxide, aluminium fluoride, flux or calcium fluoride, cryolite and sodium hexafluoroaluminate which is electrolysed using steel cathode lined with carbon and graphite anode.
Complete Step by step answer: We know that the Hall-Heroult process is an industrial technique for extracting Aluminium from its ores by melting the ores and then refining the metal from the melt. This process includes a mixture of aluminium oxide, aluminium fluoride, flux or calcium fluoride, cryolite and sodium hexafluoroaluminate, which is electrolysed using steel cathode lined with carbon and graphite anode. Molten aluminium is tapped from the bottom of this mixture after electrolysation.
The reaction can be written as:
At cathode: $A{l^{3 + }}(melt) + 3{e^ - } \to Al(l)$
At anode: $
C(s) + {O^{2 - }}(melt) \to CO(g) + 2{e^ - } \\
C(s) + 2{O^{2 - }}(melt) \to C{O_2}(g) + 4{e^ - } \\
$
So we can write the overall reaction involved as:
$2A{l_2}{O_3} + 3C \to 4Al + 3C{O_2}$
As we can see that the electrolysis of this reaction results in oxygen liberation at anode that reacts with some amount of graphite anode to form carbon dioxide and the rest of the amount disintegrates into a fine powder making the solution black. Hence, carbon is burnt and results in replacement of graphite anode periodically. And thus graphite anodes do not have a long life.
Therefore, from the above explanation we can say that the Assertion is wrong and the Reason is correct.
So the correct answer is (D).
Note: Though graphite anodes are replaced periodically in the Hall’s process but still they are used in the reaction because they prevent the oxidation of aluminium by oxygen evolved during electrolysis and they are also a good conductor of electricity thus very useful.
Complete Step by step answer: We know that the Hall-Heroult process is an industrial technique for extracting Aluminium from its ores by melting the ores and then refining the metal from the melt. This process includes a mixture of aluminium oxide, aluminium fluoride, flux or calcium fluoride, cryolite and sodium hexafluoroaluminate, which is electrolysed using steel cathode lined with carbon and graphite anode. Molten aluminium is tapped from the bottom of this mixture after electrolysation.
The reaction can be written as:
At cathode: $A{l^{3 + }}(melt) + 3{e^ - } \to Al(l)$
At anode: $
C(s) + {O^{2 - }}(melt) \to CO(g) + 2{e^ - } \\
C(s) + 2{O^{2 - }}(melt) \to C{O_2}(g) + 4{e^ - } \\
$
So we can write the overall reaction involved as:
$2A{l_2}{O_3} + 3C \to 4Al + 3C{O_2}$
As we can see that the electrolysis of this reaction results in oxygen liberation at anode that reacts with some amount of graphite anode to form carbon dioxide and the rest of the amount disintegrates into a fine powder making the solution black. Hence, carbon is burnt and results in replacement of graphite anode periodically. And thus graphite anodes do not have a long life.
Therefore, from the above explanation we can say that the Assertion is wrong and the Reason is correct.
So the correct answer is (D).
Note: Though graphite anodes are replaced periodically in the Hall’s process but still they are used in the reaction because they prevent the oxidation of aluminium by oxygen evolved during electrolysis and they are also a good conductor of electricity thus very useful.
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