
Why is cryolite added during the electrolysis of alumina?
A.It lowers the melting point of alumina
B.It increases the electrical conductivity
C.It minimizes the anode effect
D.It removes the impurities from alumina
Answer
574.2k+ views
Hint: The solution of alumina and cryolite during the electrolysis gives the aluminium at cathode and oxygen at anode. The alumina is a poor conductor of electricity and the fusion temperature of alumina is about ${2000^ \circ }C$.
Complete step by step answer:
Extraction of aluminium is done by aluminium ore which is known as bauxite, $A{l_2}{O_3}$. It is purified to yield a white powder which is known as aluminium oxide or alumina.
Aluminium oxide is an ionic compound when it melts $A{l^{3 + }}$ and ${O^{2 - }}$ ions conduct electricity and they are free to move.
Alumina and cryolite solution is less dense than aluminium metal and therefore, it falls at the bottom of the cell from where it is tapped off as pure liquid metal.
$2A{l_2}{O_3}\left( l \right) \to 4Al\left( l \right) + 3{O_2}\left( g \right)$
Oxygen is given off at positive carbon anode. Hot oxygen reacts with carbon anode and forms carbon dioxide, so carbon dioxide is given off at carbon anode.
$C\left( s \right) + {O_2}\left( g \right) \to C{O_2}\left( g \right)$
The electrolysis of alumina has some difficulties. Alumina is the poor conductor of electricity and fusses at the temperature of ${2000^ \circ }C$ and when electrolysis at this temperature is done on fused mass the metal which is formed get vaporized because the boiling point of aluminium is ${1800^ \circ }C$. Therefore, these problems are solved by using the mixture of alumina, cryolite and fluorspar. Hence, cryolite does not minimize anode effect and does not remove impurities from alumina.
So, the correct answer is Option A,B.
Note:
In the process of electrolysis:
Positive charged aluminium can gain electrons from the cathode and forms molten aluminium.
Oxide ions lose electrons at anode and form oxygen molecules.
Complete step by step answer:
Extraction of aluminium is done by aluminium ore which is known as bauxite, $A{l_2}{O_3}$. It is purified to yield a white powder which is known as aluminium oxide or alumina.
Aluminium oxide is an ionic compound when it melts $A{l^{3 + }}$ and ${O^{2 - }}$ ions conduct electricity and they are free to move.
Alumina and cryolite solution is less dense than aluminium metal and therefore, it falls at the bottom of the cell from where it is tapped off as pure liquid metal.
$2A{l_2}{O_3}\left( l \right) \to 4Al\left( l \right) + 3{O_2}\left( g \right)$
Oxygen is given off at positive carbon anode. Hot oxygen reacts with carbon anode and forms carbon dioxide, so carbon dioxide is given off at carbon anode.
$C\left( s \right) + {O_2}\left( g \right) \to C{O_2}\left( g \right)$
The electrolysis of alumina has some difficulties. Alumina is the poor conductor of electricity and fusses at the temperature of ${2000^ \circ }C$ and when electrolysis at this temperature is done on fused mass the metal which is formed get vaporized because the boiling point of aluminium is ${1800^ \circ }C$. Therefore, these problems are solved by using the mixture of alumina, cryolite and fluorspar. Hence, cryolite does not minimize anode effect and does not remove impurities from alumina.
So, the correct answer is Option A,B.
Note:
In the process of electrolysis:
Positive charged aluminium can gain electrons from the cathode and forms molten aluminium.
Oxide ions lose electrons at anode and form oxygen molecules.
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