Answer
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Hint:Electrolysis is defined as the chemical decomposition by the passage of electric current. This process is carried out in an electrolytic cell in which two electrodes are dipped in a solution containing positively and negatively charged ions.
Complete step by step answer:
Electrolysis is the process of chemical decomposition by the passage of electric current. Chemical formula of cryolite is $N{a_3}Al{F_6}$. It is used as an insecticide and pesticide. It is used to give a yellow color to fireworks. Fluorspar is the source of fluorine which is used to give non-stick coating to pans. The chemical formula of this compound is $Ca{F_2}$. In the extraction of aluminium from aluminium oxide $A{l_2}{O_3}$ melting point of electrolyte is very high$2000 - {2500^ \circ }C$. Due to which a high amount of electricity is spent in order to melt electrolytes. To lower the melting point cryolite and fluorspar are mixed with $A{l_2}{O_3}$ during electrolysis for extraction of aluminium which will be economical as well.
So, the correct answer is option C that is to lower the melting point and increase the conductivity of the electrolyte.
Note:
In electrorefining at anode oxidation reaction occurs and at cathode reduction reaction occurs. The anode is negative whereas the cathode is positive. Remember electrodes which will be consumed are made of impure metal (at the anode) on the other hand electrodes on which metal will be deposited will contain strips of pure metal (at the cathode).
Complete step by step answer:
Electrolysis is the process of chemical decomposition by the passage of electric current. Chemical formula of cryolite is $N{a_3}Al{F_6}$. It is used as an insecticide and pesticide. It is used to give a yellow color to fireworks. Fluorspar is the source of fluorine which is used to give non-stick coating to pans. The chemical formula of this compound is $Ca{F_2}$. In the extraction of aluminium from aluminium oxide $A{l_2}{O_3}$ melting point of electrolyte is very high$2000 - {2500^ \circ }C$. Due to which a high amount of electricity is spent in order to melt electrolytes. To lower the melting point cryolite and fluorspar are mixed with $A{l_2}{O_3}$ during electrolysis for extraction of aluminium which will be economical as well.
So, the correct answer is option C that is to lower the melting point and increase the conductivity of the electrolyte.
Note:
In electrorefining at anode oxidation reaction occurs and at cathode reduction reaction occurs. The anode is negative whereas the cathode is positive. Remember electrodes which will be consumed are made of impure metal (at the anode) on the other hand electrodes on which metal will be deposited will contain strips of pure metal (at the cathode).
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