During electrolytic reduction of alumina...

During electrolytic reduction of alumina, two auxillary electrolytes X and Y are added to increase the electrical conductance and lower the temperature of melt in order to making fused mixture very conducting. X and Y are :

cryolite and flourspar

cryolite and alum

alum and flourspar

flourspar and bauxite

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The reason for increase in electrical conduction of a weak electrolyte with increase in temperature

Assertion: In Hall-Heroult process aluminium is extracted by the electrolytic reduction of alumina dissolved in molten cryolite or fluorspar. Reason: The cryolite ore fluorspar lower the melting point of melt an dmake it more conducting

Conductors allow the passage of electric current through them. Metallic and electrolytic are the two type of conductors. Current carriers in metallic and electrolytic conductors are free electrons and free ions respectively. Specific conductance or consuctivity of the electrolyte solution is given by the following relation : kappa=c xx l/A where, c=1//R is the conductance and l//A is the cell constant. Molar conductance (Lambda_(m)) and equivalence conductance (Lambda_(e)) of an electrolyte solution are calculated using the following similar relations : Lambda_(m)= kappa xx 1000/M Lambda_(e)= kappa xx 1000/N Where, M and N are the molarity and normality of the solution respectively. Molar conductance of strong electrolyte depends on concentration : Lambda_(m)=Lambda_(m)^(@)-b sqrt(c) where, Lambda_(m)^(@)= molar conductance at infinite dilution c= concentration of the solution b= constant The degrees of dissociation of weak electrolytes are calculated as : alpha=Lambda_(m)/Lambda_(m)^(@)=Lambda_(e)/Lambda_(e)^(@) The correct order of equivalent conductances at infinite dilution of LiCl, NaCl and KCl is :

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