Electrolytic reduction of alumina to aluminiu | vedclass

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Question

(C) The Hall-Heroult process involves the electrolytic reduction of alumina $(Al_2O_3)$.Pure alumina has a very high melting point and is a poor condu

Vedclass · Accepted Answer

Cryolite which forms a melt with lower melting temperature

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(C) The Hall-Heroult process involves the electrolytic reduction of alumina $(Al_2O_3)$.Pure alumina has a very high melting point and is a poor conductor of electricity.To overcome this,alumina is dissolved in a fused mixture of cryolite $(Na_3AlF_6)$ and fluorspar $(CaF_2)$.This mixture serves two main purposes:$1$. It lowers the melting point of the mixture to approximately $1240 \ K$.$2$. It increases the electrical conductivity of the melt.Therefore,the correct option is $C$.

Electrolytic reduction of alumina to aluminium by the Hall-Heroult process is carried out in the presence of:

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