(b) Magnesium is extracted from its ore by electrolytic reduction process.
(c) Cryolite is used in metallurgy of aluminium in order to increase the fusibility and conductivity
(d) I Hall's process bauxite is leached with the help of ${Na} _{2}{CO} _{3}$. statements are correct. The concentrated ore is converted into metal oxide by calcination or roasting.

Pure alumina melts at about $2000^oC$ and is a bad conductor of electricity. If fused cryolite and fluorspar is added, the mixture melts at $900^0C$ and solution becomes a good conductor of electricity. 

Pure alumina melts at about $2000^oC$ and is a bad conductor of electricity. If fused cryolite and fluorspar is added, the mixture melts at $900^oC$ and solution becomes a good conductor of electricity. 

In the metallurgy of aluminium, purified $Al _2O _3$ is mixed with $Na _3AlF _6$ or $CaF _2$ which lowers the melting point of the mix and brings conductivity.

The electrolytic mixture is sprinkled with coke to prevent the anode's oxidation by the oxygen evolved

To prevent corrosion, coke powder is kept at the top which also prevents escape of heat.

Aluminium is used for making cooking utensils,the properties used.

The chemical reaction for the Hall process is given by

Both are graphite 

As Alumina has high melting point and low conductivity, Cryolite is added to reduce melting point and flour spar is added in small amount to increase electrical conductivity. 

In the extraction of aluminium by electrolytic process alumina $Al _2O _3$ is used along with molten fluorspar and cryolite.

Option A & C are correct.
But Serpeck's  Method is used to purify white bauxite.
and
Mg is extracted by electrolysis of anhydrous $MgCl _{2}$.

The cathode used in the electrolytic cell during aluminium extraction is graphite lining (carbon lining).

Pure alumina melts at about $2000^oC$ and is a bad conductor of electricity. If fused cryoliteand fluorspar is added, the mixture melts at $900^oC$ and $Al _2O _3$ becomes a good conductorof electricity.
Electrolyte: 60 parts cryolite + 20 parts fluorspar + 20 parts pure  $Al _2O _3$

In the Hall–Héroult process alumina, $Al _2O _3$, is dissolved in molten cryolite, $Na _3AlF _6$ to lower it's melting point for easier electrolysis. It also increases conductivity of the solution.

The electrolysis of pure alumina faces some difficulties. Pure alumina is a bad conductor of electricity. The fusion temperature of pure alumina is about 2000$^{\circ}$C and at this temperature when electrolysis is carried out on the fused mass, the metal formed vaporizes, as the boiling point of aluminium is 1800$^{\circ}$C. These difficulties are overcome by using a mixture containing alumina, cryolite $(Na _3,AlF _6)$, and fluorspar $(CaF _2,)$. But the addition of cryolite do not minimize the anode effect and do not remove the impurities from alumina.

Aluminium oxide has a very high melting point (over 2,000C), so it would be expensive to melt it. Instead, it is dissolved in molten cryolite, an aluminium compound with a lower melting point than aluminium oxide. The use of cryolite reduces some of the energy costs involved in extracting aluminium.

Hall's Process: (Red bauxite)

There are 7 different aluminium smelting plant in India.

A) In the furnace, first the coke reacts with oxygen in the air blast to produce carbon monoxide(oxidation):$2C+O2\rightarrow 2CO$ The carbon monoxide reduces the iron ore (hematite) to molten iron, becoming carbon dioxide in the process:$3CO + Fe _2O _3\rightarrow 2Fe + 3CO _2$
B) The HallHroult process is the major industrial process for smelting of alumina. It involves dissolving alluminium oxide(allumina) in molten cryolite, and electrolysing the molten salt bath, typically in a purpose-built cell.
C) Lead is extracted from its chief ore galena by both carbon reduction as well as self reduction,
D) Haematitic($Fe _2O _3$),cassiterite ($SnO _2$) and argentite($Ag _2S$).So Haematite and cassiterite are oxide ores but argentite is sulphide ore.Incorrect.
Hence options A,B & C are correct.

The given statement is true as $Al$ is extracted from alumina($ Al _{2}O _{3}$) by the process of electrolysis.

$Al _2O _3+2NaOH\longrightarrow 2NaAlO _2+H _2O$

The electrolyte is a molten bath of cryolite and dissolved alumina. Cryolite is a good solvent for alumina with low melting point. The molten electrolyte is maintained at high temperature inside the insulating cell by using high amount of coke powder or dust to prevent the heat loss.

Aluminium is extracted by electrolysis. The ore is first converted into pure aluminium oxide by the Bayer Process and this is then electrolysed in solution in molten cryolite another aluminium compound. The aluminium oxide has too high a melting point to electrolyse on its own. The usual aluminium ore is bauxite.

The electrolytes are used to reduce melting point and increase conductivity.

$NaAlO _2$ on hydrolysis gives sodium hydroxide and $Al _2O _3$.

The substance which is used as an electrolyte during electrolysis of aluminium is $Al _2O _3, Na _3AlF _6, CaF _2$.

Ionization of Alumina in Electrolytic refining.

Most abundant metal on the earth's crust is Aluminium.

Naturally occurring aluminium is a major component of silicates such as mica and clay. Aluminium is extracted mainly from its principal ore bauxite.

In the metallurgy of Aluminium, purified $Al _2O _3$ is mixed with $Na _3AlF _6$ or $CaF _2$ which lowers the melting point of mix and brings conductivity.

Purification of aluminium by electrolytic refining is called Hoope's process. Serpeck, Hall's and Baeyer process are chemical process.

Bayer's process is used for beneficiation of red bauxite.
Serpeck's process is used for beneficiation of white bauxite. 
Reaction to remove impurity from white bauxite is:
$SiO _2+2C\rightarrow Si+2CO$ .

$Al _2O _3 + 2NaOH + 3H _2O \rightarrow 2Na[Al(OH) _4]$
$Al _2O _3 + 6HCl \rightarrow 2AlCl _3 + 3H _2O$

Net reaction in Hall-Heroult process is

$3C\,+\,2Al _2O _3\,\rightarrow\,4Al\,+\,3CO _2$

Or $4Al^{3+}\,+\,12e^{-}\,\rightarrow \,4Al,$

Number of electrons (n) = 12

$\Delta\,G^{\circ }\,=\,3 \Delta\, G^{\circ } _f (CO _2)\,-\,2 \Delta\,G^{\circ } _f(Al _2O _3)$

$=-3\,\times\,394\,-\,2(-1520)$

$\Delta G^{\circ}\,=\, -nFE^{\circ} _{cell}$

$-E^{\circ} _{cell}\,=\, \displaystyle \frac {\Delta G^{\circ}}{nF}\, =\, \displaystyle\frac {1858\,\times\, 1000}{12\, \times\, 96500}$

$= 1.60\ V$

The electrolysis of alumina by Hall and Heroult's process is carried by using a fused mixture of alumina and cryolite  along with minor quantities of aluminum fluoride and fluorspar. The addition of cryolite and fluorspar increases the electrical conductivity of alumina and  lowers the fusion temperature. Thus, the option A is correct. Also at anode, alumina reacts with fluorine to give oxygen. The liberated oxygen reacts with carbon to form $CO$ and $CO _2$. These gases are liberated at anode.

Pure alumina melts at about $2000^oC$ and is a bad conductor of electricity. If fused cryolite and fluorspar is added, the mixture melts at $900^0C$ and $Al _2O _3$ becomes a good conductor of electricity.