Tag: extraction of metals by electrolysis

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

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.

Fluorspar is the substance act as a solvent and it increases conductivity. (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  $Al _2O _3$ 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 $900^0C$ and $Al _2O _3$ becomes a good conductor of electricity.

Electroplating is the process of plating one metal onto another, for decorative purposes or to prevent corrosion of a metal.