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The function of fluorspar in the electrolytic reduction of alumina dissolved in fused cryolite $(Na_3AlF_6)$ is :
as a catalyst
to lower the temperature of melt and to make the fused mixture very conducting
to decrease the rate of oxidation of carbon anode
none of the above
Solution
The electrolysis of alumina is carried out in a steel tank lined inside with graphite.
The graphite lining serves as a cathode.
The anode is also made of graphite rods hanging in the molten mass.
The electrolyte consists of alumina dissolved in fused cryolite(N a $_{3} \mathrm{AlF}_{6}$ ) and fluorspar ($\left.\mathrm{CaF}_{2}\right)$
Cryolite lowers the melting point of alumina to $950^{\circ} \mathrm{C}$ and fluorspar increases the fluidity of the mass, so that the liberated aluminium metal may sink at the bottom of the cell.
Therefore, it makes the fused mixture very conducting and lowers the fusion temperature of the melt.
When an electric current is passed through this mixture, the aluminium is collected at the cathode in the molten state and sinks at the bottom and is tapped off.
Similar Questions
Match List $I$ with List $II$. Choose the correct answer from the options given below:
| List $-I$ | List $-II$ |
|
$A.$ Melting point $[\mathrm{K}]$ |
$I.$ $\mathrm{Tl}>\mathrm{In}>\mathrm{Ga}>\mathrm{Al}>\mathrm{B}$ |
|
$B.$ Ionic Radius $\left[\mathrm{M}^{+3} / \mathrm{pm}\right]$ |
$II.$ $\mathrm{B}>\mathrm{Tl}>\mathrm{Al} \approx \mathrm{Ga}>\mathrm{In}$ |
| $C.$ $\Delta_{\mathrm{i}} \mathrm{H}_1 $ $ [\mathrm{~kJ} \mathrm{~mol}^{-1}]$ | $III.$ $\mathrm{Tl}>\mathrm{In}>\mathrm{Al}>\mathrm{Ga}>\mathrm{B}$ |
| $D.$ Atomic Radius $[pm]$ | $IV.$ $\mathrm{B}>\mathrm{Al}>\mathrm{Tl}>\mathrm{In}>\mathrm{Ga}$ |
