General Principles and Processes of Isolation of Elements



1   Which of the ores mentioned can be concentrated by magnetic separation method?

Solution :

Ores which are magnetic in nature can be separated from non-magnetic gangue particles by magnetic separation method. For ex: ores of iron such as haemetite $(Fe_2 O_3 ),$ magnetite $(Fe_3 O_4 ),$ siderite $(FeCO_3 )$ and iron pyrites $(FeS_2 )$ being magnetic can be separated from non-magnetic silica and other impurities by magnetic separation method.

2   What is the significance of leaching in the extraction of aluminium?

Solution :

Aluminium contains silica $(SiO_2 )$, iron oxide $(Fe_2 O_3 )$ and titanium oxide $(TiO_4 )$ as impurities. These impurities can be removed by the process of leaching. During leaching, the powdered bauxite ore is heated with a concentrated ($45$%) solution of $NaOH $ at $473-523 K,$ where alumina dissolves as sodium meta aluminate and silica as sodium silicate leaving $Fe_2 O_3 , TiO_2$ and other impurities behind:$\\$ $Al_2O_3(s)+2NaOH(aq)+3H_2O(l)\overset{473-523K}{\to} \underset{\text{Sodium meta-aluminate}}{2Na[Al(OH)_4](aq)}$$\\$ $SiO_2(l)+2NaOH(aq) \overset{473-523K}{\to} \underset{\text{Sodium silicate}}{Na_2SiO_3(aq)+H_2O(l)}$$\\$ The impurities are filtered off and solution of sodium meta-aluminate is neutralised by passing $CO_2$ when hydrated alumina separates out while sodium silicate remains in solution. The hydrated alumina thus obtained is filtered, dried and heated to give back pure alumina.$\\$ $Al_2O_3, xH_2O(s)\overset{1473k}{\to}Al_2O_3(s)+xH_2O(g)$$\\$ Thus, by leaching, pure alumina can be obtained from bauxite ore.

3   The reaction, $Cr_2 O_3 + 2 Al \to Al_2 O_3 + 2Cr (\Delta G ^\theta = -421 kJ )$ is thermodynamically feasible as is apparent from the Gibbs energy value. Why does it not take place at room temperature?

Solution :

This is explained on the basis of $K_{eq}$ , the equilibrium constant. In the given redox reaction, all reactants and products are solids at room temperature, so, there is no equilibrium between the reactants and products and hence the reactions does not occur at $RT$. At high temperature, $Cr$ melts and values of $T\Delta S $ increases. As a result, the value of $\Delta r G^\theta$ becomes more negative and hence the reaction proceeds rapidly.

4   Is it true that under certain conditions, $Mg$ can reduce $A1_2 O_3$ and $Al$ can reduce $MgO$? What are those conditions?

Solution :

Yes, below $1350^0 C, Mg$ can reduce $Al_2 O_3$ and above $1350^0 C, Al$ can reduce $MgO.$ This can be inferred from $\Delta G^o$ vs $T$ plots.

5   Copper can be extracted by hydrometallurgy but not zinc. Explain.

Solution :

Copper can be extracted by hydrometallurgy but not zinc, this is because$E^o_{Zn^{2+}/Zn}=-0.76V$ is lower than that of $E^o_{Cu^{2+}/Cu}=0.34V$.Hence, zinc can replace $Cu$ from solution of $Cu ^{+2}$ ions.$\\$ $Zn(s)+Cu^{+2}(aq) \to Zn^{+2}(aq)+Cu(s)$$\\$ In order to displace zinc from zinc solution, a more reactive metal is required, such as$\\$ $Al(E^o_{Al^{3+}/Al}=-1.66V),Mg(E^o_{Mg^{2+}/Mg}=-2.37V),$$\\$ $Ca(E^o_{Ca^{2+}/Ca}=-2.87V),K(E^o_{K^+/K}=-2.93V),$$\\$ But with water, these metals $(Al, Mg, Ca$ and $K)$ forms their corresponding ions with the evolution of $H_2$ gas. Thus, $Al, Mg, Ca, K,$ etc., cannot be used to displace zinc from zinc solution, and only copper can be extracted by hydrometallurgy but not the zinc.