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.

6   What is the role of depressant in froth-floatation process?

Solution :

The role of depressant is to prevent one type of sulphide ore particles from forming the froth with air bubbles. $NaCN$ is used as a depressant to separate lead sulphide $(PbS)$ ore from zinc sulphide $(ZnS)$ ore. $NaCN$ forms a zinc complex, $Na _2 [ Zn ( CN )_ 4] $on the surface of $ZnS$ thereby preventing it from the formation of the froth.$\\$ $4 NaCN + ZnS \to \underset{\text{Sodium tetracyno - zincate( II )}} {Na_ 2 [ Zn ( CN )_ 4 ]}+ Na_ 2 S$ $\\$ In this condition, only lead sulphide forms froth and thus can be separated from zinc sulphide ore.

7   Why is the extraction of copper from pyrites more difficult than that from its oxide ore through reduction?

Solution :

$\Delta _f G ^o$ of $Cu _2 S$ is more negative than $\Delta _ f G^ o$ of $CS_ 2 H_ 2 S$. So $Cu_ 2 S$ can not be reduced by carbon or hydrogen. $\Delta _ f G^ o$ of $CO_ 2$ is more negative than $\Delta _ f G^ o$ of $ Cu _2 O.$ So $ Cu_ 2 O$ can be reduced by carbon. So pyrites is first converted to oxide before reduction to copper.$\\$ $Cu _2 S _{( s )} +\dfrac{3}{2}O _{2 ( g )} \to Cu_ 2 O _{( s )} +SO _{2 ( g )}$ $\\$ $Cu _2 O_{ ( s )} +C _{( s )} \to 2 Cu _{( s )} +CO _{( s )}$

8   Explain:$\\$ (i) Zone refining$\\$ (ii) Column chromatography

Solution :

(i) $\text{Zone refining:}$ This method is used for production of semiconductors and other metals of very high purity, e.g., $Ge, Si, B, Ca$ and $In$. It is based on the principle that the impurities are more soluble in the molten state (melt) than in the solid state of the metal. The impure metal in the form of bar is heated at one end with a moving circular heater, as the heater is slowly moved along the length of the rod, the pure metal crystallises out of the melt whereas the impurities pass into the adjacent molten zone. This process is repeated several times till the impurities are completely driven to one end of the rod which is then cut off and discarded.$\\$ (ii)$\text{ Chromatography:}$ It is based on the principle that the different components of a mixture are adsorbed to different extents on an adsorbent. In column chromatography, an adsorbent, such as alumina $(Al _2 O_ 3 )$ or silica gel is packed in a column. This forms the stationary phase. The mixture to be separated is dissolved in a suitable solvent (mobile phase) and applied to the top of the column. The adsorbed components are extracted (eluted) from the column with a suitable solvent (eluent). The component which is more strongly adsorbed on the column takes longer time to travel through the column than a component which is weakly adsorbed. Thus, the various components of the mixture are separated as they travel through absorbent (stationary phase).

9   Out of $C $ and $CO$ which is a better reducing agent at $673K?$

Solution :

This can be explained thermodynamically, taking entropy and free energy changes into account$\\$ (a)$ C_{ ( s )} + O_ {2 ( g )} \to CO _{2 ( g )}$ $\\$ (b)$ 2 C_{ ( s )} + O_{ 2 ( g )} \to 2 CO _{( g )}$ $\\$ Case (i): Volume of $CO_ 2$ produced = Volume of $O_ 2 $ $\\$ Used. $\therefore \Delta S $ is very small and $\Delta G$ does not change with temperature.$\\$ $\therefore $ Plot of $\Delta GVsT$ is almost horizontal.$\\$ Case (ii): Volume of $CO$ produced = $2 ×$ volume of $O _2$ used.$\\$ $\therefore \Delta S$ is positive and hence $\Delta G$ becomes increasingly$\\$ Negative as the temperature increases.$\\$ $\therefore $ Plot of $\Delta ^o GVsT$ slopes downwards.$\\$

As can be seen from $\Delta ^o G VsT$ plot (Ellingham diagram), lines for the reactions,$\\$ $C\to CO_ 2$ and $C\to CO$ cross at $983 K.$ Below $983 K,$ the reaction$\\$ (a) is energetically more favourable but above $673 K,$ reaction.$\\$ (b) is favourable and preferred. Thus, below $673 K $ both $C$ and $CO$ can act as a reducing agent but since $CO $ can be more easily oxidised to $CO _2$ than $C$ to $CO _2$ , therefore, below $673K, CO$ is more effective reducing agent than carbon.

10   Name the common elements present in the anode mud in electrolytic refining of copper. Why are they so present?

Solution :

The common elements present in the anode mud are antimony, selenium, tellurium, silver, gold and platinum. These elements settle down under anode as anode mud because they are less reactive and are not effected by $ CuSO_ 4 -H _2 SO_ 4$ solution.