The sulphide ore is roasted to oxide before reduction because the $$\Delta G_f^ \circ $$ of most of the sulphides are greater than those of $$C{S_2}$$ and $${H_2}S,$$ therefore neither $$C$$ nor $$H$$ can reduce metal sulphide to metal. Further, the standard free energies of formation of oxide are much less than those of $$S{O_2}.$$ Hence oxidation of metal sulphides to metal oxide is thermodynamically favourable.
Pig iron contains about 4% $$C$$ and many other impurities like $$S,P,Si$$ and $$Mn.$$ Cast iron is made by melting pig iron with scrap iron and coke. It has slightly lower carbon content (3%) and is extremely hard and brittle.
185.
Consider the following isolation/purification processes :
(i) Heating impure metal with $${I_2}$$ at $${150^ \circ }C - {250^ \circ }C$$ and passing the resulting volatile iodide on hot tungsten filament at $${1400^ \circ }C$$ to get the pure metal.
(ii) Heating the sulphide ore in air until a part is converted to oxide and then further heating in the absence of air to let the oxide react with unchanged metal sulphide to get the metal.
(iii) Electrolysis of the molten electrolyte containing metal oxide and cryolite or fluorspar to obtain the metal.
The processes used for obtaining aluminium, titanium and lead are respectively
$$\left( {\text{i}} \right)T{i_{\left( s \right)}}\left( {{\text{impure}}} \right) + 2{I_{2\left( g \right)}}$$ \[\xrightarrow{150-{{250}^{\circ }}C}Ti{{I}_{4}}\left( \text{volatile} \right)\] \[\xrightarrow[\text{Tungsten filamen}]{{{1400}^{\circ }}C}T{{i}_{\left( s \right)}}\left( \text{pure} \right)+2{{I}_{2\left( g \right)}}\]
$$\left( {{\text{ii}}} \right)2PbS + 3{O_2} \to 2PbO + 2S{O_2}$$
\[PbS+2PbO\xrightarrow[\text{temp}\text{.}]{\text{high}}3Pb+S{{O}_{2}}\]
$$\left( {{\text{iii}}} \right){\text{Cathode :}}\,A{l^{3 + }}\left( {{\text{melt}}} \right) + 3{e^ - }$$ $$ \to A{l_{\left( l \right)}}$$
$$\,\,\,\,\,\,\,\,\,\,{\text{Anode :}}\,{C_{\left( s \right)}} + {O^{2 - }}\left( {{\text{melt}}} \right)$$ $$ \to C{O_{\left( g \right)}} + 2{e^ - }$$
$$\,\,\,\,\,\,\,\,\,\,{C_{\left( s \right)}} + 2{O^{2 - }}\left( {{\text{melt}}} \right) \to $$ $$C{O_{2\left( g \right)}} + 4{e^ - }$$
186.
In this method, a low melting metal like tin can be made to flow on a sloping surface. In this way it is separated from higher melting impurities. The method is known as
On electrolysis of aqueous solution of $$s$$ - block elements $${H_2}$$ gas discharge at cathode.
At cathode: $${H_2}O + {e^ - } \to \frac{1}{2}{H_2} + O{H^ - }$$
188.
The ore that contains the metal in the form of fluoride is:
Fused alumina $$\left( {A{l_2}{O_3}} \right)$$ is a bad conductor of electricity. Therefore, cryolite $$\left( {N{a_3}Al{F_6}} \right)$$ and fluorspar $$\left( {Ca{F_2}} \right)$$ are added to purified alumina which not only make alumina a good conductor of electricity but also reduce the melting point of the mixture to around $$1140\,K.$$