Chemistry - s Block Elements Question with Solution | TestHub

Sodium metal It is manufactured by electrolysis of a fused mixture of$\text{NaCl}\text{(40\%)}$and${\text{CaCl}}_{\text{2}}\text{(60\%)}$in Down's cell at$\text{873}\text{K}\text{.}$$\text{Na}$liberated at the cathode is collected, in kerosene oil while${\text{Cl}}_{\text{2}}$is evolved at the anode.

$\begin{array}{cccc}& \text{NaCl}\left(l\right)& \stackrel{\text{Electrolysis}}{\to }& {\text{Na}}^{+}+{\text{Cl}}^{-}\\ \text{At cathode}& {\text{Na}}^{+}+{\text{e}}^{-}& ⟶& \text{Na}\left(l\right)\\ \text{At anode}& 2{\text{Cl}}^{-}& ⟶& {\text{Cl}}_2\left(\text{g}\right)+2{\text{e}}^{-}\end{array}$

Sodium hydroxideIt is manufactured by electrolysis of an aqueous solution of$\text{NaCl}$(brine) in Castner - Kellner cell using mercury cathode and carbon anode. Sodium metal which is discharged at the cathode combines with mercury to form sodium amalgam.${\text{Cl}}_{\text{2}}\text{.}$

$\begin{array}{cccc}\text{At cathode}& {\text{Na}}^{+}+{\text{e}}^{-}& \stackrel{\text{Hg}}{⟶}& \text{Na}-\text{amalgam}\\ \text{At anode}& 2{\text{Cl}}^{-}& ⟶& {\text{Cl}}_2+2{\text{e}}^{-}\end{array}$

The sodium amalgam thus obtained is treated with water to form sodium hydroxide and hydrogen gas.

$\begin{array}{ccc}2\text{Na}-\text{amalgam}+2{\text{H}}_2\text{O}& ⟶& 2\text{NaOH}+2\text{Hg}+{\text{H}}_2\end{array}$

Sodium carbonateIt is obtained by Solvay ammonia process. When${\text{CO}}_{\text{2}}$is passed through a concentrated solution of brine (aqueous solution of$\text{NaCl}$) saturated with ammonia,${\text{NaHCO}}_{\text{3}}$gets precipitated.${\text{NaHCO}}_{\text{3}}$on subsequent heating gives${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\text{.}$

$\begin{array}{ccccc}\text{NaCl}+{\text{NH}}_3& +& {\text{CO}}_2+{\text{H}}_2\text{O}& ⟶& {\text{NaHCO}}_3\downarrow +{\text{NH}}_4\text{Cl}\\ & & 2{\text{NaHCO}}_3& \stackrel{\Delta }{⟶}& {\text{Na}}_2{\text{CO}}_3+{\text{CO}}_2+{\text{H}}_2\text{O}\end{array}$

It reacts with${\text{Cl}}_{\text{2}}$to form calcium hypochlorite,${\text{Ca(OCl)}}_{\text{2}}$

$\begin{array}{ccccccc}\text{2Ca}{\left(\text{OH}\right)}_{\text{2}}& \text{+}& {\text{2Cl}}_{\text{2}}& \text{→\hspace{0.33em}}& \stackrel{}{\underset{}{{\text{CaCl}}_{\text{2}}\text{+Ca}{\left(\text{OCl}\right)}_{\text{2}}}}& \text{+}& {\text{2H}}_{\text{2}}\text{O}\\ \text{Slaked lime}& & & & \text{Bleaching power}& & \end{array}$