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NCERT Class XI Chemistry The s-Block Elements Solutions

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Question : 16 of 32
Marks: +1, -0
Starting with sodium chloride how would you proceed to prepare
(i) sodium metal
(ii) sodium hydroxide
(iii) sodium peroxide
(iv) sodium carbonate ?
Solution:  
(i) Sodium metal is prepared by electrolysis of fused NaCl at 873 K using iron cathode and graphite anode (Down’s cell).
NaCl ⇌ Na++Cl−\mathrm{Na}^+ + \mathrm{Cl}^-
At anode : 2Cl−2\mathrm{Cl}^- → 2Cl+2e−2\mathrm{Cl} + 2\mathrm{e}^- → Cl2\mathrm{Cl}_2
At cathode : Na++e−\mathrm{Na}^+ + \mathrm{e}^- → Na
(ii) Sodium hydroxide is prepared by electrolysis of aqueous solution of NaCl using Hg cathode and carbon anode.
(Castner-Kellner cell) : NaCl ⇌ Na++Cl−\mathrm{Na}^+ + \mathrm{Cl}^-
At cathode : Na++e−\mathrm{Na}^+ + \mathrm{e}^- →Hg\xrightarrow[\mathrm{Hg}]{} Na-amalgam
At anode : Cl−\mathrm{Cl}^- → 12Cl2+e−\frac{1}{2}\mathrm{Cl}_2 + \mathrm{e}^-
Na - amalgam is treated with water to give NaOH and H2\mathrm{H}_2 :
2Na-amalgam + 2H2O2\mathrm{H}_2\mathrm{O} → 2NaOH+2Hg+H22\mathrm{NaOH} + 2\mathrm{Hg} + \mathrm{H}_2
(iii) Sodium chloride to sodium peroxide : NaCl ⇌ Na++Cl−\mathrm{Na}^+ + \mathrm{Cl}^-
At anode : 2Cl−2\mathrm{Cl}^- → 2Cl+2e−2\mathrm{Cl} + 2\mathrm{e}^- → Cl2\mathrm{Cl}_2
At cathode : Na++e−\mathrm{Na}^+ + \mathrm{e}^- → Na
Sodium is heated in excess of air.
2Na + O2→ΔNa2O2\mathrm{O}_2 \xrightarrow{\Delta} \mathrm{Na}_2\mathrm{O}_2.
(iv) Sodium hydrogen carbonate is prepared by passing CO2\mathrm{CO}_2 through NaCl solution saturated with NH3â‹…NaHCO3\mathrm{NH}_3 \cdot \mathrm{NaHCO}_3 formed is heated to give Na2CO3\mathrm{Na}_2\mathrm{CO}_3.
(Solvay process) :
NH3+H2O+CO2\mathrm{NH}_3 + \mathrm{H}_2\mathrm{O} + \mathrm{CO}_2 + NaCl → NaHCO3+NH4Cl\mathrm{NaHCO}_3 + \mathrm{NH}_4\mathrm{Cl}
2NaHCO3 Na2CO3 + H2O + CO2
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