Q23E - Chemistry (OpenStax)

Question

Describe the production of metallic aluminum by electrolytic reduction.

Step-by-Step Solution

Answer

The aluminum oxide is added to the liquid cryolite \(\left( {N{a_3}Al{F_6}} \right)\)and calcium flouride\(\left( {CaF} \right),\) which meltsaluminum oxide. The products of the electrolysis of this solution are \({\rm{Al}}\)on the cathode and \({{\rm{O}}_2},{\rm{CO}}\)and\({\rm{C}}{{\rm{O}}_2}\)on the anode.

Step 1: Electrolytic reduction

Electrolytic reduction is an electrolysis process by which an ionic substance present in a molten state tends to cause the electrode to react in a chemical manner and undergo decomposition.

 

Electrolytic reduction happens in the presence of an electric current, and the metals get accumulated at the cathode.

Step 2: Description of the production of metallic aluminum

• First, bauxite, an aluminum ore is purified.

 

• \({\rm{NaOH}}\)is added to the ore to make a soluble product containing the Al, while the rest of the impurities remain undissolved and are removed by filtration.

 

• The reaction is as shown below.

\({\rm{2Al}}({\rm{s}}) + 2{\rm{NaOH}}(aq) + 6{{\rm{H}}_2}{\rm{O}}({\rm{l}}) \to 2{\rm{Na}}\left( {{\rm{Al}}{{({\rm{OH}})}_4}} \right)(aq) + 3{H_2}\left( g \right).\)

 

• Then, to the product of the previous reaction: sodium aluminate, an acid is added to form a precipitate of aluminum hydroxide, which is also removed by filtration.

 

• The formation of aluminum hydroxide is as shown below.

\({\rm{Na}}\left( {{\rm{Al}}{{({\rm{OH}})}_4}} \right) + {{\rm{H}}_3}{{\rm{O}}^ + }(aq) \to {\rm{Al}}{({\rm{OH}})_3}(s) + {\rm{N}}{{\rm{a}}^ + }(aq) + 2{{\rm{H}}_2}{\rm{O}}({\rm{l}}).\)

 

• Aluminum hydroxide is then heated to form an aluminum oxide and the reaction is as shown below.

\(2{\rm{Al}}{({\rm{OH}})_3}(s) \to {\rm{A}}{{\rm{l}}_2}{{\rm{O}}_3}(s) + 3{{\rm{H}}_2}{\rm{O}}({\rm{l}})\)

 

• Finally, the aluminum oxide is added to the liquid cryolite \(\left( {N{a_3}Al{F_6}} \right)\)and calcium fluoride \(\left( {CaF} \right),\) which melts the oxide.
• The products of the electrolysis of this solution are \({\rm{Al}}\)on the cathode and \({{\rm{O}}_2},{\rm{CO}}\)and\({\rm{C}}{{\rm{O}}_2}\)on the anode. The overall reaction is:

\(\begin{aligned}{\underline{\phantom{xx}}}{\rm{A}}{{\rm{l}}^{{\rm{3 + }}}}{\rm{ +  3}}{{\rm{e}}^{\rm{ - }}} \to {\rm{Al}}\\{\rm{                  (metal)}}\end{aligned}\)

Production of aluminum by electrolytic reduction