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Chapter 21: Problem 41

Describe the Hall process for preparing aluminum.

Short Answer

Expert verified
The Hall process for preparing aluminum involves three main steps: preparing the raw material where bauxite is turned into alumina, electrolysis where the alumina is separated into aluminum and oxygen with the help of electricity, and the final collection and refining where the aluminum is formed into usable products.

Step by step solution

01

Raw Material Preparation

Before the Hall process begins, the main raw material, bauxite, is prepared. Bauxite is purified to produce alumina (aluminum oxide - Al2O3) by the Bayer process. In this process, bauxite is mixed with sodium hydroxide, which reacts with the aluminum in the ore to create soluble sodium aluminate, leaving the impurities behind. These are filtered out, leaving a solution from which crystalline aluminum hydroxide is precipitated out.
02

Electrolysis

The alumina is then dissolved in molten cryolite (Na3AlF6). The advantage of using cryolite is that it significantly reduces the melting point of the aluminum oxide and increases its conductivity. The solution is then placed in a carbon-lined electrolysis cell. When electricity is applied, the Al2O3 breaks down into its components: aluminum and oxygen. The oxygen reacts with the carbon anode, forming carbon dioxide, which escapes into the atmosphere. The pure aluminum metal, being denser, sinks to the bottom of the cell and is periodically drained off via a plug.
03

Collection and Refining

The liquid aluminum metal tapped from the cell is then sent to a holding furnace. From here, it is cast into the specific forms such as ingots, sheets or extruded sections, ready for use in the manufacturing industries.

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Most popular questions from this chapter

What action renders aluminum inert?

Briefly discuss the steelmaking process.

The following are two reaction schemes involving magnesium. Scheme I: When magnesium burns in oxygen, a white solid (A) is formed. A dissolves in \(1 M \mathrm{HCl}\) to give a colorless solution (B). Upon addition of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) to \(\mathrm{B},\) a white precipitate is formed(C). On heating, C decomposes to \(D\) and a colorless gas is generated (E). When \(\mathrm{E}\) is passed through limewater [an aqueous suspension of \(\left.\mathrm{Ca}(\mathrm{OH})_{2}\right], \underline{\mathrm{a}}\) white precipitate appears (F). Scheme II:Magnesium reacts with \(1 M \mathrm{H}_{2} \mathrm{SO}_{4}\) to produce a colorless solution (G). Treating G with an excess of NaOH produces a white precipitate \((\mathrm{H}) . \mathrm{H}\) dissolves in \(1 M \mathrm{HNO}_{3}\) to form a colorless solution. When the solution is slowly evaporated, a white solid (I) appears. On heating I, a brown gas is given off. Identify \(\mathrm{A}-\mathrm{I}\) and write equations representing the reactions involved.

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