Question

Compound added to lower the fusion temperature in hall heroult's

Answer 1

Answer:

CORRECT answer is

Explanation:

The Hall–Héroult process is the major industrial process for smelting aluminium. It involves dissolving aluminium oxide (alumina)(obtained most often from bauxite, aluminium's chief ore, through the Bayer process) in molten cryolite, and electrolysingthe molten salt bath, typically in a purpose-built cell. The Hall–Héroult process applied at industrial scale happens at 940–980°C and produces 99.5–99.8% pure aluminium. Recycled aluminum requires no electrolysis, thus it does not end up in this process.[1]

Process

ChallengeEdit

Elemental aluminium cannot be produced by the electrolysis of an aqueous aluminium salt because hydronium ions readily oxidizeelemental aluminium. Although a moltenaluminium salt could be used instead, aluminium oxide has a melting point of 2072 °C[2] so electrolysing it is impractical. In the Hall–Héroult process, alumina, Al2O3, is dissolved in molten synthetic cryolite, Na3AlF6, to lower its melting point for easier electrolysis.[1]

TheoryEdit



A Hall–Héroult industrial cell

In the Hall–Héroult process the following simplified reactions take place at the carbon electrodes:

Cathode:

Al3+ + 3 e− → Al

Anode:

O2- + C → CO + 2 e−

Overall:

Al2O3 + 3 C → 2 Al + 3 CO

In reality much more CO2 is formed at the anode than CO:

2 Al2O3 + 3 C → 4 Al + 3 CO2

Pure cryolite has a melting point of 1009°C ± 1 K. With a small percentage of alumina dissolved in it, its melting point drops to about 1000°C. Besides having a relatively low melting point, cryolite is used as an electrolyte because among other things it also dissolves alumina well, conducts electricity, dissociates electrolytically at higher voltage than alumina and has a lower density than aluminum at the temperatures required by the electrolysis.[1]