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In the Hall-Heroult process, a large electric current is passed through a solution of aluminum oxide (Al_2 O_3) dissolved in molten cryolite (Na_3 AlF_6), resulting in the reduction of the Al_2 O_3 to pure aluminum. Suppose a current of 9700. A is passed through a Hall-Round cell for 83.0 seconds. Calculate the mass of pure aluminum produced. Round your answer to 3 significant digits. Also, be sure your answer contains a unit symbol. [Al^3+ (aq) + 3e^- rightarrow Al(s)]

Respuesta :

Cryolite, which also has a relatively low melting point, is used as an electrolyte because, among other things, it dissolves alumina well, conducts electricity, dissociates electrolytically at a higher voltage than alumina, and has a lower density than aluminum at the temperatures needed for the electrolysis.

At 950 °C (1,750 °F), a nearly pure aluminum oxide compound called alumina is dissolved in a molten electrolyte made of aluminum, sodium, and fluorine. This electrolyte is then electrolyzed to produce aluminum metal at the cathode and oxygen gas at the anode.

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