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21 January, 06:31

In the Hall-Heroult process, a large electric current is passed through a solution of aluminum oxide dissolved in molten cryolite, resulting in the reduction of the to pure aluminum. Suppose a current of is passed through a Hall-Heroult cell for seconds. Calculate the mass of pure aluminum produced. Round your answer to significant digits. Also, be sure your answer contains a unit symbol.

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  1. 21 January, 06:50
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    The given question incomplete, the complete question is:

    In the Hall-Heroult process, a large electric current is passed through a solution of aluminum oxide (Al, 03) dissolved in molten cryolite (Na, Alts). re in the reduction of the Al, o, to pure aluminum. Suppose a current of 1800. A is passed through a Hall-Heroult cell for 37.0 seconds. Calculate the mass of pure aluminum produced Be sure your answer has a unit symbol and the correct number of significant digits.

    Answer:

    The correct answer is 6.2114 grams.

    Explanation:

    Based on the given question, the value of current or I have given is 1800 amperes, the time given is 37 seconds, and there is a need to find the mass of the pure aluminum generated in the process. Mass or weight can be determined by using Faraday's first law equation, that is, w = MIt/nF.

    Here, M is the atomic mass, w is the weight of the substance deposited, t is time, I is current, n is the number of moles of the electron, and F is the Faraday's constant, which is 96500 C. In the process mentioned in the question, aluminum oxide is reduced to give rise to pure aluminum, and in the process 3 electrons are gained. So, the value of n will be 3. The M or the atomic mass of Al is 27 gm per mole. Now putting the values in the equation we get,

    w = 27*1800*37 / 3*96500

    w = 1798200 / 289500

    w = 6.2114 grams

    Hence, pure aluminum produced in the process is 6.2114 grams.
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