An electrolysis cell for aluminum production operates at 5.0 V and a current of 3.0 × 10° A. Calculate the number of kilowatt-hours of energy required to produce 1 metric ton (1.0 x 10 kg)of aluminum. (1 kWh = 3.6 × 10® J and 1 J = 1 C.V)Energy =kWh

The half cell reaction will be as follows: Al3+(aq) + 3e- -------->Al(s). The currect, (I) =…

Answered: An electrolysis cell for aluminum…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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When a lead acid car battery is recharged by the alternator, it acts essentially as an electrolytic cell in which solid lead(II) sulfate (PbSO4) is reduced to lead at the cathode and oxidized to solid lead(II) oxide (PbO) at the anode. Suppose a current of 80.0 A is fed into a car battery for 63.0 seconds. Calculate the mass of lead deposited on the cathode of the battery. Round your answer to 3 significant digits. Also, be sure your answer contains a unit symbol. x10 0.0 X μ 010 3
O ELECTROCHEMISTRY Analyzing a galvanic cell A galvanic cell is powered by the following redox reaction: 3 Br₂(1) + 2 Cr(OH)3(s) + 10OH (aq) - → 6 Br (aq) + 2 CrO (aq) + 8 H₂0(1) Answer the following questions about this cell. If you need any electrochemical data, be sure you get it from the ALEKS Data tab. Write a balanced equation for the half-reaction that takes place at the cathode. Write a balanced equation for the half-reaction that takes place at the anode. Calculate the cell voltage under standard conditions. Round your answer to 2 decimal places. 0 0 E = OV ローロ é X Do ?
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Use standard reduction potentials to calculate the standard free energy change in kJ for the reaction: AG° 2+ Ni²+(aq) + Sn(s) → Ni(s) + Sn²+ (aq) Ni2+ (aq) + 2e → Ni(s) Fre = -0.250 V Sn2+ red (aq) + 2e¯ → Sn(s) red EO = -0.140 V kJ K for this reaction would be than one.
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An electrolysis cell for aluminum production operates at 5.0 V and a current of 3.0 × 10° A. Calculate the number of kilowatt-hours of energy required to produce 1 metric ton (1.0 × 10° kg) of aluminum. (1 kWh = 3.6 × 10° J and 1 J =1C•V) Energy = kWh