Chapter 19, Problem 72GQ

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

Chapter
Section

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

# An electrolysis cell for aluminum production operates at 5.0 V and a current of 1.0 × 105 A. Calculate the number of kilowatt-hours of energy required to produce 1 metrication (1.0 × 103 kg) of aluminum. (1 kWh = 3.6 × 106 J and 1 J = 1C · V)

Interpretation Introduction

Interpretation:

The required energy of electrolysis cell for aluminium production has to be determined.

Concept introduction:

Electrolysis:

It is a decomposition of ionic compounds by passing electricity through molten compounds or aqueous solutions of compounds.

Electricity used to produce chemical changes. The apparatus used for electrolysis is called an electrolytic cell.

Current:

Rate of the charge changing in time.

The formula is as follows.

Current (Amperes, A) = Electric charge (Coloumbs, C)time, t(seconds,s)

Simplify as follows.

Current = Electric charge time

To rearrange the above formula is as follows.

Charge = Current(A) × time(s)

Explanation

The half cell reaction of the aluminium is as follows.

Al3+(aq)Â +Â 3e-Â â†’Â Al(s)

Letâ€™s calculate the moles of required electrons:

1molÂ e-Â requiredÂ Â =Â (1.0Ã—103Ã—103Â gÂ )(1Â molÂ Al27Â gÂ Al)(3Â molÂ e-1Â molÂ Al)=Â 1.1Ã—105Â molÂ e-

Letâ€™s calculate the Quantity of charge:

ChargeÂ (C)Â =Â (1.1Ã—105Â molÂ e-)(96500Â C1Â molÂ e-)=1

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