   Chapter 18, Problem 96E

Chapter
Section
Textbook Problem

# Aluminum is produced commercially by the electrolysis of Al2O3 in the presence of a molten salt. If a plant has a continuous capacity of 1.00 million A, what mass of aluminum can be produced in 2.00 h?

Interpretation Introduction

Interpretation:

The production of aluminum by the electrolysis of Al2O3 in molten salt is given. The mass of aluminum that can be produced in 2.00h is to be calculated.

Concept introduction:

The non-spontaneous reaction takes place in an electrolytic cell in which there occurs conversion of electrical energy into chemical energy and this is used for the electrolysis of a metal.

The charge generated in the cell is calculated as,

Q=It

When electricity is passed through an electrolytic cell, at that time the amount of the substance that is liberated at an electrode is given by,

W=ZQ=ZIt

The value of Z is given as,

Z=Atomicmassn×96,485

The mass of metal is calculated by multiplying the moles of metal with its atomic molar mass. This gives the mass of the metal to be plated out in the electrolytic cell.

To determine: The mass of aluminum that can be produced in 2.00h by the electrolysis of Al2O3 in molten salt.

Explanation

Given,

The amount of current passed is 1.00millionA .

The time taken to produce a particular amount of aluminum is 2.00h .

The current in amperes is expressed as,

I=1.00millionA=106A=106C/s

The time in seconds is expressed as,

t=2.00h=2×3600sec=7200sec

When electricity is passed through an electrolytic cell, at that time the amount of the substance that is liberated at an electrode is given by,

W=ZQ=ZIt

Where,

• Q is the charge carried in the cell.
• I is the current in amperes
• t is the time for which current is passed through the cell.
• Z is the electrochemical equivalent.

The value of Z is given as,

Z=Atomicmassn×96,485

Where,

• n is the number of electrons exchanged.

Equivalent weight of aluminum is,

Z=27g/mol3e×96,485C/molee=9

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