   # The electrolysis of BiO + produces pure bismuth. How long would it take to produce 10.0 g Bi by the electrolysis of a BiO + solution using a current of 25.0 A? ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 17, Problem 94E
Textbook Problem
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## The electrolysis of BiO+ produces pure bismuth. How long would it take to produce 10.0 g Bi by the electrolysis of a BiO+ solution using a current of 25.0 A?

Interpretation Introduction

Interpretation:

The information regarding electrolysis of BiO+ to produce bismuth is given. The time taken by 25.0A of current to produce 10.0g of Bi by the electrolysis of BiO+ solution

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 time taken by 25.0A of current to produce 10.0g of Bi by the electrolysis of BiO+ solution.

### Explanation of Solution

Given,

The amount of Bi produced by electrolysis is 10.0g .

The amount of current passed through BiO+ solution is 25.0A .

The reaction taking place is,

BiO++3eBi

The conversion of each BiO+ to Bi requires 3 electrons.

The number of moles of Bi atoms present in 209g of Bi is 1mol .

The number of moles of Bi atoms present in 10g of Bi is,

10gBi×1molBi209gBi=0.0478molBi

The number of electrons required to convert to 0.0487molBi is,

0.0487×3=0.1434molBi

The quantity of charge that is carried by these electrons is,

0.1434mole×96,485mole=13.8×103C

The charge generated in the cell is calculated as,

Q=It

Where,

• Q is the charge carried in the cell

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