   Chapter 19, Problem 107IL

Chapter
Section
Textbook Problem

A “silver coulometer” (Study Question 106) was used in the past to measure the current flowing in an electrochemical cell. Suppose you found that the current flowing through an electrolysis cell deposited 0.089 g of Ag metal at the cathode after exactly 10 min. If this same current then passed through a cell containing gold(III) ion in the form of [AuCl4]−, how much gold was deposited at the cathode in that electrolysis cell?

Interpretation Introduction

Interpretation:

The mass of gold deposited at the cathode in the given electrolysis cell has to be determined.

Concept introduction:

The Faraday’s first law of electrolysis state that the mass of the substance (m) deposited at any electrode is directly proportional to the charge (Q) passed. The mathematical form of the Fraday’s first law is written as’

m=(QF)(MZ)

Here,

The symbol F is the Faraday’s constant.

The symbol M is the molar mass of the substance in grams per mol.

The symbol Z is the valence number of ions of the substance (electrons transferred per ion).

In the simple case of constant current electrolysis, Q=I×t leading to

m=(I×tF)(MZ)

Here, t is the total time the constant current (I) is applied.

Explanation

Given:

The time t is 600s.

The mass of silver deposited is 0.089g.

The molar mass of silver is 107.86gmol1.

The faraday’s constant F is equal to 96500Cmol1.

The reaction of silver ion to produce silver metal is written as,

Ag+(aq)Ag(s)+e

Here, Z=1.

Substitute these values in the equation given below to calculate the current in ten minutes.

m=(I×tF)(MZ)

0.089g=(I(600s)96500Cmol1)(107.86gmol11)I=0.1327A

Hence, the current of silver produced in ten minutes is 0.1327A.

The same current passed through a cell containing gold ion in the form of [AuCl4]

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