   # Chlorine gas is obtained commercially by electrolysis of brine (a concentrated aqueous solution of NaCl). If the electrolysis tells operate at 4.6 V and 3.0 × 10 5 A, what matt of chlorine can be produced in a 24-hour day? ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

#### Solutions

Chapter
Section ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 19, Problem 73GQ
Textbook Problem
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## Chlorine gas is obtained commercially by electrolysis of brine (a concentrated aqueous solution of NaCl). If the electrolysis tells operate at 4.6 V and 3.0 × 105 A, what matt of chlorine can be produced in a 24-hour day?

Interpretation Introduction

Interpretation:

The mass of chlorine produced in a 24 –hour day has to be determined.

Concept introduction:

According to the first law of thermodynamics, the change in internal energy of a system is equal ti the heat added to the sysytem minus the work done by the system.

The equation is as follows.

ΔU = Q - WΔU = Change in internal energyQ = Heat added to the systemW=Work done by the system

In voltaic cell, the maximum cell potential is directly related to the free energy difference between the reactants and products in the cell.

ΔG0= -nFE0n = Number of moles transferred per mole of reactant and productsF = Faradayconstant=96485C/mol  E0= Volts = Work(J)/Charge(C)

The relation between standard cell potential and equilibrium constant is as follows.

lnK = nE00.0257 at 298K

The relation between solubility product Ksp and equilibrium constant is as follows.

Ksp= e+lnK

### Explanation of Solution

Given Amp = 3.0×105A

Time = 24 hrs.

Charge (C) = Current (A) × time (s)=3.0×105 A × 24 h ×60 min h1 ×60 s.min1=2.59×1010 C

Let’s calculate the number of moles of electrons:

mol e- = (2.59 × 1010 C )(1 mol e-96500 C)=2

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