   Chapter 19, Problem 69GQ

Chapter
Section
Textbook Problem

A potential of 0.142 V is recorded (under standard conditions) for a voltaic cell constructed using the following half reactions:Cathode: Pb2+(aq) + 2 e− → Pb(s)Anode: PbCl2(s) + 2 e− → Pb(s) + 2 Cl−(aq)Net: Pb2+(aq) + 2 Cl−(aq) → PbCI2(s) (a) What is the standard reduction potential for the anode reaction? (b) Calculate the solubility product, Ksp, for PbCl2.

a)

Interpretation Introduction

Interpretation:

The standard reduction potential for the anode reaction 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

Let’s calculate the Eanodeo of reactions:

The known reduction potential values are as follows.

Ecello=  0.142 VEcathode= -0

(b)

Interpretation Introduction

Interpretation:

The solubility product,Ksp for PbCl2 has to be calculated.

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

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