Chapter 19, Problem 68GQ

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

Chapter
Section

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

# A cell is constructed using the following half-reactions:Ag+(aq) + e− → Ag(s)Ag2SO4(s) + 2 e− → 2 Ag(s) + SO42− (aq)E° = 0.653 V (a) What reactions should be observed at the anode and cathode? (b) Calculate the solubility product constant, Ksp, for Ag2SO4.

(a)

Interpretation Introduction

Interpretation:

The reactions observed at the anode and cathode has to be identified.

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

The given reactions are as follows.

Ag+(aq)Â +Â e-Â â†’Ag(s)Ag2SO4(s)Â +Â 2e-Â â†’2Ag(s)Â +Â SO42-(aq)Â Â ;Â E0=Â 0.653Â V

Letâ€™s write the reduction potential values of each reaction.

Ag+(aq)Â +Â e-Â â†’Ag(s)Â ;Â E0=0.799Â VAg2SO4(s)Â +Â 2e-Â â†’2Ag(s)Â +Â SO42-(aq)Â Â ;Â E0=Â 0

(b)

Interpretation Introduction

Interpretation:

The solubility product constant , Ksp for Ag2SO4 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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