   Chapter 19, Problem 37PS

Chapter
Section
Textbook Problem

Use standard reduction potentials (Appendix M) for the half-reactions AgBr(s) + e− → Ag(s) + Br−(aq) and Ag+(aq) + e− → Ag(s) to calculate the value of Ksp for AgBr.

Interpretation Introduction

Interpretation:

The Ksp for the AgBr by using the following half cell reactions has to be determined.

AgBr(s) + e- Ag(s) + Br-(aq)Ag+(aq) + e- Ag(s)

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 half –cell reaction are as follows.

AgBr(s) + e- Ag(s) + Br-(aq)Ag+(aq) + e- Ag(s)

Let’s calculate the Ecello of the reaction.

Ecello= ECathode0- EAnode0= 0.0731 V - 0.7994 V= -0.7281 V

Let’s calculate the equilibrium constant for the reaction.

lnK = nE00

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