   Chapter 19, Problem 39PS

Chapter
Section
Textbook Problem

Use the standard reduction potentials (Appendix M) for the half-reactions [AuCl4]−(aq) + 3 e− → Au(s) + 4 Cl−(aq) and Au3+(aq) + 3 e− → Au(s) to calculate the value of Kformation for the complex ion [AuCl4]−(aq).

Interpretation Introduction

Interpretation:

The Kformation for the complex [AuCl4]-(aq) by using the following half cell reactions has to be determined.

[AuCl4](aq) + 3e-  Au(s)+4Cl-(aq)Au3+(aq) + 3e- Au(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.

[AuCl4](aq) + 3e-  Au(s)+4Cl-(aq)Au3+(aq) + 3e- Au(s)

Let’s calculate the Ecello value of the reaction.

Ecello= ECathode0- EAnode0= 1.50 V - 1.00 V= 0.50 V

Let’s calculate the equilibrium constant for the reaction.

lnK = nE00

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