   Chapter 19, Problem 38PS

Chapter
Section
Textbook Problem

Use the standard reduction potentials (Appendix M) for the half-reactions Hg2Cl2(s) + 2 e− → 2 Hg(ℓ) + 2 Cl−(aq) and Hg2+(aq) + 2 e− → 2 Hg(ℓ) to calculate the value of Ksp for Hg2Cl2.

Interpretation Introduction

Interpretation:

The Ksp value for the Hg2Cl2 by using the following half-cell reactions has to be determined.

Hg2Cl2(s) + 2e- 2Hg(l) + 2Cl-(aq)Hg22+(aq) + 2e- 2Hg(l)

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.

Hg2Cl2(s) + 2e- 2Hg(l) + 2Cl-(aq)Hg22+(aq) + 2e- 2Hg(l)

Let’s calculate the Ecello value of the reaction.

Ecello= ECathode0- EAnode0= 0.27 V - 0.789 V= -0.519 V

Let’s calculate the equilibrium constant for the reaction.

lnK = nE00

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