Chapter 19, Problem 40PS

### 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

# Use the standard reduction potentials (Appendix M) for the half reactions [Zn(OH)4]2−(aq) + 2 e− → Zn (s) + 4 OH−(aq) and Zn2+(aq) + 2 e− → Zn(s) to calculate the value of Kformation for the complex ion [Zn(OH)4)]2−.

Interpretation Introduction

Interpretation:

The Kformation for the complex [Zn(OH)4]2- by using the following half cell reactions has to be determined.

[Zn(OH)4]2(aq) + 2e- Zn(s) + 4OH-(aq)Zn2+(aq) + 2e- Zn(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 â€“cell reaction are as follows.

[Zn(OH)4]2âˆ’(aq)Â +Â 2e-Â â†’Zn(s)Â +Â 4OH-(aq)Zn2+(aq)Â +Â 2e-Â â†’Zn(s)

Letâ€™s calculate the Ecello of the reaction.

Ecello=Â ECathode0-Â EAnode0=Â -0.763Â VÂ -Â (-1.22Â V)=Â 0.457Â V

Letâ€™s calculate the equilibrium constant for the reaction.

lnKÂ =Â nE00

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