Chapter 19, Problem 83GQ

### 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

# Two half-cells, Pt | Fe3+(aq. 0.50 M), Fe2+(aq, 1.0 × 10−5 M) and Hg2+(aq, 0.020 M) | Hg, are constructed and then linked together to form a voltaic cell. Which electrode is the anode? What will be the potential of the voltaic cell at 298 K?

Interpretation Introduction

Interpretation:

The electrode which act as anode the potential of the voltaic cell has to be determined.

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 two half cells are as follows:

PtÂ |Â Fe3+(aq,Â 0.50Â M),Fe2+(aq,Â 1.0Â Ã—Â 10-5Â M), and Hg2+(aq,Â 0.020Â M)Â |Â Hg

Here, the reduction potential for Hg2+ is higher than that of the Fe3+. Therefore, Hg2+ is reduced and it shouldÂ  be the cathode. And also Fe2+ should be oxidized and it should be anode.

The overall reaction of the given voltaic cell is as follows:

Hg2+(aq)Â +Â 2Fe2+(aq)Â â†’Hg(l)Â +Â 2Fe3+(aq)

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

Hg2+(aq)Â +Â 2e-Â â†’Hg(l)E0Â =Â +0.855Â V2Fe3+(aq)Â +Â 2e-Â â†’2Fe2+(aq)Â E0Â =Â +0.771Â V

Letâ€™s calculate the Ecell0:

Ecell0=Â Ecathodeo-Â Eanodeo=Â 0.855Â V-0.771Â V=Â 0.084Â V

Letâ€™s calculate the reduction potential of voltaic cell:

lnKÂ =Â nE00

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