   Chapter 19, Problem 83GQ

Chapter
Section
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:

Pt | Fe3+(aq, 0.50 M),Fe2+(aq, 1.0 × 10-5 M), and Hg2+(aq, 0.020 M) | Hg

Here, the reduction potential for Hg2+ is higher than that of the Fe3+. Therefore, Hg2+ is reduced and it should  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) + 2Fe2+(aq) Hg(l) + 2Fe3+(aq)

Let’s write the reduction potential values of each half reaction.

Hg2+(aq) + 2e- Hg(l)E0 = +0.855 V2Fe3+(aq) + 2e- 2Fe2+(aq) E0 = +0.771 V

Let’s calculate the Ecell0:

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

Let’s calculate the reduction potential of voltaic cell:

lnK = nE00

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