Chapter 19, Problem 86GQ

### 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

# A voltaic cell set up utilizing the reactionCu(s) + 2 Ag+(aq) → Cu2+(aq) + 2 Ag(s)has a cell potential of 0.45 V at 298 K. Describe how the potential of this cell will change as the cell is discharged. At what point does the cell potential reach a constant value? Explain your answer.

Interpretation Introduction

Interpretation:

Thet point at which the cell potential reach a constant value and its explanation has to be given.

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)

From the relation, we can state that ΔG value is negative. This means it is a spontaneous , the system spontaneously moves towards equilibrium.

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 cell potential will reach a constant value when the cell is at equilibrium. This means that Î”G value is equal to zero. And potential will also zero.

For this given system , we can calculate the ratio of concentrations of the solute ions that will be present in cell and it reaches equilibrium.

The given voltaic cell is as follows.

Cu(s)Â +Â 2Ag+(aq)Â â†’Cu2+(aq)Â +Â 2Ag(s)

Letâ€™s calculate the constant when reaches the equilibrium.

lnKÂ =Â nE00.0257Â atÂ 298K

Rearrange the above formula is as follows.

EÂ =Â E0Â -Â 0.0257nÂ lnQ

Substitute the Q expression.

EÂ =Â E0Â -Â 0

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