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Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

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BuyFindarrow_forward

Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

A voltaic cell set up utilizing the reaction

Cu(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 ΔG 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) + 2Ag+(aq) Cu2+(aq) + 2Ag(s)

Let’s calculate the constant when reaches the equilibrium.

lnK = nE00.0257 at 298K

Rearrange the above formula is as follows.

E = E0 - 0.0257n lnQ

Substitute the Q expression.

E = E0 - 0

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Chapter 19 Solutions

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Sect-19.8 P-19.11CYUSect-19.9 P-1.1ACPSect-19.9 P-1.2ACPSect-19.9 P-1.3ACPSect-19.9 P-2.1ACPSect-19.9 P-2.2ACPSect-19.9 P-2.3ACPSect-19.9 P-2.4ACPSect-19.9 P-2.5ACPCh-19 P-1PSCh-19 P-2PSCh-19 P-3PSCh-19 P-4PSCh-19 P-5PSCh-19 P-6PSCh-19 P-7PSCh-19 P-8PSCh-19 P-9PSCh-19 P-10PSCh-19 P-11PSCh-19 P-12PSCh-19 P-13PSCh-19 P-14PSCh-19 P-15PSCh-19 P-16PSCh-19 P-17PSCh-19 P-18PSCh-19 P-19PSCh-19 P-20PSCh-19 P-21PSCh-19 P-22PSCh-19 P-23PSCh-19 P-24PSCh-19 P-25PSCh-19 P-26PSCh-19 P-27PSCh-19 P-28PSCh-19 P-29PSCh-19 P-30PSCh-19 P-31PSCh-19 P-32PSCh-19 P-33PSCh-19 P-34PSCh-19 P-35PSCh-19 P-36PSCh-19 P-37PSCh-19 P-38PSCh-19 P-39PSCh-19 P-40PSCh-19 P-41PSCh-19 P-42PSCh-19 P-43PSCh-19 P-44PSCh-19 P-45PSCh-19 P-46PSCh-19 P-47PSCh-19 P-48PSCh-19 P-49PSCh-19 P-50PSCh-19 P-51PSCh-19 P-52PSCh-19 P-53PSCh-19 P-54PSCh-19 P-55PSCh-19 P-56PSCh-19 P-57GQCh-19 P-58GQCh-19 P-59GQCh-19 P-60GQCh-19 P-61GQCh-19 P-62GQCh-19 P-63GQCh-19 P-64GQCh-19 P-65GQCh-19 P-66GQCh-19 P-67GQCh-19 P-68GQCh-19 P-69GQCh-19 P-70GQCh-19 P-71GQCh-19 P-72GQCh-19 P-73GQCh-19 P-74GQCh-19 P-75GQCh-19 P-76GQCh-19 P-77GQCh-19 P-78GQCh-19 P-79GQCh-19 P-80GQCh-19 P-81GQCh-19 P-82GQCh-19 P-83GQCh-19 P-84GQCh-19 P-85GQCh-19 P-86GQCh-19 P-87GQCh-19 P-88GQCh-19 P-89GQCh-19 P-90GQCh-19 P-91GQCh-19 P-92GQCh-19 P-93GQCh-19 P-94GQCh-19 P-95GQCh-19 P-96GQCh-19 P-97GQCh-19 P-98GQCh-19 P-99GQCh-19 P-100GQCh-19 P-101GQCh-19 P-102GQCh-19 P-103GQCh-19 P-104GQCh-19 P-105ILCh-19 P-106ILCh-19 P-107ILCh-19 P-108ILCh-19 P-109ILCh-19 P-110ILCh-19 P-111SCQCh-19 P-112SCQCh-19 P-113SCQCh-19 P-114SCQCh-19 P-115SCQ

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