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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

Iron(II) ion undergoes a disproportionation reaction to give Fe(s) and the iron(III) ion. That is, iron(II) ion is both oxidized and reduced within the same reaction.

3 Fe2+(aq) ⇄ Fe(s) + 2 Fe3+(aq)

  1. (a) What two half-reactions make up the disproportionation reaction?
  2. (b) Use the values of the standard reduction potentials for the two half-reactions in part (a) to determine whether this disproportionation reaction is product-favored at equilibrium.
  3. (c) What is the equilibrium constant for this reaction?

(a)

Interpretation Introduction

Interpretation:

The two half reactions which make up the disproportination reaction has to be determined.

Concept introduction:

Voltaic cell or Galvanic cell:

The device to produce electricity by using chemical reactions. In these divces are redox chemical reactions are occured.

A voltaic cell converts chemical energy into electrical energy.

It consists of two half cells. Each half cell consists of a metal and a solution of a salt of metal. Two half cells are connected by salt bridge.

The chemical reaction in the half cell is an oxidation reduction (redox)reactions.

For example:

Cell diagram of voltaic or galvanic cell is as follows.

                 Salt bridge                        Cu(s)|Cu2+(aq)  ||  Ag+(aq)|Ag(s)____________     ___________                                       Half cell             Half cell

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.

Explanation

From the given information, Iron (II) ion undergoes a disproportionaion reaction to give Fe(s) and the iron(III) ion. That is iron (II) ion as both oxidized and reduced within the same reaction.

The half reactions are as follows.

Fe2+(aq) + 2e- Fe (s)2Fe2+(aq) 2Fe3+(aq) + 2e-

By adding these two reactions we get , overall reaction.

Fe2+(aq) + 2e- Fe

(b)

Interpretation Introduction

Interpretation:

It has to be identified whether the disproportination reaction is a product-favored at equilibrium or not using the standard reduction potentials.

Concept introduction:

Voltaic cell or Galvanic cell:

The device to produce electricity by using chemical reactions. In these divces are redox chemical reactions are occured.

A voltaic cell converts chemical energy into electrical energy.

It consists of two half cells. Each half cell consists of a metal and a solution of a salt of metal. Two half cells are connected by salt bridge.

The chemical reaction in the half cell is an oxidation reduction (redox)reactions.

For example:

Cell diagram of voltaic or galvanic cell is as follows.

                 Salt bridge                        Cu(s)|Cu2+(aq)  ||  Ag+(aq)|Ag(s)____________     ___________                                       Half cell             Half cell

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

(c)

Interpretation Introduction

Interpretation:

The equilbrium constant of the reaction has to be calculated.

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

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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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