prof. t.l. heise, che 116 1 chapter twenty copyright © tyna l. heise 2001 - 2002 all rights...

Prof. T.L. Heise, CHE 116

1 Chapter TwentyChapter Twenty

Copyright © Tyna L. Heise 2001 - 2002Copyright © Tyna L. Heise 2001 - 2002

All Rights ReservedAll Rights Reserved


2 ElectrochemistryElectrochemistryRedox reactionsRedox reactions

Oxidation is the loss of electronsOxidation is the loss of electrons

Reduction is the gain of electronsReduction is the gain of electrons

The transfer of electrons that occurs is The transfer of electrons that occurs is exothermic and can be used to exothermic and can be used to

produce electricityproduce electricity


3 ElectrochemistryElectrochemistryAt other times the reactions you want At other times the reactions you want to occur may be nonspontaneous and to occur may be nonspontaneous and require the addition of electricity to require the addition of electricity to

drive the reaction.drive the reaction.

Electrochemistry is the branch of Electrochemistry is the branch of chemistry that deals with the chemistry that deals with the

relationships between electricity and relationships between electricity and chemical reactionschemical reactions


4Oxidation - Oxidation - Reductions Reductions ReactionReactionDetermination of oxidation reduction Determination of oxidation reduction

reactions occurs by examining the reactions occurs by examining the following:following:

assign oxidation numbersassign oxidation numberslook for the change in oxidation look for the change in oxidation statesstatesdon’t forget, something that don’t forget, something that oxidizes is also the reducing agent, oxidizes is also the reducing agent, something that reduces is also the something that reduces is also the oxidizing agentoxidizing agent


5Oxidation - Oxidation - Reductions Reductions ReactionReactionSample Exercise: Identify the oxidizing Sample Exercise: Identify the oxidizing

and reducing agents in the following and reducing agents in the following oxidation-reduction equation:oxidation-reduction equation:

2H2H22O(l) + Al(s) + MnOO(l) + Al(s) + MnO44--(aq) (aq)

Al(OH)Al(OH)44--(aq) + MnO(aq) + MnO22(s)(s)




+1 -2 0 +7 -2+1 -2 0 +7 -2


+3 -2 +1 +4 -2 +3 -2 +1 +4 -2



8Balancing Oxidation Balancing Oxidation

- Reductions - Reductions ReactionReactionWhenever we balance a chemical Whenever we balance a chemical

equation, we must obey the conservation equation, we must obey the conservation of mass.of mass.

As we balance redox reactions we will see As we balance redox reactions we will see an additional rule: the balancing of charge an additional rule: the balancing of charge as well.as well.



- Reductions - Reductions ReactionReaction1. Assign oxidation numbers to all 1. Assign oxidation numbers to all

elementselements

2. Bridge the elements that are changing 2. Bridge the elements that are changing in oxidation statein oxidation state

3. Identify the number of electrons being 3. Identify the number of electrons being lost and gained and balance them.lost and gained and balance them.

4. Balance all remaining elements besides 4. Balance all remaining elements besides H and OH and O

5. Balance O by adding H5. Balance O by adding H22O to the O to the deficient sidedeficient side



- Reductions - Reductions ReactionReaction6. Balance H by adding H6. Balance H by adding H++ to the deficient to the deficient

sideside

7. Check to make sure your charge for 7. Check to make sure your charge for each side is the sameeach side is the same



- Reductions - Reductions ReactionReactionSample Exercise: Complete and balance Sample Exercise: Complete and balance

the following oxidation-reduction the following oxidation-reduction equations in acidic solutions.equations in acidic solutions.

A) Cu(s) + NOA) Cu(s) + NO33--(aq) (aq) Cu Cu2+2+(aq) + NO(aq) + NO22(g)(g)





A) A) 0 +5 -2 +2 +4 -20 +5 -2 +2 +4 -2

Cu(s) + NOCu(s) + NO33--(aq) (aq) Cu Cu2+2+(aq) + NO(aq) + NO22(g)(g)





A) A) Leo = RA 2e -Leo = RA 2e -

0 +5 -2 +2 +4 -20 +5 -2 +2 +4 -2

Cu(s) + NOCu(s) + NO33--(aq) (aq) Cu Cu2+2+(aq) + NO(aq) + NO22(g)(g)

Ger = OA 1e -Ger = OA 1e -






0 +5 -2 +2 +4 -20 +5 -2 +2 +4 -2

Cu(s) + Cu(s) + 22NONO33--(aq)(aq)CuCu2+2+(aq) + (aq) + 22NONO22(g)(g)

Ger = OA 2(1e -)Ger = OA 2(1e -)



- Reductions - Reductions ReactionReactionSample Exercise: Complete and balance the Sample Exercise: Complete and balance the

following oxidation-reduction equations in following oxidation-reduction equations in acidic solutions.acidic solutions.


0 +5 -2 +2 +4 -20 +5 -2 +2 +4 -2

Cu(s)+Cu(s)+22NONO33--(aq)(aq)CuCu2+2+(aq)+(aq)+22NONO22(g)(g)+2H+2H22OO

Ger = OA 2(1e -)Ger = OA 2(1e -)



- Reductions - Reductions ReactionReactionSample Exercise: Complete and balance the Sample Exercise: Complete and balance the

following oxidation-reduction equations in following oxidation-reduction equations in acidic solutions.acidic solutions.


0 +5 -2 +2 +4 -20 +5 -2 +2 +4 -2

Cu +Cu +22NONO33--(aq)(aq)+4H+4H++CuCu2+2+(aq)+(aq)+22NONO22(g)(g)+2H+2H22OO

Ger = OA 2(1e -)Ger = OA 2(1e -)





B) MnB) Mn2+2+ + NaBiO + NaBiO33 Bi Bi3+3+ + MnO + MnO44--





B) B) +2 +5 -2 +3 +7 -2 +2 +5 -2 +3 +7 -2

MnMn2+2+ + BiO + BiO33-- Bi Bi3+3+ + MnO + MnO44

--





B) B) Leo = RA 5e -Leo = RA 5e -

+2 +5 -2 +3 +7 -2+2 +5 -2 +3 +7 -2

MnMn2+2+ + BiO + BiO33 - - Bi Bi3+3+ + MnO + MnO44--






B) B) Leo = RA 2(5e -)Leo = RA 2(5e -)

+2 +5 -2 +3 +7 -2+2 +5 -2 +3 +7 -2

22MnMn2+2+ + BiO + BiO33 -- Bi Bi3+3+ + + 22MnOMnO44--






B) B) Leo = RA 2(5e -)Leo = RA 2(5e -)

+2 +5 -2 +3 +7 -2+2 +5 -2 +3 +7 -2

22MnMn2+2+ + + 55BiOBiO33 -- 55BiBi3+3+ + + 22MnOMnO44--

Ger = OA 5(2e -)Ger = OA 5(2e -)





B)B)

22MnMn2+ 2+ + + 55BiOBiO33- - + + 14H14H++

55BiBi3+ 3+ + + 22MnOMnO44- - + + 7H7H22OO



- Reductions - Reductions ReactionReactionAt times the solution in which a redox At times the solution in which a redox

reaction occurs is a basic one. The reaction occurs is a basic one. The balancing of an equation in a basic balancing of an equation in a basic solution is slightly different. Instead of solution is slightly different. Instead of balancing with Hbalancing with H22O and HO and H++, you must , you must balance with Hbalance with H22O and OHO and OH--..



- Reductions - Reductions ReactionReactionSample exercise: Complete and balance Sample exercise: Complete and balance

the following equations for oxidation and the following equations for oxidation and reduction reactions that occur in basic reduction reactions that occur in basic solutions.solutions.

A) NOA) NO22-- + Al + Al NH NH33 + Al(OH) + Al(OH)44

--





A) A) +3 -2 0 -3 +1 +3 -2 +1 +3 -2 0 -3 +1 +3 -2 +1

NONO22-- + Al + Al NH NH33 + Al(OH) + Al(OH)44

--





A) A) Ger = OA 6e - Ger = OA 6e -

+3 -2 0 -3 +1 +3 -2 +1+3 -2 0 -3 +1 +3 -2 +1

NONO22-- + Al + Al NH NH33 + Al(OH) + Al(OH)44

--

Leo = RA 3e -Leo = RA 3e -





A) A) Ger = OA 6e - Ger = OA 6e -

+3 -2 0 -3 +1 +3 -2 +1+3 -2 0 -3 +1 +3 -2 +1

NONO22-- + + 22Al Al NH NH33 + + 22Al(OH)Al(OH)44

--

Leo = RA 2(3e -)Leo = RA 2(3e -)





A) A)

NONO22-- + + 22Al + Al + 5H5H22OO + + OHOH--

NH NH33 + + 22Al(OH)Al(OH)44--





B) Cr(OH)B) Cr(OH)33 + ClO + ClO-- CrO CrO442-2- + Cl + Cl22





B) B) +3 -2 +1 +1 -2 +6 -2 0+3 -2 +1 +1 -2 +6 -2 0

Cr(OH)Cr(OH)33 + ClO + ClO-- CrO CrO442-2- + Cl + Cl22






+3 -2 +1 +1 -2 +6 -2 0+3 -2 +1 +1 -2 +6 -2 0

Cr(OH)Cr(OH)33 + ClO + ClO-- CrO CrO442-2- + Cl + Cl22







+3 -2 +1 +1 -2 +6 -2 0+3 -2 +1 +1 -2 +6 -2 0

Cr(OH)Cr(OH)33 + + 66ClOClO-- CrO CrO442-2- + + 33ClCl22

Ger = OA 6(1e -)Ger = OA 6(1e -)





B) B) Leo = RA 2(3e -)Leo = RA 2(3e -)

+3 -2 +1 +1 -2 +6 -2 0+3 -2 +1 +1 -2 +6 -2 0

22Cr(OH)Cr(OH)33 + + 66ClOClO-- 22CrOCrO442-2- + + 33ClCl22

Ger = OA 6(1e -)Ger = OA 6(1e -)





B) B) 22Cr(OH)Cr(OH)33 + + 66ClOClO-- 22CrOCrO44

2-2- + + 33ClCl2 2 + 2H+ 2H22O + 2OHO + 2OH--


35Voltaic CellsVoltaic Cells

The energy released in a spontaneous The energy released in a spontaneous redox reaction can be used to perform redox reaction can be used to perform electrical work.electrical work.

This task is accomplished in a voltaic cell, This task is accomplished in a voltaic cell, a device in which the transfer of electrons a device in which the transfer of electrons takes place through an external pathway takes place through an external pathway rather than directly between reactants.rather than directly between reactants.


36 Voltaic CellsVoltaic Cells

The The spontaneous spontaneous reaction reaction occurs in a occurs in a single cell, single cell, but splitting but splitting the cell into the cell into two halves two halves allows for allows for the external the external pathway of pathway of electrons to electrons to be created.be created.



Parts of Voltaic Cell:Parts of Voltaic Cell:the two solid metals connected by the the two solid metals connected by the external circuit are called external circuit are called electrodeselectrodes..the electrode at which oxidation occurs the electrode at which oxidation occurs is called the anodeis called the anodethe electrode at which reduction occurs the electrode at which reduction occurs is called the cathode.is called the cathode.



Parts of Voltaic Cell:Parts of Voltaic Cell:the anode metal is the substance that the anode metal is the substance that undergoes oxidation to form more undergoes oxidation to form more positive ions.positive ions.the positive metallic ion in the beaker the positive metallic ion in the beaker that contains the cathode is the substance that contains the cathode is the substance that undergoes reduction to form more of that undergoes reduction to form more of the metal electrode.the metal electrode.



Parts of Voltaic Cell:Parts of Voltaic Cell:As the oxidation half cell gets a larger As the oxidation half cell gets a larger concentration of positive ions, negative concentration of positive ions, negative ions flow through the semipermeable ions flow through the semipermeable membrane that forms the salt bridge to membrane that forms the salt bridge to balance the charges in the beaker.balance the charges in the beaker.As the reduction half cell gets a smaller As the reduction half cell gets a smaller concentration of positive ions, positive concentration of positive ions, positive ions flow through the membrane to ions flow through the membrane to replace the loss.replace the loss.



Parts of Voltaic Cell:Parts of Voltaic Cell:in any voltaic cell the electrons flow in any voltaic cell the electrons flow from the anode (-) to the cathode (+) from the anode (-) to the cathode (+) through the external circuitthrough the external circuitThe collection of lost electrons at the The collection of lost electrons at the anode causes it to be negative, and the anode causes it to be negative, and the cathode is positive so it can attract the cathode is positive so it can attract the electrons and pull them through the wire.electrons and pull them through the wire.



Sample exercise: The two half-reactions in Sample exercise: The two half-reactions in a voltaic cell area voltaic cell are

Zn(s) Zn(s) Zn Zn2+2+(aq) + 2e(aq) + 2e--

ClOClO33--(aq) + 6H(aq) + 6H++ + 6e + 6e-- Cl Cl--(aq) + 3H(aq) + 3H22O(l)O(l)

a) Indicate which reaction occurs at the a) Indicate which reaction occurs at the anodeanode and the and the cathodecathode






b) Which electrode is consumed first in the b) Which electrode is consumed first in the cell reaction?cell reaction?






b) Which electrode is consumed first in the b) Which electrode is consumed first in the cell reaction?cell reaction?

The anode is consumedThe anode is consumed






c) Which electrode is positive?c) Which electrode is positive?






c) Which electrode is positive?c) Which electrode is positive?

The cathode is positiveThe cathode is positive



An atomic view of what occurs in a single An atomic view of what occurs in a single cell in which both oxidation and reduction cell in which both oxidation and reduction are occurring may give better insight.are occurring may give better insight.



A molecular view allows a better A molecular view allows a better understanding of what is occurring in each understanding of what is occurring in each half cell.half cell.


52Cell EMFCell EMF

The reason that electrons flow The reason that electrons flow spontaneously is do to the “driving force” spontaneously is do to the “driving force” behind certain redox reactions.behind certain redox reactions.

The ‘driving force’ is a difference in The ‘driving force’ is a difference in potential energy.potential energy.The anode has a higher potential The anode has a higher potential energy, and electrons flow to the cathode energy, and electrons flow to the cathode to achieve a lower potential energy.to achieve a lower potential energy.As in many natural wonders, the flow As in many natural wonders, the flow from higher concentration to lower from higher concentration to lower continuescontinues


53Cell EMFCell EMF

The difference in potential energy per The difference in potential energy per charge is measured in volts.charge is measured in volts.

1 V (volt) is the potential difference 1 V (volt) is the potential difference required to impart 1 J (joule) of energy to required to impart 1 J (joule) of energy to a charge of 1 C (coulomb)a charge of 1 C (coulomb)the potential difference is called an the potential difference is called an electromotive force or emfelectromotive force or emfemf, or Eemf, or Ecellcell, is also called the cell , is also called the cell potentialpotential


54Cell EMFCell EMF

The difference in potential energy per The difference in potential energy per charge is measured in volts.charge is measured in volts.

The emf of a particular cell is The emf of a particular cell is dependent on the reaction that is dependent on the reaction that is occurring there.occurring there.We will always assume standard We will always assume standard conditions, and the cell potential should conditions, and the cell potential should be denoted as such, Ebe denoted as such, E°°

cellcell


55Cell EMFCell EMF

Standard reduction potential:Standard reduction potential:the cell potential will depend on the the cell potential will depend on the particular cathode and anode usedparticular cathode and anode usedthe cell potential is the difference the cell potential is the difference between two electrode potentialsbetween two electrode potentials

EEcellcell = E = Ecathodecathode + E + Eanodeanode

the amount of moles of each DOESthe amount of moles of each DOES NOT change the potentials of cellsNOT change the potentials of cells


56Cell EMFCell EMF

Standard reduction potential:Standard reduction potential:when finding the potentials, a reduction when finding the potentials, a reduction table will be used.table will be used.Finding the potential for cathodes is Finding the potential for cathodes is straightforward, simply read the tablestraightforward, simply read the tableFinding the potential for anodes means Finding the potential for anodes means reading the table BACKWARDS and the reading the table BACKWARDS and the sign of the potential should be switchedsign of the potential should be switchedTable 20.1 is a standard reduction table Table 20.1 is a standard reduction table


57Cell EMFCell EMF

Sample exercise: Calculate the standard Sample exercise: Calculate the standard emf for a cell that employs the following emf for a cell that employs the following overall reaction:overall reaction:

0 +3 +2 +20 +3 +2 +2

Ni(s) + 2FeNi(s) + 2Fe3+3+(aq) (aq) 2Fe 2Fe2+2+(aq) + Ni(aq) + Ni+2+2(aq) (aq)


58Cell EMFCell EMF


0 +3 +2 +20 +3 +2 +2

Ni(s) + 2FeNi(s) + 2Fe3+3+(aq) (aq) 2Fe 2Fe2+2+(aq) + Ni(aq) + Ni+2+2(aq) (aq)


59Cell EMFCell EMF

Sample exercise: Calculate the standard Sample exercise: Calculate the standard emf for a cell that employs the following emf for a cell that employs the following overall reaction: overall reaction: Ger = cathodeGer = cathode

0 +3 +2 +20 +3 +2 +2

Ni(s) + 2FeNi(s) + 2Fe3+3+(aq) (aq) 2Fe 2Fe2+2+(aq) + Ni(aq) + Ni+2+2(aq)(aq)

Leo = anodeLeo = anode


60Cell EMFCell EMF

Sample exercise: Calculate the standard Sample exercise: Calculate the standard emf for a cell that employs the following emf for a cell that employs the following overall reaction: overall reaction: Ger = cathodeGer = cathode

0 +3 +2 +20 +3 +2 +2





61Cell EMFCell EMF

Sample exercise: Calculate the standard Sample exercise: Calculate the standard emf for a cell that employs the following emf for a cell that employs the following overall reaction: overall reaction: Ger = cathode Ger = cathode

0 +3 +2 +20 +3 +2 +2




= +0.77 + +0.28= +0.77 + +0.28

= +1.05 V= +1.05 V


62Cell EMFCell EMF


2Al(s) + 3I2Al(s) + 3I22(s) (s) 2Al 2Al3+3+(aq) + 6I(aq) + 6I--(aq) (aq)


63Cell EMFCell EMF


0 0 +3 -10 0 +3 -1



64Cell EMFCell EMF


leo = anodeleo = anode

0 0 +3 -10 0 +3 -1


ger = cathodeger = cathode


65Cell EMFCell EMF


leo = anodeleo = anode

0 0 +3 -10 0 +3 -1


ger = cathodeger = cathode

Ecell = +0.54 + +1.66 = 2.20 VEcell = +0.54 + +1.66 = 2.20 V


66Cell EMFCell EMF

The more positive the potential difference, The more positive the potential difference, the greater the driving force.the greater the driving force.


67 Spontaneity of Spontaneity of Redox ReactionsRedox Reactions

Any reaction that can occur in a voltaic cell Any reaction that can occur in a voltaic cell to produce a positive emf must be to produce a positive emf must be spontaneous.spontaneous.

EEcellcell = E = Eredred + E + Eoxox

a + emf is spontaneousa + emf is spontaneousa - emf is not spontaneousa - emf is not spontaneous



Sample exercise: Using the standard Sample exercise: Using the standard reduction potentials, determine which of the reduction potentials, determine which of the following reactions are spontaneous.following reactions are spontaneous.

A) IA) I22(s) + 5Cu(s) + 5Cu2+2+(aq) + 6H(aq) + 6H22O(l) O(l)

2IO2IO33--(aq) + 5Cu(s) + 12H(aq) + 5Cu(s) + 12H++(aq)(aq)






Red: CuRed: Cu2+2+(aq) (aq) Cu(s) Cu(s)

Ox: IOx: I22(s) + 6H(s) + 6H22O O 2IO 2IO33--(aq) + 12H(aq) + 12H++(aq)(aq)








Ecell = +0.34 + -1.195Ecell = +0.34 + -1.195








Ecell = +0.34 + -1.195 = -0.855 NOTEcell = +0.34 + -1.195 = -0.855 NOT




B) HgB) Hg2+2+(aq) + 2I(aq) + 2I--(aq) (aq) Hg(l) + I Hg(l) + I22(s)(s)





Red: HgRed: Hg2+2+(aq) (aq) Hg(l) Hg(l)

Ox: 2IOx: 2I--(aq) (aq) I I22(s) (s)






Ox: 2IOx: 2I--(aq) (aq) I I22(s) (s)

Ecell = +0.854 + -0.536 Ecell = +0.854 + -0.536






Ox: 2IOx: 2I--(aq) (aq) I I22(s) (s)

Ecell = +0.854 + -0.536 = +0.318 SPONEcell = +0.854 + -0.536 = +0.318 SPON



Looking at the standard reduction table, a Looking at the standard reduction table, a general rule can be established:general rule can be established:

any metal in the series is able to any metal in the series is able to spontaneously react with any ion beneath spontaneously react with any ion beneath ititthere is also a relationship between emf there is also a relationship between emf and Gibbs Free Energy and Gibbs Free Energy (remember a negative (remember a negative G is spontaneous)G is spontaneous)

G = -nFEG = -nFE



G = -nFEG = -nFE

n = number of electrons n = number of electrons transferred in reactiontransferred in reaction

F = Faraday’s constantF = Faraday’s constant

1 F = 96,500 J/V-mol1 F = 96,500 J/V-mol

E = emf of cellE = emf of cell

units of units of G are J/molG are J/mol



Sample exercise: Consider the following Sample exercise: Consider the following reaction:reaction:

3Ni3Ni2+2+(aq) + 2Cr(OH)(aq) + 2Cr(OH)33(s) + 10OH(s) + 10OH-- 3Ni(s) + 2CrO 3Ni(s) + 2CrO44

2-2-(aq) + 8H(aq) + 8H22O(l)O(l)

a) what is the value of n for this reaction?a) what is the value of n for this reaction?





2-2-(aq) + 8H(aq) + 8H22O(l)O(l)

a) what is the value of n for this reaction?a) what is the value of n for this reaction?

Each NiEach Ni2+2+ gains 2e gains 2e-- and there are 3 and there are 3 ions, so 6e- are involvedions, so 6e- are involved

n = 6n = 6





2-2-(aq) + 8H(aq) + 8H22O(l)O(l)

b) Calculate b) Calculate GG

G = -nFEG = -nFE





2+2+(aq) + 8H(aq) + 8H22O(l)O(l)


G = -nFE n = 6G = -nFE n = 6

F = 96,500 J/V-molF = 96,500 J/V-mol

E = -0.28 + +0.13= -0.15 VE = -0.28 + +0.13= -0.15 V





2+2+(aq) + 8H(aq) + 8H22O(l)O(l)


G = -nFE G = -nFE -(6)(96,500 J/V-mol)(-0.15 V) -(6)(96,500 J/V-mol)(-0.15 V)+86850 J/mol+86850 J/mol

NOT SponNOT Spon


83Effect of Effect of

Concentration on Concentration on Cell EMFCell EMFAs a voltaic cell is discharged, the reactants As a voltaic cell is discharged, the reactants

of the reaction are consumed and the of the reaction are consumed and the products are generated, so the products are generated, so the concentrations of these substances change.concentrations of these substances change.

The emf progressively drops until E = 0, at The emf progressively drops until E = 0, at which point we say the cell is dead. At that which point we say the cell is dead. At that point the concentrations in the cell are at point the concentrations in the cell are at equilibrium.equilibrium.



Concentration on Concentration on Cell EMFCell EMFThe dependence of cell emf on concentration The dependence of cell emf on concentration

can be obtained from the dependence of the can be obtained from the dependence of the free energy change on concentrationfree energy change on concentration

Nernst Equation: E = ENernst Equation: E = E°° - RT (lnQ) - RT (lnQ) nF nF

= E= E°° - 0.0592 V (log Q) - 0.0592 V (log Q) n n



Concentration on Concentration on Cell EMFCell EMFThe dependence of cell emf on concentration The dependence of cell emf on concentration

can be obtained from the dependence of the can be obtained from the dependence of the free energy change on concentrationfree energy change on concentration

Nernst Equation: E = ENernst Equation: E = E°° - RT (lnQ) - RT (lnQ) nF nF

* Q is the reaction* Q is the reaction = E= E°° - 0.0592 V (log Q) - 0.0592 V (log Q)quotient quotient n n



Concentration on Concentration on Cell EMFCell EMFSample exercise: Calculate the emf Sample exercise: Calculate the emf

generated by the cellgenerated by the cell

2Al(s) + 3I2Al(s) + 3I22(s) (s) 2Al 2Al3+3+(aq) + 6I(aq) + 6I--(aq)(aq)

when [Alwhen [Al3+3+] = 4.0 x 10] = 4.0 x 10-3-3M and [IM and [I--] = 0.010 M.] = 0.010 M.







E = EE = E°° - 0.0592 V (log Q) - 0.0592 V (log Q) n n








EE°° = +0.536 + +1.66 = 2.22 V = +0.536 + +1.66 = 2.22 V








n = 6 en = 6 e--








Q = [AlQ = [Al33+]+]22[I[I--]]66 = [4.0 x 10 = [4.0 x 10-3-3]]22[0.010][0.010]66 = 1.6 x 10= 1.6 x 10-17-17








= 2.22 - 0.0592 (log 1.6 x 10= 2.22 - 0.0592 (log 1.6 x 10-17-17)) 6 6








= 2.22 - 0.0592 (log 1.6 x 10= 2.22 - 0.0592 (log 1.6 x 10-17-17)) 6 6

= 2.36 V = 2.36 V



Concentration on Concentration on Cell EMFCell EMFConcentration cells: a voltaic cell with an Concentration cells: a voltaic cell with an

emf of ZERO can be constructed if the same emf of ZERO can be constructed if the same species is used as both the anode and the species is used as both the anode and the cathode if concentrations are the same; cathode if concentrations are the same; using two different concentrations of the using two different concentrations of the same species will create a non-zero cell.same species will create a non-zero cell.

A cell based solely on the emf generated A cell based solely on the emf generated because of a difference in a concentration is because of a difference in a concentration is called a concentration cellcalled a concentration cell



Concentration on Concentration on Cell EMFCell EMF



Concentration on Concentration on Cell EMFCell EMFThe standard emf for this cell would have to The standard emf for this cell would have to

zero, but the cell operates due to the zero, but the cell operates due to the differences in concentration.differences in concentration.

The driving force for the cell is provided by The driving force for the cell is provided by the difference in concentration.the difference in concentration.

Operation of the cell proceeds to equalize Operation of the cell proceeds to equalize the concentrations: the more dilute the concentrations: the more dilute concentration acts as the anode, the solid concentration acts as the anode, the solid metal electrode will oxidize to turn into metal electrode will oxidize to turn into more ions.more ions.



Concentration on Concentration on Cell EMFCell EMFThe more concentrated solution acts as the The more concentrated solution acts as the

cathode, where the larger number of ions cathode, where the larger number of ions reduce to plate the electrode and thereby reduce to plate the electrode and thereby reduce there number. reduce there number.

The nernst equation can also be used here to The nernst equation can also be used here to calculate the Ecalculate the Ecellcell under non-standard under non-standard conditions.conditions.

Anode: X(s) Anode: X(s) X X++(dilute) + e(dilute) + e--

Cathode: XCathode: X++(conc.) + e(conc.) + e-- X(s) X(s)



Concentration on Concentration on Cell EMFCell EMFAnode: X(s) Anode: X(s) X X++(dilute) + e(dilute) + e--

Cathode: XCathode: X++(conc.) + e(conc.) + e-- X(s) X(s)

XX++(conc.) + X(s) (conc.) + X(s) X X++(dilute) + X(s)(dilute) + X(s)

Q = Q = X X++(dilute) / X(dilute) / X++(conc.) (conc.)



Concentration on Concentration on Cell EMFCell EMFSample exercise: A concentration cell is Sample exercise: A concentration cell is

constructed with two Zn(s) constructed with two Zn(s) Zn Zn2+2+(aq) half-(aq) half-cells. The first cell has [Zncells. The first cell has [Zn2+2+] = 1.35 M and ] = 1.35 M and the second cell has [Znthe second cell has [Zn2+2+] = 3.75 x 10] = 3.75 x 10-4-4 M. M.

a) Which half cell is the anode of the cell?a) Which half cell is the anode of the cell?





a) Which half cell is the anode of the cell?a) Which half cell is the anode of the cell?

The cell with [ZnThe cell with [Zn2+2+] = 3.75 x 10] = 3.75 x 10-4-4 M is M is the anode.the anode.





b) What is the emf of the cell?b) What is the emf of the cell?







= 0 - 0.0592 (log 3.75x10= 0 - 0.0592 (log 3.75x10-4-4/1.35)/1.35) 2 2







= 0 - 0.0592 (log 3.75x10= 0 - 0.0592 (log 3.75x10-4-4/1.35)/1.35) 2 2

= 0.105 V= 0.105 V



Concentration on Concentration on Cell EMFCell EMFCell emf and Chemical Equilibrium: the Cell emf and Chemical Equilibrium: the

nernst equation helps us understand why nernst equation helps us understand why the emf drops as the cell is discharged: as the emf drops as the cell is discharged: as the reactants are converted into products, the reactants are converted into products, the value of Q increases, so the value of E the value of Q increases, so the value of E decreases.decreases.

The cell emf eventually reaches 0.The cell emf eventually reaches 0.

When When G = 0, E = 0, and an EG = 0, E = 0, and an Ecellcell = 0 shows a = 0 shows a cell at equilibrium.cell at equilibrium.



Concentration on Concentration on Cell EMFCell EMFIf 0 = EIf 0 = E°° - 0.0592 V (log Q) - 0.0592 V (log Q) n n

Then log K = nEThen log K = nE°° 0.0592 0.0592



Concentration on Concentration on Cell EMFCell EMFSample exercise: Using standard reduction Sample exercise: Using standard reduction

potentials, calculate the equilibrium potentials, calculate the equilibrium constant at 25°C for the reactionconstant at 25°C for the reaction

BrBr22(l) + 2Cl(l) + 2Cl--(aq) (aq) Cl Cl22(g) + 2Br(g) + 2Br--(aq)(aq)






log K = nElog K = nE°° 0.0592 0.0592






log K = nElog K = nE°° 0.0592 0.0592

= 2(+1.065 + -1.359)= 2(+1.065 + -1.359) 0.0592 0.0592






log K = nElog K = nE°° 0.0592 0.0592

= 2(+1.065 + -1.359)= 2(+1.065 + -1.359) 0.0592 0.0592

= -9.93= -9.93






K = 10K = 10(-9.93)(-9.93)

= 1.2 x 10= 1.2 x 10-10-10


111BatteriesBatteries

A battery is a portable, self-contained A battery is a portable, self-contained electrochemical power source that consists electrochemical power source that consists of one or more voltaic cells.of one or more voltaic cells.

The common 1.5 V battery is a single voltaic The common 1.5 V battery is a single voltaic cell, where a 12 V battery uses multiple cell, where a 12 V battery uses multiple voltaic cells in one case.voltaic cells in one case.

When cells are connected in series, with the When cells are connected in series, with the cathode of one attached to the anode to cathode of one attached to the anode to another, the battery produces a voltage that another, the battery produces a voltage that is the sum of the emfs of the single cells.is the sum of the emfs of the single cells.



Batteries connected in series...



Commerical batteries that has specific Commerical batteries that has specific performance characteristics can require performance characteristics can require considerable ingenuity.considerable ingenuity.

The emf of a battery is determined by the The emf of a battery is determined by the substances used as the cathode and anode, substances used as the cathode and anode, and the usable life of the battery depends on and the usable life of the battery depends on the quantities of the substances packaged.the quantities of the substances packaged.

Keep in mind the anode and cathode need to Keep in mind the anode and cathode need to be separated by a porous barrier similar to be separated by a porous barrier similar to a salt bridge.a salt bridge.



The materials used to construct the battery The materials used to construct the battery must be stable under the conditions in which must be stable under the conditions in which it is to be used and must be chosen to it is to be used and must be chosen to minimize health and environmental minimize health and environmental concerns upon use and disposal.concerns upon use and disposal.

Different applications require batteries with Different applications require batteries with different properties.different properties.



Lead-Acid BatteryLead-Acid Battery

Battery used in cars is one of the most Battery used in cars is one of the most common batteries, with over 100 million common batteries, with over 100 million produced annually.produced annually.

A 12-V battery consists of six voltaic cells in A 12-V battery consists of six voltaic cells in series, each producing 2-V.series, each producing 2-V.

Cathode: lead dioxide packed on a Cathode: lead dioxide packed on a metal gridmetal grid

Anode: leadAnode: lead




Both electrodes are immersed in sulfuric Both electrodes are immersed in sulfuric acid.acid.

Both reactants are solids so there is no need Both reactants are solids so there is no need to separate the cell into anode and cathode to separate the cell into anode and cathode compartments.compartments.

To keep the electrodes from touching, wood To keep the electrodes from touching, wood or fiber glass spacers are placed between or fiber glass spacers are placed between themthem




Using solids offers another advantage, no Using solids offers another advantage, no concentration changes occur and the emf concentration changes occur and the emf stays mostly constant, fluctuations occuring stays mostly constant, fluctuations occuring with variations in [Hwith variations in [H22SOSO44].].

One major advantage with this battery is One major advantage with this battery is the ability to be rechargedthe ability to be recharged



Alkaline Battery:Alkaline Battery:

The most common non-rechargeable battery The most common non-rechargeable battery is the alkaline battery. is the alkaline battery. More than 10More than 101010 are produced annually. are produced annually.

The anode of this battery consists of The anode of this battery consists of powdered zinc metal immobilized in a gel in powdered zinc metal immobilized in a gel in contact with a concentrated solution of contact with a concentrated solution of KOH.KOH.

The cathode is a mixture of MnOThe cathode is a mixture of MnO22 and and graphitegraphite



Nickel-Cadmium, Nickel-Metal-Hydride, Nickel-Cadmium, Nickel-Metal-Hydride, and Lithium-Ion Batteriesand Lithium-Ion Batteries

All of these batteries are lightweight, easily All of these batteries are lightweight, easily rechargeable batteries for cell phones, rechargeable batteries for cell phones, notebook computers and video recorders.notebook computers and video recorders.



Nickel-CadmiumNickel-Cadmium

Cadmium is oxidized while nickel Cadmium is oxidized while nickel oxyhydroxide is reducedoxyhydroxide is reduced

The solid reaction products adhere to the The solid reaction products adhere to the electrodes, which permits the reactions to be electrodes, which permits the reactions to be reversed during charging.reversed during charging.

Drawbacks: cadmium is a toxic heavy metalDrawbacks: cadmium is a toxic heavy metal



Nickel-Metal-HydrideNickel-Metal-Hydride

Nickel oxyhydroxide is still reduced, but Nickel oxyhydroxide is still reduced, but now the anode is a metal alloy that has the now the anode is a metal alloy that has the ability to absorb hydrogen atoms.ability to absorb hydrogen atoms.

The hydrogen atoms are released as waterThe hydrogen atoms are released as water



Lithium-Ion BatteriesLithium-Ion Batteries

Lithium is a very light-weight element , and Lithium is a very light-weight element , and the technology is unique.the technology is unique.

Li+ ions insert themselves into certain Li+ ions insert themselves into certain layered solids; during discharge lithium ions layered solids; during discharge lithium ions migrate between two different layered migrate between two different layered materials that serve as the anode and materials that serve as the anode and cathodecathode



Fuel CellsFuel Cells

The direct production of electricity from The direct production of electricity from fuels by a voltaic cell could, in principle, fuels by a voltaic cell could, in principle, yield a higher rate of conversion of the yield a higher rate of conversion of the chemical energy of the reaction.chemical energy of the reaction.

Voltaic cells that perform this conversion Voltaic cells that perform this conversion using conventional fuels are called fuel cells.using conventional fuels are called fuel cells.

Fuel cells are not true batteries because they Fuel cells are not true batteries because they are not self-contained.are not self-contained.



Fuel CellsFuel Cells

The most promising The most promising fuel cell system fuel cell system involves the reaction of involves the reaction of HH22 and O and O22 to form to form HH22O(l) as the only O(l) as the only product.product.

Cathode: 4e- + OCathode: 4e- + O22 + 2H + 2H22O O 4OH- 4OH-

Anode: 2HAnode: 2H22 + 4OH- + 4OH- 4H 4H22O + 4e- O + 4e-


127CorrosionCorrosion

Undesirable redox reactions that lead to the Undesirable redox reactions that lead to the corrosion of metals are the other side to corrosion of metals are the other side to spontaneous reactions.spontaneous reactions.

For nearly all metals, oxidation is a For nearly all metals, oxidation is a thermodynamically favorable process in air thermodynamically favorable process in air at room temperature.at room temperature.

With certain metals, when oxidation occurs With certain metals, when oxidation occurs a protective oxide layer is formed - example: a protective oxide layer is formed - example: aluminumaluminum



Corrosion of ironCorrosion of iron

From an economic standpoint, this is a From an economic standpoint, this is a major problem: 20% of the iron produced is major problem: 20% of the iron produced is used to replace iron objects that have been used to replace iron objects that have been discarded because of rust damage.discarded because of rust damage.

The rusting of iron requires both oxygen The rusting of iron requires both oxygen and water.and water.

Other factors can accelerate rusting - pH, Other factors can accelerate rusting - pH, salts, contact with metals and stresssalts, contact with metals and stress



Corrosion of ironCorrosion of iron

Corrosion can occur anywhere, but the best Corrosion can occur anywhere, but the best place is the spot where the iron has the most place is the spot where the iron has the most access to oxygen.access to oxygen.



Preventing the Corrosion of IronPreventing the Corrosion of Iron

Covering with paint or another metal can Covering with paint or another metal can protect the iron from corrosion.protect the iron from corrosion.

Paint offers a layer against the iron coming Paint offers a layer against the iron coming into contactinto contact

A second metal that is easier to oxidize then A second metal that is easier to oxidize then iron will corrode first offering protection.iron will corrode first offering protection.


133ElectrolysisElectrolysis

It is possible to use electrical energy to It is possible to use electrical energy to cause nonspontaneous redox reactions to cause nonspontaneous redox reactions to occur.occur.

Processes which are driven by an outside Processes which are driven by an outside source of electrical energy are called source of electrical energy are called electrolysis reactions and take place in electrolysis reactions and take place in electrolytic cells.electrolytic cells.



Electrolytic cellsElectrolytic cellsconsists of two electrodes in a molten consists of two electrodes in a molten salt or a solution.salt or a solution.driven by a battery or some other driven by a battery or some other source of direct electrical current.source of direct electrical current.the battery acts as an electron pump, the battery acts as an electron pump, pushing electrons into one electrode and pushing electrons into one electrode and pulling them from the otherpulling them from the otherthe electrodes are inert, acting only as the electrodes are inert, acting only as the site for oxidation and reductionthe site for oxidation and reduction



Electrolytic cellsElectrolytic cells

reduction reduction still occurs still occurs

at the at the

cathodecathode

oxidation oxidation still occurs still occurs

at the at the

anodeanode



Electrolytic cellsElectrolytic cells

Na+ Na+ reduces at reduces at

cathode to cathode to form Naform Na

Cl- oxidizes Cl- oxidizes at anode to at anode to form Clform Cl22



Electrolytic cellsElectrolytic cellsthe electrode connected to the negative the electrode connected to the negative terminal is now the cathode due to the terminal is now the cathode due to the battery supplying the electrons for battery supplying the electrons for reductionreductionthe electrode connected to the positive the electrode connected to the positive terminal is now the anode to draw terminal is now the anode to draw electrons off the metal and cause the electrons off the metal and cause the metal to oxidize and form ionsmetal to oxidize and form ions



Electrolysis of Aqueous SolutionsElectrolysis of Aqueous Solutionsbecause of the high melting points of because of the high melting points of ionic substances, the electrolysis of ionic substances, the electrolysis of molten salts requires very high molten salts requires very high temperaturestemperatureswe can produce ions from soluble salts we can produce ions from soluble salts at room temperatures by dissolving the at room temperatures by dissolving the salt in watersalt in waterthe electrolysis of an aqueous solution the electrolysis of an aqueous solution is complicated by the presence of wateris complicated by the presence of water



Electrolysis of Aqueous SolutionsElectrolysis of Aqueous Solutionswe must consider if the water is we must consider if the water is oxidized or reducedoxidized or reducedif the water is oxidized, Oif the water is oxidized, O22 and H and H++ ions ions are formedare formedif the water is reduced, Hif the water is reduced, H22 and OH and OH-- ions are formedions are formedthe more positive the value of Ethe more positive the value of E°°

redred, the , the more favorable the reduction ismore favorable the reduction is



Electrolysis with Active ElectrodesElectrolysis with Active Electrodesseveral practical applications of several practical applications of electrochemistry are based on active electrochemistry are based on active electrodeselectrodeselectroplating involves using electroplating involves using electrolysis to deposit a thin layer of one electrolysis to deposit a thin layer of one metal on top of another in order to metal on top of another in order to improve beauty or resistance to improve beauty or resistance to corrosioncorrosion



Electrolysis with Active ElectrodesElectrolysis with Active Electrodes



Quantitative Aspects of ElectrolysisQuantitative Aspects of Electrolysisfor any half reaction the amount of a for any half reaction the amount of a substance that is reduced or oxidized in substance that is reduced or oxidized in an electrolytic cell is directly an electrolytic cell is directly proportional to the number of electrons proportional to the number of electrons passed into the cellpassed into the cellFig 20.31Fig 20.31



Sample exercise: The half reaction for Sample exercise: The half reaction for formation of magnesium metal upon formation of magnesium metal upon electrolysis of molten MgClelectrolysis of molten MgCl22 is is MgMg2+2+ + 2e + 2e-- Mg. Calculate the mass of Mg. Calculate the mass of magnesium formed upon passage of magnesium formed upon passage of current of 60.0 A for a period of current of 60.0 A for a period of 4.00 x 104.00 x 1033 s. s.



Sample exercise: MgSample exercise: Mg2+2+ + 2e + 2e-- Mg. Calculate Mg. Calculate the mass of magnesium formed upon the mass of magnesium formed upon passage of current of 60.0 A for a period of passage of current of 60.0 A for a period of 4.00 x 104.00 x 1033 s. s.

Coulombs= A x sCoulombs= A x s

= 60.0 A x 4.00 x 10= 60.0 A x 4.00 x 1033 s s

= 2.4 x 10= 2.4 x 1055 C C



Sample exercise: MgSample exercise: Mg2+2+ + 2e + 2e-- Mg. Calculate the Mg. Calculate the mass of magnesium formed upon passage of mass of magnesium formed upon passage of current of 60.0 A for a period of 4.00 x 10current of 60.0 A for a period of 4.00 x 1033 s. s.

2.4 x 102.4 x 1055 C 1 mole e C 1 mole e-- 1 mole Mg 24.305 g Mg 1 mole Mg 24.305 g Mg 96,500 C 2 mole e 96,500 C 2 mole e-- 1 mole Mg 1 mole Mg



Sample exercise: MgSample exercise: Mg2+2+ + 2e + 2e-- Mg. Calculate the Mg. Calculate the mass of magnesium formed upon passage of mass of magnesium formed upon passage of current of 60.0 A for a period of 4.00 x 10current of 60.0 A for a period of 4.00 x 1033 s. s.

2.4 x 102.4 x 1055 C 1 mole e C 1 mole e-- 1 mole Mg 24.305 g Mg 1 mole Mg 24.305 g Mg 96,500 C 2 mole e 96,500 C 2 mole e-- 1 mole Mg 1 mole Mg

30.2 g Mg30.2 g Mg



Work done by electrical current can be Work done by electrical current can be calculated now based on the same formula as calculated now based on the same formula as Gibbs Free Energy.Gibbs Free Energy.

wwmaxmax = -nFE * voltaic cells = -nFE * voltaic cells

w = nFEw = nFEextext *electrolytic cells*electrolytic cells

prof. t.l. heise, che 116 1 chapter twenty copyright © tyna l. heise 2001 - 2002 all rights...

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