Why Study CHAPTER 20 Electrochemistry? · ions in acidic solution. ... peroxide is reduced to hydroxide ions. Write ... Quantitative Aspects of Electrochemistry Solution (a)

1

CHAPTER 20

� Electron Transfer Reactions

Why Study Electrochemistry?Why Study Why Study Electrochemistry?Electrochemistry?

�� BatteriesBatteries

�� CorrosionCorrosion

�� Industrial production Industrial production

of chemicals such as of chemicals such as

ClCl22, , NaOHNaOH, F, F22 and Aland Al

�� Biological Biological redoxredox

reactionsreactions

The The hemeheme groupgroup

TRANSFER REACTIONSTRANSFER REACTIONS

Atom/Group transferAtom/Group transfer

HClHCl + H+ H22O O ------> > ClCl-- + H+ H33OO

++

Electron transferElectron transfer

Cu(s) + 2 Cu(s) + 2 AgAg++(aq(aq) ) ------> Cu> Cu2+2+(aq) + 2 Ag(s) (aq) + 2 Ag(s)

4

Oxidation-Reduction Reactions

� Rules for assigning oxidation numbers

� Alkali (+1), Alkaline Earth (+2), Al family (+3)

� Halides (-1) w/out oxygen

� Oxygen family (-2) (except peroxide,

superoxide)

� H (+1) (except with metal -1, hydride)

� Transition metals (by difference)

Oxidation-Reduction Reactions

� Oxidation – increase in oxidation #

� Loss of electron

� Fe(III) � Fe(II) + e-

� Reduction – decrease in oxidation #

� Gain of electron

� MnO4- + 5 e- � Mn(II)

5

Review of Terminology for Redox ReactionsReview of Terminology Review of Terminology for for RedoxRedox ReactionsReactions

� OXIDATION—loss of electron(s) by a species;

increase in oxidation number.

� REDUCTION—gain of electron(s); decrease in

oxidation number.

� OXIDIZING AGENT—electron acceptor;

species is reduced.

� REDUCING AGENT—electron donor; species is

oxidized.

�� OXIDATIONOXIDATION——loss of electron(s) by a species; loss of electron(s) by a species;

increase in oxidation number.increase in oxidation number.

�� REDUCTIONREDUCTION——gain of electron(s); decrease in gain of electron(s); decrease in

oxidation number.oxidation number.

�� OXIDIZING AGENTOXIDIZING AGENT——electron acceptor; electron acceptor;

species is reduced.species is reduced.

�� REDUCING AGENTREDUCING AGENT——electron donor; species is electron donor; species is

oxidized.oxidized.

7

The Half-Reaction Method

� Half reaction method rules:

1. Write the unbalanced reaction.

2. Break the reaction into 2 half reactions:

One oxidation half-reaction and

One reduction half-reaction

Each reaction must have complete formulas for

molecules and ions.

3. Mass balance each half reaction by adding

appropriate stoichiometric coefficients. To balance H and O we can add:

� H+ or H2O in acidic solutions.

� OH- or H2O in basic solutions.8


4. Charge balance the half reactions by adding appropriate numbers of electrons.� Electrons will be products in the oxidation half-

reaction.

� Electrons will be reactants in the reduction half-reaction.

5. Multiply each half reaction by a number to make the number of electrons in the oxidation half-reaction equal to the number of electrons reduction half-reaction.

6. Add the two half reactions.

7. Eliminate any common terms and reduce coefficients to smallest whole numbers.

Tips on Balancing EquationsTips on Balancing Equations

�� Never add ONever add O22, O atoms, or , O atoms, or

OO22-- to balance oxygen.to balance oxygen.

�� Never add HNever add H22 or H atoms to or H atoms to

balance hydrogen.balance hydrogen.

�� Be sure to write the correct Be sure to write the correct

charges on all the ions.charges on all the ions.

�� Check your work at the end to Check your work at the end to

make sure mass and charge make sure mass and charge

are balanced.are balanced.

�� PRACTICE!PRACTICE!

Balancing EquationsBalancing Equations

Cu + AgCu + Ag++ --------> Cu> Cu2+2+ + Ag+ Ag


Step 1:Step 1: Divide the reaction into halfDivide the reaction into half--reactions, reactions,

one for oxidation and the other for reduction.one for oxidation and the other for reduction.

OxOx Cu Cu ------> Cu> Cu2+2+

RedRed AgAg++ ------> Ag> Ag

Step 2:Step 2: Balance each for mass. Already done Balance each for mass. Already done

in this case.in this case.

Step 3:Step 3: Balance each halfBalance each half--reaction for charge reaction for charge

by adding electrons.by adding electrons.

OxOx Cu Cu ------> Cu> Cu2+2+ + + 2e2e--

RedRed AgAg++ + + ee-- ------> Ag> Ag


Step 4:Step 4: Multiply each halfMultiply each half--reaction by a factor so that reaction by a factor so that

the reducing agent supplies as many electrons as the the reducing agent supplies as many electrons as the

oxidizing agent requires.oxidizing agent requires.

Reducing agentReducing agent Cu Cu ------> Cu> Cu2+2+ + 2e+ 2e--

Oxidizing agentOxidizing agent 22 AgAg++ + + 22 ee-- ------> > 22 AgAg

Step 5:Step 5: Add halfAdd half--reactions to give the overall reactions to give the overall

equation.equation.

Cu + 2 AgCu + 2 Ag++ ------> Cu> Cu2+2+ + 2Ag+ 2Ag

The equation is now balanced for both charge and mass.The equation is now balanced for both charge and mass.

13


� Example: Tin (II) ions are oxidized to tin (IV)

by bromine.

-+4

2

+2 BrSnBrSn

Reaction Starting

+→+

Reduction of VOReduction of VO22++ with Znwith Zn

Balance the following in Balance the following in acidacid solutionsolution——

VOVO22++ + Zn + Zn ------> VO> VO2+ 2+ + Zn+ Zn2+2+

15


� Example: Dichromate ions oxidize iron (II) ions to

iron (III) ions and are reduced to chromium (III)

ions in acidic solution. Write and balance the net

ionic equation for the reaction.

reaction starting FeCrFeOCr +3+3+22

72 +→+−

16


� Example: In basic solution hydrogen

peroxide oxidizes chromite ions, Cr(OH)4-, to

chromate ions, CrO42-. The hydrogen

peroxide is reduced to hydroxide ions. Write

and balance the net ionic equation for this

reaction.

17

Stoichiometry of Redox Reactions

� Just as we have done stoichiometry with

acid-base reactions, it can also be done with

redox reactions.

� Example: What volume of 0.200 M KMnO4 is

required to oxidize 35.0 mL of 0.150 M HCl?

The balanced reaction is:

OH 8Cl 5MnCl 2+KCl 2HCl 16 KMnO 2 2224 ++→+

You do it!You do it!

18

Stoichiometry of Redox Reactions

� Example: A volume of 40.0 mL of iron

(II) sulfate is oxidized to iron (III) by

20.0 mL of 0.100 M potassium

dichromate solution. What is the

concentration of the iron (II) sulfate

solution? The balanced equation is:

OH 7Cr 2Fe 6H 14OCrFe 6 2

+3+3+2

72

+2++→++

−

19

Electrochemistry

� Electrochemical reactions are oxidation-reduction

reactions.

� The two parts of the reaction are physically separated.

� The oxidation reaction occurs in one cell.

� The reduction reaction occurs in the other cell.

� There are two kinds electrochemical cells.

1. Electrochemical cells containing in nonspontaneous

chemical reactions are called electrolytic cells.

2. Electrochemical cells containing spontaneous chemical

reactions are called voltaic or galvanic cells.

20

Electrical Conduction

� In ionic or electrolytic conduction ionic

motion transports the electrons.

� Positively charged ions, cations, move toward

the negative electrode.

� Negatively charged ions, anions, move toward

the positive electrode.

21

Electrodes

� The following convention for electrodes is

correct for either electrolytic or voltaic cells:

� The cathode is the electrode at which

reduction occurs.� The cathode is negative in electrolytic cells and

positive in voltaic cells.

� The anode is the electrode at which

oxidation occurs.� The anode is positive in electrolytic cells and negative

in voltaic cells.22

Electrolytic Cells

� Electrical energy is used to force

nonspontaneous chemical reactions to

occur.

� The process is called electrolysis.

� Two examples of commercial electrolytic

reactions are:

1. The electroplating of jewelry and auto parts.

2. The electrolysis of chemical compounds.

23

Electrolytic Cells

� Electrolytic cells consist of:

1. A container for the reaction mixture.

2. Two electrodes immersed in the reaction

mixture.

3. A source of direct current.

24

The Electrolysis of Molten Potassium Chloride

� Liquid potassium is produced at one electrode.

� Indicates that the reaction K+(l) + e

- → K(l) occurs at this electrode.

� Is this electrode the anode or cathode?

� Gaseous chlorine is produced at the other electrode.

� Indicates that the reaction 2 Cl- → Cl2(g) + 2 e-

occurs at this electrode.

� Is this electrode the anode or cathode?

25

The Electrolysis of Molten Potassium Chloride

Diagram of this electrolytic cell.

Porous barrier

e-

K+ + e- → K(l)

cathode reaction

2Cl- → Cl2 (g) + 2e-

anode reaction

Generator-source

of DC

- electrode + electrode

e-

molten KCl

26

The Electrolysis of Aqueous Potassium Chloride

2 H2O + 2e- → H2 (g) + 2 OH

-

cathode reaction

2Cl- → Cl2 (g) + 2e-

anode reaction

Cell diagram Battery, a source

of direct currente- flow

- electrode + electrode

e- flow

aqueous KCl

Cl2 gasH2 gas

+ pole of battery- pole of battery

Electrolysis of Molten Electrolysis of Molten NaClNaCl

Figure 20.18Figure 20.18

Electrolysis of Molten Electrolysis of Molten NaClNaCl

Anode (+) Anode (+)

2 2 ClCl-- ------> Cl> Cl22(g) + 2e(g) + 2e--

Cathode (Cathode (--) )

NaNa++ + e+ e-- ------> Na> Na

BATTERY

+

Na+Cl-

Anode Cathode

electrons

EEoo for cell (in water) = Efor cell (in water) = E˚̊cc -- EE˚̊aa

= = -- 2.71 V 2.71 V –– (+1.36 V)(+1.36 V)

= = -- 4.07 V (in water)4.07 V (in water)

External energy needed because External energy needed because EEoo is (is (--). ).

Electrolysis of Aqueous Electrolysis of Aqueous NaClNaCl

Anode (+) Anode (+)

2 2 ClCl-- ------> >

ClCl22(g) + 2e(g) + 2e--

Cathode (Cathode (--) )

2 H2 H22O + 2eO + 2e-- ------> >

HH22 + 2 OH+ 2 OH--

EEoo for cell = for cell = --2.19 V2.19 V

Note that HNote that H22O is more O is more

easily reduced easily reduced

than Nathan Na++. .

BATTERY

+

Na+Cl -

Anode Cathode

H2O

electrons

Also, Cl- is oxidized in preference to H2O because of kinetics.

Also, Also, ClCl-- is oxidized in is oxidized in

preference to Hpreference to H22O because of O because of

kinetics.kinetics.

Michael FaradayMichael Faraday17911791--18671867

Originated the terms anode, Originated the terms anode,

cathode, anion, cathode, anion, cationcation, ,

electrode.electrode.

Discoverer of Discoverer of

�� electrolysiselectrolysis

�� magnetic props. of mattermagnetic props. of matter

�� electromagnetic inductionelectromagnetic induction

�� benzene and other organic benzene and other organic

chemicalschemicals

Was a popular lecturer.Was a popular lecturer.

31

Counting Electrons:

Coulometry and Faraday’s Law of Electrolysis

� Faraday’s Law - The amount of substance

undergoing chemical reaction at each electrode

during electrolysis is directly proportional to the

amount of electricity that passes through the

electrolytic cell.

� A faraday is the amount of electricity that reduces

one equivalent of a species at the cathode and

oxidizes one equivalent of a species at the anode.

-23 e 106.022y electricit offaraday 1 ×≡

32

Counting Electrons:

Coulometry and Faraday’s Law of Electrolysis

� A coulomb is the amount of charge that passes a

given point when a current of one ampere (A) flows

for one second.

� 1 amp = 1 coulomb/second

coulombs 487,96e 106.022faraday 1 -23 ≡×≡

Quantitative Aspects of Quantitative Aspects of ElectrochemistryElectrochemistry

Consider electrolysis of aqueous silver ion.Consider electrolysis of aqueous silver ion.

AgAg++ ((aqaq) + e) + e-- ------> Ag(s)> Ag(s)

1 mol e1 mol e-- ------> 1 mol Ag> 1 mol Ag

If we could measure the moles of eIf we could measure the moles of e--, we could , we could

know the quantity of Ag formed.know the quantity of Ag formed.

But how to measure moles of eBut how to measure moles of e--??

Current = charge passing

time

I (amps) = coulombs

seconds

But how is charge related to moles of But how is charge related to moles of electrons?electrons?

Current = charge passing

timeI (amps) =

coulombs

seconds

Quantitative Aspects of Quantitative Aspects of ElectrochemistryElectrochemistry

= = 96,500 C/mol e96,500 C/mol e--= = 1 Faraday1 Faraday

Michael FaradayMichael Faraday

17911791--18671867

Quantitative Aspects of ElectrochemistryQuantitative Aspects of Electrochemistry

1.50 amps flow thru a 1.50 amps flow thru a AgAg++(aq(aq) solution for 15.0 ) solution for 15.0

min. What mass of Ag metal is deposited?min. What mass of Ag metal is deposited?

SolutionSolution

(a)(a) Calc. chargeCalc. charge

Charge (C) = current (A) x time (t)Charge (C) = current (A) x time (t)

= (1.5 amps)(15.0 min)(60 s/min) = 1350 C= (1.5 amps)(15.0 min)(60 s/min) = 1350 C

I (amps) = coulombs

seconds


SolutionSolution

(a)(a) Charge = 1350 CCharge = 1350 C

(b)(b) Calculate moles of eCalculate moles of e-- usedused

I (amps) = coulombs

seconds

1350 C • 1 mol e -

96,500 C==== 0.0140 mol e -

0.0140 mol e - • 1 mol Ag

1 mol e -==== 0.0140 mol Ag or 1.51 g Ag

1.50 amps flow thru a 1.50 amps flow thru a AgAg++(aq(aq) solution for 15.0 min. What ) solution for 15.0 min. What

mass of Ag metal is deposited?mass of Ag metal is deposited?

(c)(c) Calc. quantity of AgCalc. quantity of Ag


The anode reaction in a lead storage battery isThe anode reaction in a lead storage battery is

Pb(sPb(s) + HSO) + HSO44--(aq) (aq) ------> PbSO> PbSO44(s) + (s) + HH++(aq(aq) + 2e) + 2e--

If a battery delivers 1.50 amp, and you have 454 g of If a battery delivers 1.50 amp, and you have 454 g of PbPb, ,

how long will the battery last?how long will the battery last?

SolutionSolution

a)a) 454 g 454 g PbPb = 2.19 mol = 2.19 mol PbPb

b)b) Calculate moles of eCalculate moles of e--

2.19 mol Pb • 2 mol e -

1 mol Pb = 4.38 mol e -

c)c) Calculate chargeCalculate charge

4.38 mol e4.38 mol e-- •• 96,500 C/mol e96,500 C/mol e-- = 423,000 C= 423,000 C


The anode reaction in a lead storage battery isThe anode reaction in a lead storage battery is

Pb(sPb(s) + HSO) + HSO44--(aq) (aq) ------> PbSO> PbSO44(s) + (s) + HH++(aq(aq) + 2e) + 2e--

If a battery delivers 1.50 amp, and you have 454 g of If a battery delivers 1.50 amp, and you have 454 g of PbPb, how long will the battery last?, how long will the battery last?

SolutionSolution

a)a) 454 g 454 g PbPb = 2.19 mol = 2.19 mol PbPb

b)b) Mol of eMol of e-- = 4.38 mol= 4.38 mol

c)c) Charge = 423,000 CCharge = 423,000 C

Time (s) = Charge (C)

I (amps)

Time (s) = 423, 000 C

1.50 amp = 282, 000 s About 78 hoursAbout 78 hours

d)d) Calculate timeCalculate time

39

Counting Electrons:

Coulometry and Faraday’s Law of Electrolysis� Example: Calculate the mass of palladium produced by the

reduction of palladium (II) ions during the passage of 3.20 amperes of current through a solution of palladium (II) sulfate for 30.0 minutes.

40

Counting Electrons:

Coulometry and Faraday’s Law of Electrolysis� Example: Calculate the mass of palladium produced by the

reduction of palladium (II) ions during the passage of 3.20 amperes of current through a solution of palladium (II) sulfate for 30.0 minutes.

( )

Cathode: Pd + 2e Pd

1 mol 2 mol 1 mol

106 g 2(96,500) 106 g

3.20 amp = 3.20

g = 30.0 min60 s

min

C

s

g Pd

2 96,500 C g Pd

2+ - 0

Cs

→

× × × =?.

.320 106

316

41

Commercial Applications of Electrolytic Cells

Electrolytic Refining and Electroplating of Metals

� Impure metallic copper can be purified

electrolytically to ≈ 100% pure Cu.

� The impurities commonly include some active metals

plus less active metals such as: Ag, Au, and Pt.

� The cathode is a thin sheet of copper metal

connected to the negative terminal of a direct

current source.

� The anode is large impure bars of copper.

Oxidation:Oxidation: Zn(s) Zn(s) ------> Zn> Zn2+2+(aq) + (aq) + 2e2e--

Reduction:Reduction: CuCu2+2+(aq) + (aq) + 2e2e-- ------> Cu(s)> Cu(s)

----------------------------------------------------------------------------------------------------------------

CuCu2+2+(aq) + Zn(s) (aq) + Zn(s) ------> Zn> Zn2+2+(aq) + Cu(s)(aq) + Cu(s)

Zn metal

Cu2+ ions

Zn metal

Cu2+ ionsElectrons are transferred from Zn to

Cu2+, but there is no useful electric

current.

Electrons are transferred from Zn to

Cu2+, but there is no useful electric

current.

With time, Cu plates out onto Zn metal strip, and Zn strip

“disappears.”

With time, Cu plates out onto Zn metal strip, and Zn strip

“disappears.”

Chemical Change�� Electrical

Current

��To obtain a useful current, To obtain a useful current,

we separate the oxidizing we separate the oxidizing

and reducing agents so that and reducing agents so that

electron transfer occurs thru electron transfer occurs thru

an external wire. an external wire.

Zn

Zn2+ ions

Cu

Cu2+ ions

wire

saltbridge

elect rons

This is accomplished in a This is accomplished in a GALVANICGALVANIC or or

VOLTAICVOLTAIC cell.cell.

A group of such cells is called a A group of such cells is called a batterybattery..

Chemical Change�� Electrical

Current

44

Voltaic or Galvanic Cells

� Electrochemical cells in which a spontaneous

chemical reaction produces electrical energy.

� Cell halves are physically separated so that

electrons (from redox reaction) are forced to travel

through wires and creating a potential difference.

� Examples of voltaic cells include:

batteries calculator andComputer

batteries Flashlight

batteries Auto

45

The Construction of Simple Voltaic Cells

� Voltaic cells consist of two half-cells which

contain the oxidized and reduced forms of an

element (or other chemical species) in

contact with each other.

� A simple half-cell consists of:

� A piece of metal immersed in a solution of its

ions.

� A wire to connect the two half-cells.

� And a salt bridge to complete the circuit, maintain

neutrality, and prevent solution mixing.

Terms Used for Voltaic CellsTerms Used for Voltaic Cells


47

Standard Electrode Potential

� To measure relative

electrode potentials, we

must establish an arbitrary

standard.

� That standard is the

Standard Hydrogen

Electrode (SHE).

� The SHE is assigned an

arbitrary voltage of

0.000000… V

48

The Zinc-SHE Cell

� For this cell the

components are:

1. A Zn strip

immersed in 1.0

M zinc (II) sulfate.

2. The other

electrode is the

Standard

Hydrogen

Electrode.

3. A wire and a salt

bridge to

complete the

circuit.

� The initial cell

voltage is 0.763

volts.

49

The Zinc-SHE Cell

( )

( ) V 763.0E H+Zn2H+Znreaction Cell

V 0.000 He 2+H 2reaction Cathode

V 0.763 e 2+Zn Znreaction Anode

E

0

cellg2

+2+0

g2

-+

-+20

0

=→

→

→

� The cathode is the Standard Hydrogen Electrode.

� In other words Zn reduces H+ to H2.

� The anode is Zn metal.

� Zn metal is oxidized to Zn2+ ions.

Volts

Zn

H2

Salt Bridge

Zn2+ H+

Zn Zn2+ + 2e- OXIDATION ANODE

2 H+ + 2e- H2REDUCTIONCATHODE

-+

Overall reaction is reduction of HOverall reaction is reduction of H++ by Zn metal.by Zn metal.

Zn(sZn(s) + 2 H) + 2 H++ ((aqaq) ) ----> Zn> Zn2+2+ + H+ H22(g)(g) EEoo = +0.76 V= +0.76 V

Therefore, Therefore, EEoo for for Zn Zn ------> Zn> Zn2+2+ ((aqaq) + 2e) + 2e-- is is +0.76 V+0.76 V

Zn is a Zn is a (better) (poorer)(better) (poorer) reducing agent than Hreducing agent than H22..

Volts

Cu

H2

Salt Bridge

Cu2+ H+

Cu2+ + 2e- Cu REDUCTION CATHODE

H2 2 H+ + 2e-

OXIDATION ANODE

-+

Cu/CuCu/Cu2+2+ and Hand H22/H/H++ CellCell

EEoo = +0.34 V= +0.34 V

Acceptor Acceptor

of of

electronselectrons

Supplier Supplier

of of

electronselectrons

CuCu2+2+ + 2e+ 2e-- ----> Cu> Cu

ReductionReduction

CathodeCathode

HH22 ----> 2 H> 2 H++ + 2e+ 2e--

OxidationOxidation

AnodeAnode

PositivePositive NegativeNegative

Cu/CuCu/Cu2+2+ and Hand H22/H/H++ CellCell

Overall reaction is reduction of CuOverall reaction is reduction of Cu2+2+ by Hby H22 gas.gas.

CuCu2+2+ ((aqaq) + H) + H22(g) (g) ------> Cu(s) + 2 > Cu(s) + 2 HH++(aq(aq))

Measured Measured EEoo = +0.34 V= +0.34 V

Therefore, Therefore, EEoo for Cufor Cu2+2+ + 2e+ 2e-- ------> Cu> Cu isis

Volts

Cu

H2

Salt Bridge

Cu2+ H+

Cu2+ + 2e- Cu REDUCTION CATHODE

H2 2 H+ + 2e-

OXIDATION ANODE

-+

+0.34 V+0.34 V

53

Uses of Standard Electrode Potentials

� Electrodes that force the SHE to act as an anode are assigned

positive standard reduction potentials.

� Electrodes that force the SHE to act as the cathode are assigned

negative standard reduction potentials.

� Standard electrode (reduction) potentials tell us the tendencies of

half-reactions to occur as written.

� For example, the half-reaction for the standard potassium

electrode is:

V -2.925=E K e K 00→+ −+

The large negative value tells us that this reaction

will occur only under extreme conditions.

54

Uses of Standard Electrode Potentials

� Compare the potassium half-reaction to fluorine’s half-

reaction:

� The large positive value denotes that this reaction

occurs readily as written.

� Positive E0 values denote that the reaction tends to

occur to the right.

� The larger the value, the greater the tendency to occur

to the right.

� It is the opposite for negative values of Eo.

F + 2 e 2 F E = +2.87 V20 - - 0→

Using Standard Potentials, Using Standard Potentials, EEoo

Table 20.1Table 20.1

� Which is the best oxidizing agent:

O2, H2O2, or Cl2? _________________

� Which is the best reducing agent:

Hg, Al, or Sn? ____________________

Standard Standard RedoxRedox Potentials, Potentials, EEoo

Any substance on the right will Any substance on the right will

reduce any substance higher reduce any substance higher

than it on the left.than it on the left.

�� Zn can reduce HZn can reduce H++ and Cuand Cu2+2+..

�� HH22 can reduce Cucan reduce Cu2+2+ but not Znbut not Zn2+2+

�� Cu cannot reduce HCu cannot reduce H++ or Znor Zn2+2+..

Eo (V)

Cu2+ + 2e- Cu +0.34

2 H+ + 2e- H2 0.00

Zn2+ + 2e- Zn -0.76

oxidizingability of ion

reducing abilityof element


Cu2+ + 2e- --> Cu +0.34

+2 H + 2e- --> H2 0.00

Zn2+ + 2e- --> Zn -0.76

NorthwestNorthwest--southeast rule:southeast rule: productproduct--favored favored

reactions occur between reactions occur between

•• reducing agent at southeast corner reducing agent at southeast corner

•• oxidizing agent at northwest corneroxidizing agent at northwest corner

Any substance on the right will reduce any substance Any substance on the right will reduce any substance

higher than it on the left.higher than it on the left.

Ox. agentOx. agent

Red. agentRed. agent

Using Standard Potentials, Using Standard Potentials, EEooTable 20.1Table 20.1

�� In which direction do the following reactions go?In which direction do the following reactions go?

�� Cu(s) + 2 Cu(s) + 2 AgAg++(aq(aq) ) ------> Cu> Cu2+2+(aq) + 2 Ag(s)(aq) + 2 Ag(s)

�� Goes right as writtenGoes right as written

�� 2 Fe2 Fe2+2+(aq) +(aq) + Sn2+(aq) ---> 2 Fe3+(aq) + Sn(s)

�� Goes LEFT opposite to direction writtenGoes LEFT opposite to direction written

� What is Eonet for the overall reaction?


EE˚̊netnet = = ““distancedistance”” from from ““toptop”” halfhalf--reaction reaction

(cathode)(cathode) to to ““bottombottom”” halfhalf--reaction reaction (anode)(anode)

EE˚̊netnet = E= E˚̊cathodecathode -- EE˚̊anodeanode

EEoonetnet for Cu/Agfor Cu/Ag++ reaction = +0.46 Vreaction = +0.46 V

Zn

Zn2+ ions

Cu

Cu2+ ions

wire

saltbridge

elect rons

•Electrons travel thru external wire.

�Salt bridge allows anions and cations to move

between electrode compartments.

••Electrons travel thru external wire.Electrons travel thru external wire.

��Salt bridge Salt bridge allows anions and allows anions and cationscations to move to move

between electrode compartments.between electrode compartments.

Zn Zn ----> Zn> Zn2+2+ + 2e+ 2e-- CuCu2+2+ + 2e+ 2e-- ----> Cu> Cu

<<----AnionsAnions

CationsCations---->>

Oxidation

Anode

Negative

OxidationOxidation

AnodeAnode

NegativeNegative

Reduction

CathodePositive

ReductionReduction

CathodeCathode

PositivePositive

CELL POTENTIAL, ECELL POTENTIAL, E

�� Electrons are Electrons are ““drivendriven”” from anode to cathode by an from anode to cathode by an electromotive forceelectromotive force or or

emfemf..

�� For Zn/Cu cell, this is indicated by a voltage of 1.10 V at 25 For Zn/Cu cell, this is indicated by a voltage of 1.10 V at 25 ˚̊C and when C and when

[Zn[Zn2+2+] and [Cu] and [Cu2+2+] = 1.0 M.] = 1.0 M.

�� STANDARD CELL POTENTIAL, STANDARD CELL POTENTIAL, EEoo

�� ——a quantitative measure of the tendency of reactants to proceed ta quantitative measure of the tendency of reactants to proceed to o

products when all are in their standard states at 25 products when all are in their standard states at 25 ˚̊C. C.

Zn and ZnZn and Zn2+2+,,

anodeanodeCu and CuCu and Cu2+2+,,

cathodecathode

Zn

Zn2+ ions

Cu

Cu 2+ ions

wire

saltbridge

e lect rons

1.10 V1.10 V

1.0 M1.0 M 1.0 M1.0 M

Calculating Cell VoltageCalculating Cell Voltage

�� Balanced halfBalanced half--reactions can be reactions can be

added together to get overall, added together to get overall,

balanced equation. balanced equation.

Zn(s) ---> Zn2+(aq) + 2e-Cu2+(aq) + 2e- ---> Cu(s)--------------------------------------------

Cu2+(aq) + Zn(s) ---> Zn2+(aq) + Cu(s)

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

CuCu2+2+(aq) + 2e(aq) + 2e-- ------> Cu(s)> Cu(s)

----------------------------------------------------------------------------------------

CuCu2+2+(aq) + Zn(s) (aq) + Zn(s) ------> Zn> Zn2+2+(aq) + Cu(s)(aq) + Cu(s)

If we know If we know EEoo for each halffor each half--reaction, we reaction, we

could get could get EEoo for net reaction.for net reaction.

Uses of Eo ValuesUses of Uses of EEoo ValuesValues

Organize halfOrganize half--reactions by reactions by relative ability to act as relative ability to act as oxidizing agentsoxidizing agents

Zn

Zn2+ ions

Cu

Cu2+ ions

wire

saltbridge

elect rons

CuCu2+2+(aq) + 2e(aq) + 2e-- ------> Cu(s)> Cu(s) EEoo = +0.34 V= +0.34 V

ZnZn2+2+(aq) + 2e(aq) + 2e-- ------> Zn(s) > Zn(s) EEoo = = ––0.76 V0.76 V

Zn(s) Zn(s) ------> Zn> Zn2+2+(aq) + 2e(aq) + 2e-- EEoo = +0.76 V= +0.76 V

CuCu2+2+(aq) + 2e(aq) + 2e-- ------> Cu(s)> Cu(s) EEoo = +0.34 V= +0.34 V

------------------------------------------------------------------------------------------------------------------------------

CuCu2+2+(aq) + Zn(s) (aq) + Zn(s) ------> Zn> Zn2+2+(aq) + Cu(s) (aq) + Cu(s)

EEoo ((calccalc’’dd) = +1.10 V) = +1.10 V

� In all voltaic cells, electrons flow spontaneously from the

negative electrode (anode) to the positive electrode (cathode).

65

The Zinc-Copper Cell

� There is a commonly used short hand notation for voltaic cells.

� The Zn-Cu cell provides a good example.

Zn/Zn2+(1.0 M ) || Cu2+(1.0 M )/Cu

species (and concentrations)in contact with electrode surfaces

electrode surfaces

salt bridge

Volts

Cd Salt Bridge

Cd2+

Fe

Fe2+

CdCd ----> Cd> Cd2+2+ + 2e+ 2e--

oror

CdCd2+2+ + 2e+ 2e-- ----> > CdCd

Fe Fe ----> Fe> Fe2+2+ + 2e+ 2e--

oror

FeFe2+2+ + 2e+ 2e-- ----> Fe> Fe

EEoo for a Voltaic Cellfor a Voltaic Cell

All ingredients are present. Which way does All ingredients are present. Which way does

reaction proceed? What is the value of Ereaction proceed? What is the value of E00cellcell??

67

Effect of Concentrations (or Partial Pressures) on Electrode Potentials

The Nernst Equation

� Standard electrode potentials, those

compiled in appendices, are determined at

thermodynamic standard conditions.

� Reminder of standard conditions. 1.00 M solution concentrations

1.00 atm of pressure for gases

All liquids and solids in their standard thermodynamic

states.

Temperature of 250 C.68

The Nernst Equation

� The value of the cell potentials change if conditions are nonstandard.

� The Nernst equation describes the electrode potentials at nonstandard conditions.

� The Nernst equation is:

( )

quotientreaction =Q

e mol J/V 487,96

VCJ 1)e C/mol (96,487=faraday the=F

ed transferrelectrons ofnumber =n

Kin etemperatur=T

K molJ 8.314=constant gas universal=R

conditions standardunder potentialE

interest ofcondition under potential=E

Q log nF

2.303RT-E=E

-

.-

0

0

=

×

=

69

The Nernst Equation

� Substitution of the values of the constants into the

Nernst equation at 25o C gives:

Q log n

0.0592-E=E Thus

0592.0e mol J/V 96,487

K 298314.8303.2

F

RT 2.303

0

-

K molJ

=××

=

70

The Nernst Equation

� For this half-reaction:

� The corresponding Nernst equation

is:

V +0.153=ECueCu 0-+2 ←→+ +

[ ][ ]

E = E -0.0592

1 log

Cu

Cu

0

+

2+

71

The Nernst Equation

� Substituting E0 into the above expression gives:

� If [Cu2+] and [Cu+] are both 1.0 M, i.e. at standard

conditions, then E = E0 because the concentration

term equals zero.

[ ][ ]

E = 0.153 V -0.0592

1 log

Cu

Cu

+

2+

E = 0.153 V -0.0592

1 log

1

172

The Nernst Equation

� The Nernst equation can also be used to calculate

the potential for a cell that consists of two

nonstandard electrodes.

� Example: Calculate the initial potential of a cell that

consists of an Fe3+/Fe2+ electrode in which

[Fe3+]=1.0 x 10-2 M and [Fe2+]=0.1 M connected to

a Sn4+/Sn2+ electrode in which [Sn4+]=1.0 M and

[Sn2+]=0.10 M . A wire and salt bridge complete

the circuit.

73

Relationship of E0cell to

∆∆∆∆G0 and K� From previous chapters we know the

relationship of ∆G0 and K for a reaction.

K log RT 303.2G

or

lnK -RTG

0

0

−=∆

=∆

e ofnumber n

e mol J/V 96,487 F where

E F-n G

-

-

0

cell

0

=

=

=∆

EEoo and and ∆∆GGoo

∆∆GGoo = = -- n F n F EEoo

For a For a productproduct--favoredfavored reactionreaction

Reactants Reactants --------> Products> Products

∆∆GGo o < 0 and so < 0 and so EEoo > 0> 0

EEoo is positiveis positive

For a For a reactantreactant--favoredfavored reactionreaction

Reactants <Reactants <-------- ProductsProducts

∆∆GGo o > 0 and so > 0 and so EEoo < 0< 0

EEoo is negativeis negative

75

Relationship of E0cell to ∆∆∆∆G

0

and K

� Example: Calculate the standard Gibbs free

energy change, ∆G0 , at 250C for the

following reaction.

Cu PbCuPb 22 +→+ ++

76

Relationship of E0cell to ∆∆∆∆G

0

and K

� Example: Calculate the Gibbs Free Energy

change, ∆G and the equilibrium constant at

250C for the following reaction with the

indicated concentrations.

( ) ( )MM 0.50 Zn Ag2 0.30 Ag 2 Zn 2++ +→+

Dry Cell BatteryDry Cell Battery

Anode (Anode (--))

Zn Zn ------> Zn> Zn2+2+ + 2e+ 2e--

Cathode (+)Cathode (+)

2 NH2 NH44++ + 2e+ 2e-- ------> >

2 NH2 NH33 + H+ H22

Primary batteryPrimary battery —— uses uses

redoxredox reactions that cannot reactions that cannot

be restored by recharge.be restored by recharge.

Nearly same reactions as in common dry Nearly same reactions as in common dry

cell, but under basic conditions.cell, but under basic conditions.

Alkaline BatteryAlkaline Battery

Anode (Anode (--): ): Zn + 2 OHZn + 2 OH-- ------> > ZnOZnO + H+ H22O + 2eO + 2e--

Cathode (+): Cathode (+): 2 MnO2 MnO22 + H+ H22O + 2eO + 2e-- ------> >

MnMn22OO33 + 2 OH+ 2 OH--

79

The Lead Storage Battery

� Diagram of the lead storage battery.

Lead Storage BatteryLead Storage Battery

Anode (Anode (--) ) EEoo = +0.36 V= +0.36 V

PbPb + HSO+ HSO44-- ------> PbSO> PbSO44 + H+ H++ + 2e+ 2e--

Cathode (+) Cathode (+) EEoo = +1.68 V= +1.68 V

PbOPbO22 + HSO+ HSO44-- + 3 H+ 3 H++ + 2e+ 2e--

------> PbSO> PbSO44 + 2 H+ 2 H22OO

NiNi--Cad BatteryCad BatteryAnode (Anode (--))

CdCd + 2 OH+ 2 OH-- ------> Cd(OH)> Cd(OH)22 + 2e+ 2e--

Cathode (+) Cathode (+)

NiO(OHNiO(OH) + H) + H22O + eO + e-- ------> Ni(OH)> Ni(OH)22 + OH+ OH--

Fuel Cells: HFuel Cells: H22 as a Fuelas a Fuel

••Fuel cellFuel cell -- reactants are reactants are

supplied continuously supplied continuously

from an external source.from an external source.

••Cars can use electricity Cars can use electricity

generated by Hgenerated by H22/O/O22 fuel fuel

cells.cells.

••HH22 carried in tanks or carried in tanks or

generated from generated from

hydrocarbons.hydrocarbons.

HydrogenHydrogen——Air Fuel CellAir Fuel Cell


HH22 as a Fuelas a Fuel

Comparison of the volumes of substances Comparison of the volumes of substances

required to store 4 kg of hydrogen relative to required to store 4 kg of hydrogen relative to

car size. car size. (Energy, p. 290)(Energy, p. 290)

Storing HStoring H22 as a Fuelas a Fuel

One way to store HOne way to store H22 is to adsorb the gas is to adsorb the gas

onto a metal or metal alloy. onto a metal or metal alloy. (Energy, p. 290)(Energy, p. 290)

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