1 CHAPTER 20 PRINCIPLES OF REACTIVITY: ELECTRON TRANSFER REACTIONS 1-11 + all bold numbered...

11

CHAPTER 20CHAPTER 20PRINCIPLES OF REACTIVITY: PRINCIPLES OF REACTIVITY:

ELECTRON TRANSFER ELECTRON TRANSFER REACTIONSREACTIONS

CHAPTER 20CHAPTER 20PRINCIPLES OF REACTIVITY: PRINCIPLES OF REACTIVITY:

ELECTRON TRANSFER ELECTRON TRANSFER REACTIONSREACTIONS

1-11 + all bold numbered problems1-11 + all bold numbered problems

22

ELECTROCHEMISTRYELECTROCHEMISTRYELECTROCHEMISTRYELECTROCHEMISTRY

Electric automobile

2

33

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

• BatteriesBatteries• CorrosionCorrosion• Industrial Industrial

production of production of chemicalschemicals such such as Clas Cl22, NaOH, , NaOH, FF22 and Al and Al

• Biological Biological redox reactionsredox reactions

The heme group

A rusted car.

44

20.1 OXIDATION-REDUCTION 20.1 OXIDATION-REDUCTION REACTIONSREACTIONS

• Redox reactions make up Redox reactions make up

electrochemical cells. electrochemical cells.

• Those that Those that produceproduce energy are called energy are called

batteries or batteries or galvanic cellsgalvanic cells (voltaic). (voltaic).

• Those that Those that requirerequire energy to operate are energy to operate are

called called electrolysis cellselectrolysis cells..

55

TRANSFER REACTIONSTRANSFER REACTIONSTRANSFER REACTIONSTRANSFER REACTIONSAtom transferAtom transfer

HOAc + HHOAc + H22O ---> OAcO ---> OAc-- + H + H33OO++

no change in oxidation state of any of the elementsno change in oxidation state of any of the elements

Electron transferElectron transfer

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

Cu looses 2 eCu looses 2 e-- and and AuAu++ gains 1 e gains 1 e--

OIL OIL RIGRIG

66

Electron Transfer ReactionsElectron Transfer ReactionsElectron Transfer ReactionsElectron Transfer Reactions

• Electron transfer reactions are Electron transfer reactions are oxidation-reduction or oxidation-reduction or redoxredox reactions.reactions.

• Redox reactions can result in the Redox reactions can result in the generation of an electric current generation of an electric current or be caused by imposing an or be caused by imposing an electric current. electric current.

• Therefore, this field of chemistry Therefore, this field of chemistry is often called is often called ELECTROCHEMISTRY.ELECTROCHEMISTRY.

77

CHAPTER OVERVIEWCHAPTER OVERVIEW

• This chapter examines electron This chapter examines electron transfer reactions, transfer reactions,

redoxredoxReduction/OxidationReduction/Oxidation

Note: Oil Rig = Note: Oil Rig =

Oxidation is Loss of an electronOxidation is Loss of an electron

Reduction is Gain of an electronReduction is Gain of an electron

88

Review of Terminology for Review of Terminology for Redox ReactionsRedox Reactions

Review of Terminology for Review of Terminology for Redox ReactionsRedox Reactions

• OXIDATIONOXIDATION—loss of electron(s) by —loss of electron(s) by a species; increase in oxidation a species; increase in oxidation number.number.

• REDUCTIONREDUCTION—gain of electron(s); —gain of electron(s); decrease in oxidation number.decrease in oxidation number.

• OXIDIZING AGENTOXIDIZING AGENT—electron —electron acceptor; species is reduced.acceptor; species is reduced.

• REDUCING AGENTREDUCING AGENT—electron —electron donor; species is oxidized.donor; species is oxidized.

• OXIDATIONOXIDATION—loss of electron(s) by —loss of electron(s) by a species; increase in oxidation a species; increase in oxidation number.number.

• REDUCTIONREDUCTION—gain of electron(s); —gain of electron(s); decrease in oxidation number.decrease in oxidation number.

• OXIDIZING AGENTOXIDIZING AGENT—electron —electron acceptor; species is reduced.acceptor; species is reduced.

• REDUCING AGENTREDUCING AGENT—electron —electron donor; species is oxidized.donor; species is oxidized.

99

Solid interacting with aqueous ions

1010

OXIDATION-REDUCTION OXIDATION-REDUCTION REACTIONSREACTIONS


Direct Redox ReactionDirect Redox ReactionOxidizing and reducing agents in direct Oxidizing and reducing agents in direct

contact.contact.Cu(s) + 2 AgCu(s) + 2 Ag++(aq) ---> Cu(aq) ---> Cu2+2+(aq) + 2 Ag(s)(aq) + 2 Ag(s)

1111

Electrochemical CellsElectrochemical CellsElectrochemical CellsElectrochemical Cells• An apparatus that allows An apparatus that allows

a redox reaction to occur a redox reaction to occur by transferring electrons by transferring electrons through an external through an external connector.connector.

• Product favored reaction Product favored reaction ---> voltaic or galvanic cell ---> voltaic or galvanic cell ----> electric current----> electric current

• Reactant favored reaction Reactant favored reaction ---> electrolytic cell ---> ---> electrolytic cell ---> electric current used to electric current used to cause chemical change.cause chemical change.

Batteries are voltaic Batteries are voltaic cellscells

1212



Indirect Redox ReactionIndirect Redox ReactionA battery functions by transferring A battery functions by transferring electrons through an external wire electrons through an external wire

from the reducing agent to the from the reducing agent to the oxidizing agent.oxidizing agent.

1313

CHEMICAL CHANGE -->CHEMICAL CHANGE -->ELECTRIC CURRENTELECTRIC CURRENT


Cu2+ ions

Zn metal

Cu2+ ions

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.”

Zn is oxidized Zn is oxidized and is the reducing agent and is the reducing agent Zn(s) ---> ZnZn(s) ---> Zn2+2+(aq) + 2e(aq) + 2e--

CuCu2+2+ is reduced is reduced and is the oxidizing agentand is the oxidizing agentCuCu2+2+(aq) + 2e(aq) + 2e-- ---> Cu(s) ---> Cu(s)

1414



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

Reduction: CuReduction: Cu2+2+(aq) + 2e(aq) + 2e-- ---> Cu(s) ---> Cu(s)----------------------------------------------------------------------------------------------------------------CuCu2+2+(aq) + Zn(s) ---> Zn(aq) + Zn(s) ---> Zn2+2+(aq) + Cu(s)(aq) + Cu(s)

Electrons 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.

Cu2+ ions

Zn metal

Cu2+ ions

1515



•To obtain a useful current, To obtain a useful current, we separate the oxidizing we separate the oxidizing and reducing agents so and reducing agents so that electron transfer that electron transfer occurs thru an external occurs thru an external wire. wire.

•This is accomplished in a This is accomplished in a GALVANICGALVANIC or or VOLTAICVOLTAIC cell.cell.

•A group of such cells is A group of such cells is

called a called a batterybattery..

Zn

Zn2+ ions

Cu

Cu2+ ions

wire

saltbridge

e le c t ro ns

1616

••Electrons travel thru external wire.Electrons travel thru external wire.•Salt bridge Salt bridge allows anions and cations to move allows anions and cations to move between electrode compartments.between electrode compartments.•This maintains electrical This maintains electrical neutralityneutrality..

••Electrons travel thru external wire.Electrons travel thru external wire.•Salt bridge Salt bridge allows anions and cations to move allows anions and cations to move between electrode compartments.between electrode compartments.•This maintains electrical This maintains electrical neutralityneutrality..

1717

Electrochemical Electrochemical Cell or BatteryCell or Battery

Electrochemical Electrochemical Cell or BatteryCell or Battery

Electrons move from anode to cathode in the wire.Anions & cations move thru the salt bridge.

1818

Balancing Redox EquationsBalancing Redox EquationsOXIDATION-REDUCTION REACTIONSOXIDATION-REDUCTION REACTIONS

• To be useful, the reactions for these To be useful, the reactions for these cells must be balanced. cells must be balanced.

• The simplest method of balancing is The simplest method of balancing is the ion-electron or half-reaction the ion-electron or half-reaction method. method.

• We learned this method in the first We learned this method in the first half of General Chemistry, but a quick half of General Chemistry, but a quick review is in order. review is in order.

1919

OXIDATION-REDUCTION REACTIONSOXIDATION-REDUCTION REACTIONS

• Oxidation is the loss of electron, Oxidation is the loss of electron, and reducing agents are oxidized. and reducing agents are oxidized.

• Reduction is the gain of electrons, Reduction is the gain of electrons, and oxidizing agents are reduced.and oxidizing agents are reduced.

• The simple steps in balancing a The simple steps in balancing a redox equation are:redox equation are:

2020

BALANCING REDOX REACTIONSBALANCING REDOX REACTIONS1. Separate the reaction into half reactions.1. Separate the reaction into half reactions.

2. Perform a mass balance for each element 2. Perform a mass balance for each element beginning with the element being oxidized beginning with the element being oxidized or reduced. Next balance the oxygen or reduced. Next balance the oxygen atoms with water and the hydrogen atoms atoms with water and the hydrogen atoms with hydrogen ions.with hydrogen ions.

3. Perform a charge balance by adding 3. Perform a charge balance by adding electrons to balance the charge.electrons to balance the charge.

2121

BALANCING REDOX REACTIONSBALANCING REDOX REACTIONS

4. If the reaction is in basic 4. If the reaction is in basic solution, add as many hydroxide solution, add as many hydroxide ions to both sides as there are ions to both sides as there are hydrogen ions and simplify by hydrogen ions and simplify by forming water. Eliminate the forming water. Eliminate the excess water, if any occurs on excess water, if any occurs on both sides of the reaction.both sides of the reaction.

5. Multiply each half-reaction by 5. Multiply each half-reaction by the appropriate factor so the the appropriate factor so the electrons are eliminated when the electrons are eliminated when the half-reactions are added.half-reactions are added.

2222

Balancing Equations for Balancing Equations for Redox ReactionsRedox Reactions

Balancing Equations for Balancing Equations for Redox ReactionsRedox Reactions

Some redox reactions have equations Some redox reactions have equations that must be balanced by special that must be balanced by special techniques.techniques.

MnOMnO44-- + 5 Fe + 5 Fe2+2+ + 8 H + 8 H++

---> Mn---> Mn2+2++ 5 Fe+ 5 Fe3+3+ + 4 H + 4 H22OO

22

2323

Balancing EquationsBalancing EquationsBalancing EquationsBalancing Equations Consider the reduction of AgConsider the reduction of Ag++

ions with copper metal.ions with copper metal.

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

2424

Balancing EquationsBalancing EquationsBalancing EquationsBalancing EquationsStep 1:Step 1: Divide the reaction into half-Divide the reaction into half-

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

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

RedRed Ag Ag++ ---> Ag ---> AgStep 2:Step 2: Balance each for mass. Already Balance each for mass. Already

done in this case.done in this case.Step 3:Step 3: Balance each half-reaction for Balance each half-reaction for

charge by adding electrons.charge by adding electrons.OxOx Cu ---> Cu Cu ---> Cu2+2+ + + 2e2e--

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

2525

Balancing EquationsBalancing EquationsBalancing EquationsBalancing EquationsStep 4:Step 4: Multiply each half-reaction by a Multiply each half-reaction by a

factor that means the reducing agent factor that means the reducing agent supplies as many electrons as the supplies as many electrons as the oxidizing agent requires.oxidizing agent requires.

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

Oxidizing agentOxidizing agent 22 Ag Ag++ + + 22 e e-- ---> ---> 22 AgAg

Step 5:Step 5: Add half-reactions to give the Add half-reactions to give the overall equation.overall equation.

Cu + 2 AgCu + 2 Ag++ ---> Cu ---> Cu2+2+ + 2Ag + 2AgThe equation is now balanced for The equation is now balanced for

both charge and mass.both charge and mass.

2626

Balancing EquationsBalancing EquationsBalancing EquationsBalancing EquationsBalance the following in acid Balance the following in acid

solution—solution—

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

Step 1:Step 1: Write the half-reactionsWrite the half-reactions

OxOx Zn ---> ZnZn ---> Zn2+2+

RedRed VOVO22++ ---> VO ---> VO2+2+

Step 2:Step 2: Balance each half-reaction Balance each half-reaction for mass.for mass.


2727

Balancing EquationsBalancing EquationsBalancing EquationsBalancing EquationsBalance the following in acid solution—Balance the following in acid solution— VOVO22

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

Step 1:Step 1: Write the half-reactionsWrite the half-reactionsOxOx Zn ---> ZnZn ---> Zn2+2+

RedRed VOVO22++ ---> VO ---> VO2+2+

Step 2:Step 2: Balance each half-reaction for Balance each half-reaction for mass.mass.


RedRed VO VO22++ ---> VO ---> VO2+2+ + + HH22OO2 H2 H++

++ Add HAdd H22O on O-deficient side and add HO on O-deficient side and add H++ on other side on other side for H-balance.for H-balance.

2828

Balancing EquationsBalancing EquationsBalancing EquationsBalancing EquationsStep 3:Step 3: Balance half-reactions for charge.Balance half-reactions for charge.

OxOx Zn ---> ZnZn ---> Zn2+2+ + + 2e 2e--

RedRed ee-- + 2 H+ 2 H++ + VO + VO22++ ---> VO ---> VO2+2+ + H + H22OO

Step 4:Step 4: Multiply by an appropriate factor.Multiply by an appropriate factor.

OxOx Zn ---> ZnZn ---> Zn2+2+ + + 2e2e--

RedRed 22ee-- + + 44 H H++ + + 22 VO VO22++

---> ---> 22 VO VO2+2+ + + 22 H H22OO

Step 5:Step 5: Add half-reactionsAdd half-reactions

Zn + 4 HZn + 4 H++ + 2 VO + 2 VO22++

---> Zn ---> Zn2+2+ + 2 VO + 2 VO2+2+ + 2 H + 2 H22OO

2929

Tips on Balancing EquationsTips on Balancing Equations• Never add ONever add O22, O atoms, or , O atoms, or

OO2-2- to balance oxygen. to balance oxygen.• Never add HNever add H22 or H atoms or H atoms

to balance hydrogen.to balance hydrogen.• Be sure to write the Be sure to write the

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

• Check your work at the Check your work at the end to make sure mass end to make sure mass and charge are balanced.and charge are balanced.

MORE PRACTICE!!

3030

20.2 CHEMICAL CHANGES LEADING 20.2 CHEMICAL CHANGES LEADING TO ELECTRIC CURRENTTO ELECTRIC CURRENT

• An electrochemical cell has two An electrochemical cell has two electrodes, the electrodes, the anodeanode where where oxidation oxidation occursoccurs, and the , and the cathodecathode where where reductionreduction occurs. occurs.

• The compartment containing the anode The compartment containing the anode is called the is called the anodic compartmentanodic compartment

• The compartment containing the The compartment containing the cathode is called the cathode is called the cathodic cathodic compartmentcompartment. .

• These two half-cells are frequently These two half-cells are frequently connected by a porous barrier or a connected by a porous barrier or a salt salt

bridgebridge. .

3131

CHEMICAL CHANGES LEADING CHEMICAL CHANGES LEADING TO ELECTRIC CURRENT TO ELECTRIC CURRENT

• The electrical current is carried by The electrical current is carried by electrons in the metals present, and electrons in the metals present, and by ions in the solution. by ions in the solution.

• Ions carrying charge are called Ions carrying charge are called electrolytes.electrolytes.

• Electrons travel from the anode (-) to Electrons travel from the anode (-) to the cathode (+)the cathode (+)

3232

Reduction

Oxidation

Cathodiccompartment Anodic

compartment

3333

Illustrates a general cell

3434

Abbreviated Cell NotationAbbreviated Cell NotationAnode | Anodic sol’n || Cathodic sol’n | CathodeAnode | Anodic sol’n || Cathodic sol’n | CathodeThis simple notation is very efficient

2Fe3+(aq) + H2(g) 2Fe2+(aq) + 2H+(aq)Pt|H2(1atm)|H+(1M)||Fe2+(1M),Fe3+(1M)|Pt

3535

Cu|Cu2+(1.0 M)||Ag+(1.0 M)|Ag

Anode | Anodic sol’n || Cathodic sol’n | CathodeAnode | Anodic sol’n || Cathodic sol’n | Cathode

3636

20.3 ELECTROCHEMICAL 20.3 ELECTROCHEMICAL CELLS AND POTENTIALSCELLS AND POTENTIALS

• The electrical potential is called the The electrical potential is called the electromotive force or emf, and is given electromotive force or emf, and is given the symbol E. the symbol E.

• The unit of charge is called the The unit of charge is called the coulomb coulomb

• emf is measured in joules/coulomb or emf is measured in joules/coulomb or volts, Vvolts, V

• Under standard conditions, 1.0 M Under standard conditions, 1.0 M for solutes, 1.0 bar for gases, and for solutes, 1.0 bar for gases, and 298.15 K, the emf is given the symbol 298.15 K, the emf is given the symbol EEoo, and called the standard potential., and called the standard potential.

3737

EEoo vs. vs. GGoo

• The electrical work is nFE, where F is The electrical work is nFE, where F is Faraday's constant, 96, 485 coulombs/ Faraday's constant, 96, 485 coulombs/ mole electrons or J/V mole, and n is mole electrons or J/V mole, and n is the moles of electrons in the balanced the moles of electrons in the balanced equation. equation.

3838

EEoo vs. vs. GGoo

This leads to the following equation This leads to the following equation between E and between E and G: G: G = - nFE.G = - nFE.

• This can be derived from: This can be derived from:

G = H -TS = E + PG = H -TS = E + PV - TV - TSS

• Since Since E = q + w and w = - PE = q + w and w = - PV – wV – wextext

• If the process is done reversibly at If the process is done reversibly at constant temperature and pressure, q constant temperature and pressure, q = T= TS.S.

G = q - PG = q - PV – wV – wext ext + P+ PV - TV - TSS

• This leads toThis leads toG = - wG = - wextext. .

3939

EEoo vs. vs. GGoo

In the standard state, In the standard state, GGoo = - nFE = - nFEoo..• Now we can use E and ENow we can use E and Eoo to predict to predict

reaction spontaneity, i.e. whether a reaction spontaneity, i.e. whether a reaction is product-favored. reaction is product-favored.

• Calculations of Calculations of GGoo from E from Eoo and visa and visa versa are algebraic.versa are algebraic.

4040

Calculating the Potential ECalculating the Potential Eoo of of an Electrochemical Cellan Electrochemical Cell

• All standard potentials are bases All standard potentials are bases on a relative scale where the on a relative scale where the hydrogen half-cell is assigned a hydrogen half-cell is assigned a value to 0.00 V at all temperature, value to 0.00 V at all temperature, and is called the standard and is called the standard hydrogen electrode, SHE. hydrogen electrode, SHE.

• All electrode potentials are written All electrode potentials are written as reductions and derive their as reductions and derive their values from direct or indirect values from direct or indirect measurements vs. the SHE. measurements vs. the SHE.

4141

Calculating the Potential ECalculating the Potential Eoo of of an Electrochemical Cellan Electrochemical Cell

• Demonstrates this measure for the zinc Demonstrates this measure for the zinc

electrode potential.electrode potential.

4242

Demonstrates

this measure for

the copper

electrode

potential.

4343

CELL POTENTIAL, ECELL POTENTIAL, ECELL POTENTIAL, ECELL POTENTIAL, E

• Electrons are “driven” from anode to cathode by Electrons are “driven” from anode to cathode by an electromotive force or an electromotive force or emfemf..

• For Zn/Cu cell, this is indicated by a voltage of For Zn/Cu cell, this is indicated by a voltage of 1.10 V at 25 1.10 V at 25 C and when [ZnC and when [Zn2+2+] and [Cu] and [Cu2+2+] = 1.0 M.] = 1.0 M.

Zn and ZnZn and Zn2+2+,,anodeanode

Cu and CuCu and Cu2+2+,,cathodecathode

43

4444

CELL CELL POTENTIAL, EPOTENTIAL, E

CELL CELL POTENTIAL, EPOTENTIAL, E

• For Zn/Cu cell, voltage is 1.10 V at 25 For Zn/Cu cell, voltage is 1.10 V at 25 C C and when [Znand when [Zn2+2+] and [Cu] and [Cu2+2+] = 1.0 M.] = 1.0 M.

• This is the This is the STANDARD CELL STANDARD CELL POTENTIAL, EPOTENTIAL, Eoo

• -- a quantitative measure of the tendency -- a quantitative measure of the tendency of reactants to proceed to products when of reactants to proceed to products when all are in their standard states at 25 all are in their standard states at 25 C. C.

4545

Calculating Cell VoltageCalculating Cell VoltageCalculating Cell VoltageCalculating Cell Voltage• Balanced half-reactions can be added Balanced half-reactions can be added

together to get overall, balanced equation. together to get overall, balanced equation.

2 I- ---> I2 + 2e-

2 H2O + 2e- ---> 2 OH- + H2

-------------------------------------------------2 I- + 2 H2O --> I2 + 2 OH- + H2

2 I- ---> I2 + 2e-

2 H2O + 2e- ---> 2 OH- + H2

-------------------------------------------------2 I- + 2 H2O --> I2 + 2 OH- + H2

If we know Eo for each half-reaction, we could get Eo for net reaction.

4646

CELL POTENTIALS, ECELL POTENTIALS, EooCELL POTENTIALS, ECELL POTENTIALS, Eoo

• Can’t measure 1/2 reaction ECan’t measure 1/2 reaction Eoo directly. directly. Therefore, measure it relative to a STANDARD Therefore, measure it relative to a STANDARD HALF CELL, HALF CELL, SHESHE..

2 H2 H++(aq, 1 M) + 2e(aq, 1 M) + 2e-- -->--> H H22(g, 1 atm)(g, 1 atm)2 H2 H++(aq, 1 M) + 2e(aq, 1 M) + 2e-- -->--> H H22(g, 1 atm)(g, 1 atm)

EEoo = 0.0 V = 0.0 V

Typically made of Pt or Au, are inert, and importantTypically made of Pt or Au, are inert, and importantbecause SHE’s can act as either an anode orbecause SHE’s can act as either an anode orcathode, difficult to maintain, not used often cathode, difficult to maintain, not used often

4747

Other Reference Electrodes, know EOther Reference Electrodes, know Eoo, easy to make, easy to makeSelf-contained, only make once, sealed, last a longtimeSelf-contained, only make once, sealed, last a longtime

||Hg||Hg22ClCl22(sat),KCl(xM)|Hg(sat),KCl(xM)|Hg

||AgCl(sat), KCl(xM)|Ag||AgCl(sat), KCl(xM)|Ag||TlCl(sat),KCl(xM)|Tl||TlCl(sat),KCl(xM)|Tl

4848

Volts

ZnH2

Salt Bridge

Zn 2+ H+

Zn Zn 2+ + 2e-

OXIDATION ANODE

2 H+ + 2e- H2

REDUCTIONCATHODE

- +

Zn/Zn2+ half-cell hooked to a SHE. Eo for the cell = +0.76 V

Zn/Zn2+ half-cell hooked to a SHE. Eo for the cell = +0.76 V

Remember ERemember Eoo for SHE = 0.0 Vfor SHE = 0.0 V

4949

Overall reaction is reduction of H+ by Zn metal.

Zn(s) + 2 H+ (aq) --> Zn2+ + H2(g) Eo = + 0.76 V

Therefore, Eo for Zn ---> Zn2+ (aq) + 2e- is

+ 0.76 V+ 0.76 V. Reduction potential is - 0.76 V- 0.76 V.

Zn is a betterbetter reducing agent (Zn gets oxidized)

than) H2.

Volts

ZnH2

Salt Bridge

Zn2+ H+

Zn Zn2+

+ 2 e-

OXIDATION ANODE

2 H+ + 2 e- H2

REDUCTIONCATHODE

-+

Remember ERemember Eoo for SHE = 0.0 Vfor SHE = 0.0 V

5050

Cu/CuCu/Cu2+2+ and H and H22/H/H++ Cell CellCu/CuCu/Cu2+2+ and H and H22/H/H++ Cell Cell

Volts

CuH2

Salt Bridge

Cu2+ H+

Cu2+ + 2e- Cu REDUCTION CATHODE

H2 2 H+ + 2e-OXIDATION ANODE

-+

Volts

CuH2

Salt Bridge

Cu2+ H+



-+

EEoo = +0.34 V = +0.34 V

5151

Cu/CuCu/Cu2+2+ and H and H22/H/H++ Cell CellCu/CuCu/Cu2+2+ and H and H22/H/H++ Cell Cell

• Overall reaction is reduction of CuOverall reaction is reduction of Cu2+2+ by H by H22 gas. gas.• CuCu2+2+ (aq) + H (aq) + H22(g) ---> Cu(s) + 2 H(g) ---> Cu(s) + 2 H++(aq)(aq)• Measured EMeasured Eoo = + 0.34 V = + 0.34 V• Therefore, ETherefore, Eoo for Cu for Cu2+2+ + 2e + 2e-- ---> Cu is ---> Cu is

+ 0.34 V + 0.34 V so Cu is a better Oxidizing agent (Cu gets so Cu is a better Oxidizing agent (Cu gets reduced) the Hreduced) the H++

Volts

CuH2

Salt Bridge

Cu2+ H+



-+

5252

• Lets Combine the work we just did Lets Combine the work we just did with the Zinc and Copper Cells with the Zinc and Copper Cells and combine them into ONE cell and combine them into ONE cell and see the net effectand see the net effect

COMBINING Voltaic CELLSCOMBINING Voltaic CELLS

Zn2+(aq) + 2e- ---> Zn(s) Eo = - 0.76 VCu2+(aq) + 2e- ---> Cu(s) Eo = + 0.34 V

Cu is a better Oxidizing agent than HCu is a better Oxidizing agent than H++

Zn is a better better reducing agent than H+

From this info we know Zn From this info we know Zn MUSTMUST be oxidized be oxidized and that Cu and that Cu MUSTMUST be reduced be reduced

5353

Zn/Cu Electrochemical CellZn/Cu Electrochemical CellZn/Cu Electrochemical CellZn/Cu Electrochemical Cell

Zn(s) ---> ZnZn(s) ---> Zn2+2+(aq) + 2e(aq) + 2e-- EEoo = + 0.76 V = + 0.76 VCuCu2+2+(aq) + 2e(aq) + 2e-- ---> Cu(s) ---> Cu(s) EEoo = + 0.34 V = + 0.34 V------------------------------------------------------------------------------------------------------------------------------CuCu2+2+(aq) + Zn(s) ---> Zn(aq) + Zn(s) ---> Zn2+2+(aq) + Cu(s) (aq) + Cu(s)

EEoo (calculated) = + 1.10 V (calculated) = + 1.10 V

Zn

Zn2+ ions

Cu

Cu2+ ions

wire

saltbridge

electrons

Zn

Zn2+ ions

Cu

Cu2+ ions

wire

saltbridge

electrons

Cathode, positive, sink for electrons(reduction)

Anode, negative, source of electrons(oxidation)

5454

Uses of EUses of Eoo Values ValuesUses of EUses of Eoo Values ValuesThese experiments show we canThese experiments show we can

a) decide on relative ability of a) decide on relative ability of elements to act as reducing elements to act as reducing agents (or oxidizing agents)agents (or oxidizing agents)

b) assign a voltage to a half-b) assign a voltage to a half-reaction that reflects this ability.reaction that reflects this ability.

Zn

Zn2+ ions

Cu

Cu2+ ions

wire

saltbridge

electrons

Zn

Zn2+ ions

Cu

Cu2+ ions

wire

saltbridge

electrons

5555

USING STANDARD POTENTIALSUSING STANDARD POTENTIALS

EEoocellcell = E = Eoo

cathodiccathodic - E - Eooanodicanodic

EEoocellcell = E = Eoo

reductionreduction + E + Eoooxidationoxidation

• The first equation uses the values right The first equation uses the values right out of the table of standard electrode out of the table of standard electrode potentials. potentials.

• The table can also give us information The table can also give us information about which reactions will be about which reactions will be product-product-favoredfavored (produce current) in the (produce current) in the standard state since these reactions standard state since these reactions with have a positive Ewith have a positive Eoo

cellcell..

5959

USING STANDARD POTENTIALSUSING STANDARD POTENTIALS

• The following is an The following is an abbreviated table form the abbreviated table form the previous tables to give previous tables to give practice at using them.practice at using them.

6060

TABLE OF STANDARD TABLE OF STANDARD POTENTIALSPOTENTIALS

TABLE OF STANDARD TABLE OF STANDARD POTENTIALSPOTENTIALS

Eo (V)

Cu2+ + 2e- Cu +0.34

2 H+ + 2e- H2 0.00

Zn2+ + 2e- Zn -0.76

OxidizingOxidizingability of Ion ability of Ion

reducing abilityreducing abilityof elementof element

6161

Standard Redox Potentials, EStandard Redox Potentials, EooStandard Redox Potentials, EStandard Redox Potentials, Eoo

• Any substance on Any substance on the right will reduce the right will reduce any substance any substance higher than it on the higher than it on the left.left.

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

• HH22 can reduce Cu can reduce Cu2+2+ but not Znbut not Zn2+2+

• Cu cannot reduce Cu cannot reduce HH++ or Zn or 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

Eo (V)

Cu2+ + 2e- Cu +0.34

2 H+ + 2e- H2 0.00

Zn2+ + 2e- Zn -0.76



6262

Standard Redox Potentials, EStandard Redox Potentials, EooStandard Redox Potentials, EStandard Redox Potentials, Eoo

Eo (V)

Cu2+ + 2e- Cu +0.34

2 H+ + 2e- H2 0.00

Zn2+ + 2e- Zn -0.76



Eo (V)

Cu2+ + 2e- Cu +0.34

2 H+ + 2e- H2 0.00

Zn2+ + 2e- Zn -0.76



• Zn(s) ---> ZnZn(s) ---> Zn2+2+(aq) + 2e-(aq) + 2e- EEoo = + 0.76 V = + 0.76 V• CuCu2+2+(aq) + 2e- ---> Cu(s)(aq) + 2e- ---> Cu(s) EEoo = + 0.34 V = + 0.34 V• ------------------------------------------------------------------------------------------------------------------------------• CuCu2+2+(aq) + Zn(s) ---> Zn(aq) + Zn(s) ---> Zn2+2+(aq) + Cu(s) (aq) + Cu(s) • EEoo (calculated) = + 1.10 V (calculated) = + 1.10 V

6363

Using Standard Potentials, EUsing Standard Potentials, EooUsing Standard Potentials, EUsing Standard Potentials, Eoo

Which is the best oxidizing agent:Which is the best oxidizing agent:OO22, H, H22OO22, or Cl, or Cl22? _________________? _________________

• Which is the best reducing agent:Which is the best reducing agent:

Hg, Al, or Sn? ____________________Hg, Al, or Sn? ____________________

• Which is the correct direction for the following Which is the correct direction for the following

reaction to produce a favorable potential?reaction to produce a favorable potential?

2Al(s) + 3 Cl2Al(s) + 3 Cl22(aq) ---> AlCl(aq) ---> AlCl33(aq) (aq)

AlClAlCl33(aq) ---> 2Al(s) + 3 Cl(aq) ---> 2Al(s) + 3 Cl22(aq) (aq)

See next slideSee next slide

6565

EEoo for a Voltaic Cell for a Voltaic CellEEoo for a Voltaic Cell for a Voltaic Cell

Volts

Cd Salt Bridge

Cd2+

Fe

Fe2+

Volts

Cd Salt Bridge

Cd2+

Fe

Fe2+

Cd --> Cd2+ + 2e-

orCd2+ + 2e- --> Cd

Fe --> Fe2+ + 2e-

orFe2+ + 2e- --> Fe

See previous slideSee previous slide

6666

From the table, you see From the table, you see •• Fe is a better reducing Fe is a better reducing

agent than Cdagent than Cd•• CdCd2+2+ is a better is a better

oxidizing agent than oxidizing agent than FeFe2+2+

Overall reactionOverall reactionFe + CdFe + Cd2+2+ ------

> Cd + Fe> Cd + Fe2+2+

EEoo = + 0.04 V = + 0.04 V

Volts

Cd Salt Bridge

Cd2+

Fe

Fe2+

Volts

Cd Salt Bridge

Cd2+

Fe

Fe2+

EEoo for a Voltaic Cell for a Voltaic CellEEoo for a Voltaic Cell for a Voltaic Cell

6767

Examples of Cells that NEED electricity Examples of Cells that NEED electricity (opposite of Voltaic or Galvanic cells(opposite of Voltaic or Galvanic cells

Production of Production of Oxygen and Oxygen and HydrogenHydrogen

6868

Electrolysis of SnClElectrolysis of SnCl22(aq)(aq)

SnSn2+2+(aq) + 2Cl(aq) + 2Cl--(aq) (aq) Sn(s) and Cl Sn(s) and Cl22(g)(g)

Anode (+)Anode (+) Cathode (-)Cathode (-)

6969

NaINaI

phenolphthaleinphenolphthalein

More on this later, industrialapplications

OH- detected

7070

EEoo and and GGooEEoo and and GGoo

EEoo is related to is related to GGoo, the free , the free energy change for the energy change for the reaction.reaction.

GGoo = - n F E = - n F Eoo where F = Faraday constant where F = Faraday constant

= 9.6485 x 10= 9.6485 x 1044 J/V•molJ/V•mol

and n is the number of moles and n is the number of moles of electrons transferredof electrons transferred

Michael Faraday1791-1867

7171

Michael FaradayMichael Faraday1791-18671791-1867

Michael FaradayMichael Faraday1791-18671791-1867

Originated the terms anode, Originated the terms anode, cathode, anion, cation, cathode, anion, cation, 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

chemicalschemicalsWas a popular lecturer.Was a popular lecturer.

7272

EEoo and and GGooEEoo and and GGoo

GGoo = - n F E = - n F Eoo

For a For a product-favoredproduct-favored reaction reaction Reactants ----> ProductsReactants ----> Products

GGo o < 0 and so E < 0 and so Eo o > 0 > 0EEoo is positive, is positive, results in results in

currentcurrentFor a For a reactant-favoredreactant-favored reaction reaction Reactants <---- ProductsReactants <---- Products

GGo o > 0 and so E > 0 and so Eo o < 0 < 0EEoo is negative is negative , needs current, needs current

7373

ELECTROCHEMICAL CELLS AT ELECTROCHEMICAL CELLS AT NON STANDARD CONDITIONS NON STANDARD CONDITIONS

• We learned the relationship between We learned the relationship between G and G and GGoo in chapter 19. in chapter 19.

G = G = GGoo + RT + RT lnln Q. Q. • This yields: -nFE = -nFEThis yields: -nFE = -nFEoo + RT + RT lnln Q Q • At 298.15 K, we solve for EAt 298.15 K, we solve for E

E = EE = Eoo - [0.0592 V/ - [0.0592 V/nn] ] loglog Q or Q or

E = EE = Eoo - [0.0257 V/ - [0.0257 V/nn]] lnln Q Q

7474

20.5 E20.5 Eoo and the Equilibrium Constant and the Equilibrium ConstantNernst EquationsNernst Equations

• At At equilibriumequilibrium for the cell reaction, for the cell reaction, E = 0, (equivalent to saying the cell is E = 0, (equivalent to saying the cell is dead) Therefore:dead) Therefore:

EEoo = [0.0592 V/ = [0.0592 V/nn] ] loglog K or K or

EEoo = [0.0257 V/ = [0.0257 V/nn] ] lnln K K

• These equations can be solved for K These equations can be solved for K using the inverse functions.using the inverse functions.

• nn is the number of moles of e is the number of moles of e--

7575

Calculating KCalculating Kspsp, K, Kff, and other K's , and other K's from cell potentialsfrom cell potentials

• In the laboratory we will be calculating KIn the laboratory we will be calculating Kspsp's and 's and KKff 's for reactions using electrochemical cells. 's for reactions using electrochemical cells.

• Determine the KDetermine the Kff for AlF for AlF66-3-3 from the cell below: E from the cell below: E

= 2.465 V= 2.465 V

Al |Al |10.0 mL 0.10 M10.0 mL 0.10 M Al(NOAl(NO33))33,,40.0 mL 1.0 M40.0 mL 1.0 M NaF NaF ||||0.10 M0.10 M CuSOCuSO44||CuCu

2 Al + 3 Cu2 Al + 3 Cu+2 +2 ===> 2 Al ===> 2 Al+3+3 + 3 Cu + 3 Cu AlAl+3+3 + 6F + 6F-- <===> AlF <===> AlF66

-3-3

Given: [FGiven: [F--] = 0.68 M, [AlF] = 0.68 M, [AlF66-3-3] = 0.020 M] = 0.020 M

Note: no reaction between AlNote: no reaction between Al3+3+ or Cu or Cu2+2+ and SO and SO44-2-2

7676

2 Al + 3 Cu+2 ===> 2 Al+3 + 3 Cu

Eo = 1.676 + .337 = 2.013 VEcell = 2.465 V (given), 0.10 M Cu+2

Ecell = Eo - [0.0257 V/n] ln K

2.465 = 2.013 - [0.0257/6] Ln (Al+3)2/(.1)3

solved for Al+3, note that as soon as Al+3 forms it immediately reacts with F-, so Al+3 is NOT .1M

Which gives: [Al+3] = 4.0x10-25 M

Given: [F-] = 0.68 M, [AlF6-3] = 0.020 M

Sub these values into Kf

Kf = 5.0 x 1023

7777

21.6 BATTERIES AND FUEL CELLS21.6 BATTERIES AND FUEL CELLS

• There are two types of batteries: There are two types of batteries:

• Primary, which are not Primary, which are not rechargeable rechargeable

• Secondary, or storage batteries Secondary, or storage batteries which are easily recharged.which are easily recharged.

7878

Primary BatteriesPrimary Batteries

• 3 types: Dry, Alkaline, Button (Hg, Li)3 types: Dry, Alkaline, Button (Hg, Li)• The The dry celldry cell produces H produces H22 gas gas• The battery has a zinc anode and The battery has a zinc anode and

reduces ammonium ion to ammonia and reduces ammonium ion to ammonia and hydrogen gas at a graphite cathode. hydrogen gas at a graphite cathode.

• The net-reaction does not contain the The net-reaction does not contain the hydrogen gas since MnOhydrogen gas since MnO22 is added to is added to oxidize the hydrogen gas to hydrogen oxidize the hydrogen gas to hydrogen ion.ion.

• This battery has a voltage of 1.5 Volts This battery has a voltage of 1.5 Volts when new. when new.

7979

Dry Cell BatteryDry Cell BatteryDry Cell BatteryDry Cell Battery

Anode (-)Anode (-)

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

Cathode (+)Cathode (+)

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

2 NH2 NH33 + H + H22

Common dry cell

79

1.5 Volts1.5 Volts

8080

+ -4 3 2

2+ -

2NH + 2e 2NH ( ) + H ( )

Zn Zn + 2e

g g

Problem alkaline & button: gasses formedProblem alkaline & button: gasses formed

3 2The gasses NH ( ) + H ( ) are removed:g g

2+ -3 3 2 2

2 2 2 3 2

Zn + 2NH ( ) + 2Cl Zn(NH ) Cl ( )

2MnO ( ) + H ( ) Mn O ( ) + H O( )

g s

and

s g s l

anode

cathode

8181

Primary Batteries: No HPrimary Batteries: No H22 Gas produced Gas produced

• The The alkaline batteryalkaline battery has a zinc anode has a zinc anode and a graphite cathode, the MnOand a graphite cathode, the MnO22 is is

reduced directly to Mnreduced directly to Mn22OO33 in a basic in a basic

environment.environment.

• It's voltage is 1.54 Volts. It's voltage is 1.54 Volts.

• The mercury and lithium batteries are The mercury and lithium batteries are buttonbutton shaped and have voltages of shaped and have voltages of 1.35 V and 3.0 V respectively.1.35 V and 3.0 V respectively.

• 50% stronger then a dry cell50% stronger then a dry cell

8282

Nearly same reactions as in Nearly same reactions as in common dry cell, but under basic common dry cell, but under basic conditions (1.54V)conditions (1.54V)

Alkaline BatteryAlkaline BatteryAlkaline BatteryAlkaline Battery

8383

Mercury BatteryMercury BatteryMercury BatteryMercury BatteryAnode:Anode:

Zn is reducing agent under basic conditionsZn is reducing agent under basic conditionsCathode:Cathode:

HgO + HHgO + H22O + 2eO + 2e-- ---> Hg + 2 OH ---> Hg + 2 OH--

Primary, Hg (1.35V, Li (3.0V)Primary, Hg (1.35V, Li (3.0V)

8484

- -2

- -2 2 2

Zn( ) + 2OH ZnO + 2H O( ) + 2e

2MnO + H O( ) + 2e MnO + 2OH

s l

l

Note OHNote OH--, either KOH or NaOH, either KOH or NaOH

anode

cathode

Also note that Zn here is the cathode, it is the anode in a dry cell

8585

Primary

At anode

8686Primary: MnO replaced with-Hg/Li (1.35V, Li (3.0V)Primary: MnO replaced with-Hg/Li (1.35V, Li (3.0V)

8787

Secondary BatteriesSecondary Batteries

• The net-reaction turns lead and The net-reaction turns lead and lead(IV)oxide, in the presence of lead(IV)oxide, in the presence of sulfuric acid, into lead(II)sulfate and sulfuric acid, into lead(II)sulfate and waterwater

• The nickel-cadmium battery is a The nickel-cadmium battery is a rechargeable battery. rechargeable battery.

Fuel CellsFuel Cells• Fuel cells convert hydrogen and Fuel cells convert hydrogen and

oxygen gasoxygen gas into water and electrical into water and electrical energy without combustion.energy without combustion.

8989

Lead Storage BatteryLead Storage BatteryLead Storage BatteryLead Storage BatteryAnode (-) Anode (-) EEoo = +0.36 V = +0.36 VPb + HSOPb + HSO44

-- ---> PbSO ---> PbSO44 + H + H++ + 2e + 2e--

Cathode (+) Cathode (+) EEoo = +1.68 V = +1.68 VPbOPbO22 + HSO + HSO44

-- + 3 H + 3 H++ + 2e + 2e-- ---> PbSO ---> PbSO44 + 2 H + 2 H22OO

9090

Ni-Cad BatteryNi-Cad BatteryNi-Cad BatteryNi-Cad BatteryAnode (-)Anode (-)

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

Cathode (+) Cathode (+)

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

9191

CORROSION: REDOX REACTIONS CORROSION: REDOX REACTIONS IN THE ENVIRONMENT IN THE ENVIRONMENT

• Corrosion of iron produces Corrosion of iron produces iron(III)oxide and other iron oxides iron(III)oxide and other iron oxides depending on the conditions, depending on the conditions, moisture and oxygen levels. moisture and oxygen levels.

• Other active metals also corrode Other active metals also corrode and form oxides, but frequently this and form oxides, but frequently this is in the form of a stable oxide is in the form of a stable oxide coating on the metal surface. coating on the metal surface.

• The oxides of iron flake off easily, The oxides of iron flake off easily, exposing fresh iron surfaces for exposing fresh iron surfaces for additional oxidation. additional oxidation.

9292

CORROSION: REDOX REACTIONS CORROSION: REDOX REACTIONS IN THE ENVIRONMENT IN THE ENVIRONMENT

• These processes are true These processes are true electrochemical processes and a cathode electrochemical processes and a cathode and an anode can be identified. and an anode can be identified.

• The presence of conducting ions The presence of conducting ions enhances the process. (Salt water spray, enhances the process. (Salt water spray, salt water from ice suppression, etc.)salt water from ice suppression, etc.)

• One protection process is called anodic One protection process is called anodic inhibition and involves changing the inhibition and involves changing the surface of the metal. surface of the metal.

• Paint and other coatings are examples. Paint and other coatings are examples.

9393

CORROSIONCORROSION

• A more effective method is A more effective method is cathodic protection which places cathodic protection which places the metal in electrical contact with the metal in electrical contact with a more active metal like a more active metal like magnesium or zinc. magnesium or zinc.

• The latter with iron is called The latter with iron is called galvanization and the former with galvanization and the former with iron is called a sacrificial anode. iron is called a sacrificial anode.

9494

Cathodic Protection

9595

Fe corrosion to rust FeFe corrosion to rust Fe22OO33

9696

Why Aluminum does not rustWhy Aluminum does not rust

9797

20.8 ELECTROLYSIS: 20.8 ELECTROLYSIS: CHEMICAL CHANGE FROM CHEMICAL CHANGE FROM ELECTRIC ENERGYELECTRIC ENERGY• When a reaction is driven by an When a reaction is driven by an

external electrical energy source external electrical energy source the process is called electrolysis. the process is called electrolysis.

• The anode and cathode of such The anode and cathode of such cells are the sites of oxidation and cells are the sites of oxidation and reduction as in a galvanic cell, but reduction as in a galvanic cell, but the polarity of the electrodes is the polarity of the electrodes is reversed. reversed.

• The anode is now (+) and The anode is now (+) and the cathode (-) . the cathode (-) .

9898

ELECTROLYSISELECTROLYSIS

• For the electrolysis of molten NaCl, NaFor the electrolysis of molten NaCl, Na++ is is reduced to Na, and Clreduced to Na, and Cl-- is oxidized to Cl is oxidized to Cl22. .

• In aqueous solution, HIn aqueous solution, H22O is reduced to HO is reduced to H2 2

and OHand OH-- and Cl and Cl-- is oxidized to Cl is oxidized to Cl22. .

• These decisions are made by consulting These decisions are made by consulting the standard electrode potential table to the standard electrode potential table to determine which species are most easily determine which species are most easily oxidized and reduced. oxidized and reduced.

9999

ELECTROLYSISELECTROLYSIS• The oxidation and reduction of water The oxidation and reduction of water

can be a little tricky since the table is a can be a little tricky since the table is a Standard electrode potential table Standard electrode potential table which means the all reactants and which means the all reactants and products are present at standard products are present at standard concentrations, 1M and 1 atm. concentrations, 1M and 1 atm.

• Reduction of water to form HReduction of water to form H2(g)2(g) occurs occurs at - 0.414 V when concentration is at - 0.414 V when concentration is taken into account, and oxidation of taken into account, and oxidation of water to form Owater to form O2(g)2(g) occurs at - 0.816 V. occurs at - 0.816 V.

• Over-voltages for hydrogen and Over-voltages for hydrogen and oxygen gases also complicate matters. oxygen gases also complicate matters.

100100

Electrolysis of Aqueous NaOHElectrolysis of Aqueous NaOHElectrolysis of Aqueous NaOHElectrolysis of Aqueous NaOH

Anode (+) Anode (+) EEoo = - 0.40 V = - 0.40 V

4 OH4 OH-- ---> O ---> O22(g) + 2 H(g) + 2 H22O + 2eO + 2e--

Cathode (-) Cathode (-) EEoo = - 0.83 V = - 0.83 V

4 H4 H22O + 4eO + 4e-- ---> 2 H ---> 2 H22 + 4 OH + 4 OH--

EEoo for cell = - 1.23 V for cell = - 1.23 V

Electric Energy ----> Chemical ChangeElectric Energy ----> Chemical Change

101101

ElectrolysisElectrolysisElectric Energy ---> Chemical ChangeElectric Energy ---> Chemical Change

ElectrolysisElectrolysisElectric Energy ---> Chemical ChangeElectric Energy ---> Chemical Change

BATTERY

+

Na+Cl-

Anode Cathode

electrons

BATTERY

+

Na+Cl-

Anode Cathode

electrons

Electrolysis of molten NaCl.Here a battery “pumps” electrons from Cl- to Na+.

102102

Electrolysis of Molten NaClElectrolysis of Molten NaClElectrolysis of Molten NaClElectrolysis of Molten NaCl


2 Cl2 Cl-- ---> Cl ---> Cl22(g) + 2e(g) + 2e--

Cathode (-) Cathode (-) EEoo = - 2.71 V = - 2.71 V

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


External energy needed External energy needed because Ebecause Eoo is (-). is (-).

Note that signs of electrodes Note that signs of electrodes are reversed from batteries.are reversed from batteries.

BATTERY

+

Na+Cl-

Anode Cathode

electrons

BATTERY

+

Na+Cl-

Anode Cathode

electrons

103103

Simpler process for making Na(s)?Simpler process for making Na(s)?

• We just saw the molten process of We just saw the molten process of taking NaCl and by applying a taking NaCl and by applying a voltage (electrolysis) we can make voltage (electrolysis) we can make both Na(s) and Clboth Na(s) and Cl22(g)(g)

• Wouldn’t it be easier to take sea-Wouldn’t it be easier to take sea-water, perform electrolysis on the water, perform electrolysis on the “brine” to get Na(s) and Cl“brine” to get Na(s) and Cl22(g)(g)

104104

What are the possible reactionsWhat are the possible reactions

At the anodeNa+ + e- Na(s)2H2O(l) + 2e- H2(g) + 2OH-

At the cathode2Cl- Cl2 + 2e-

2H2O(l) O2(g) + 4H+ + 4e-

E = -2.714E = -0.8277

E = -1.360E = -1.229

Once a potential is applied the process with the lowest potential will start FIRST!

105105

At the anodeNa+ + e- Na(s)2H2O(l) + 2e- H2(g) + 2OH-

At the cathode2Cl- Cl2 + 2e-

2H2O(l) O2(g) + 4H+ + 4e-

E = -2.714E = -0.8277

E = -1.360E = -1.229

Once a potential is applied the process with the lowest potential will start FIRST!

The process with the lowest potential:4H2O(l) + 4e- 2H2(g) + 4OH-

2H2O(l) O2(g) + 4H+ + 4e-

2H2O(l) O2(g) + 2H2

E = -0.8277E = -1.229Ecell = -2.057V

106106

Electrolysis of Aqueous Electrolysis of Aqueous NaClNaCl

Electrolysis of Aqueous Electrolysis of Aqueous NaClNaCl

Since Na(s) can not be made from sea-water due to hydrogen and oxygen gas are made another technique is used on sea-water

107107

“Brine Process” for making NaOH from sea-water

108108

Electrolysis of Aqueous NaClElectrolysis of Aqueous NaClElectrolysis of Aqueous NaClElectrolysis of Aqueous NaClCells like these are the source of Cells like these are the source of

NaOH and ClNaOH and Cl22..

In 1995In 1995

25.1 x 1025.1 x 1099 lb Cl lb Cl22

26.1 x 1026.1 x 1099 lb NaOH lb NaOH

Also the source of NaOCl for use in bleach.

109109

Electrolysis of Aqueous CuClElectrolysis of Aqueous CuCl22Electrolysis of Aqueous CuClElectrolysis of Aqueous CuCl22


2 Cl2 Cl-- ---> Cl ---> Cl22(g) + 2e(g) + 2e--

Cathode (-) Cathode (-) EEoo = + 0.34 V = + 0.34 V

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


Note that CuNote that Cu+2+2 is more is more easily reduced than easily reduced than either Heither H22O or NaO or Na++..

BATTERY

+

Cu2+Cl-

Anode Cathode

electrons

H2O

BATTERY

+

Cu2+Cl-

Anode Cathode

electrons

H2OThe water process potential:4H2O(l) + 4e- 2H2(g) + 4OH-

2H2O(l) O2(g) + 4H+ + 4e-

2H2O(l) O2(g) + 2H2

E = -0.8277E = -1.229Ecell = -2.057V

110110

• AluminumAluminum• Chlorine, sodium hydroxide, Chlorine, sodium hydroxide,

and hydrogenand hydrogen• These chemicals are extremely These chemicals are extremely

important to the economy. important to the economy. • The power required for these The power required for these

processes can be enormous. processes can be enormous.

THE COMMERCIAL PRODUCTION THE COMMERCIAL PRODUCTION OF CHEMICALS BY OF CHEMICALS BY ELECTROCHEMICAL METHODSELECTROCHEMICAL METHODS

111111

112112

Producing AluminumProducing AluminumProducing AluminumProducing Aluminum2 Al2 Al22OO33 + 3 C ---> 4 Al + 3 CO + 3 C ---> 4 Al + 3 CO22

Charles Hall (1863-1914) at age of 19 developed the Al electrolysis process. Founded Alcoa!

113113

COMMERCIAL PRODUCTION COMMERCIAL PRODUCTION OF CHEMICALSOF CHEMICALS

• Power is measured in watts and Power is measured in watts and has basic units of has basic units of joules/second. joules/second.

• Power is calculated from the Power is calculated from the equationequation

P = V I, (J/C) (C/S) = J/SP = V I, (J/C) (C/S) = J/S• A kilowatt-hour is an energy A kilowatt-hour is an energy

unit and is equal to unit and is equal to 3.60x103.60x1066J:J:3600 S/hr times 1000 W/kW3600 S/hr times 1000 W/kW

114114

20.8 COUNTING ELECTRONS20.8 COUNTING ELECTRONS• Current, I, is measured in amperes and Current, I, is measured in amperes and

has basic units of coulombs/second. has basic units of coulombs/second. • Faraday's Law tells us that the current Faraday's Law tells us that the current

times the time can be used to calculate the times the time can be used to calculate the mass of material oxidized or reduced. mass of material oxidized or reduced.

• FF, Faraday's constant, is 9.65x10, Faraday's constant, is 9.65x1044C/mol eC/mol e--. . • These unit factor problems change: These unit factor problems change:

current x time ==> coulombs ==> current x time ==> coulombs ==> moles emoles e- - ==> moles oxidized or reduced ==> moles oxidized or reduced species ==> massspecies ==> mass

115115

• The balanced half-reaction The balanced half-reaction provides the conversion from provides the conversion from moles electrons to moles of moles electrons to moles of species sought. species sought.

• Calculate the grams of aluminum Calculate the grams of aluminum formed at the cathode if a current formed at the cathode if a current of 4.40 A flows for 1.50 hours.of 4.40 A flows for 1.50 hours.

–Let’s do an example first.Let’s do an example first.

COUNTING ELECTRONSCOUNTING ELECTRONS

116116

Quantitative Aspects of Quantitative Aspects of ElectrochemistryElectrochemistry


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

AgAg++ (aq) + e (aq) + e-- ---> Ag(s) ---> Ag(s)1 mol e1 mol e-- ---> 1 mol Ag---> 1 mol AgIf we could measure the moles of If we could measure the moles of

ee--, we could know the quantity of , we could know the quantity of Ag formed.Ag formed.

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

117117



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

AgAg++ (aq) + e (aq) + e-- ---> Ag(s) ---> Ag(s)1 mol e1 mol e-- ---> 1 mol Ag---> 1 mol AgIf we could measure the moles of eIf we could measure the moles of e--, we , we

could know the quantity of Ag formed.could know the quantity of Ag formed.But how to measure moles of eBut how to measure moles of e--??

Current = charge passingtime

I (amps) = coulombsseconds

118118

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

Charge on 1 mol of eCharge on 1 mol of e-- = = (1.60 x 10(1.60 x 10-19 -19 C/eC/e--)(6.02 x 10)(6.02 x 102323 e e--/mol)/mol)

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

Current = charge passing

time




119119

Quantitative Aspects of ElectrochemistryQuantitative Aspects of ElectrochemistryQuantitative Aspects of ElectrochemistryQuantitative Aspects of Electrochemistry

1.50 amps flow thru a Ag1.50 amps flow thru a Ag++(aq) (aq) solution for 15.0 min. What mass solution for 15.0 min. What mass of Ag metal is deposited?of Ag metal is deposited?

SolutionSolution(a)(a) Calculate the chargeCalculate the charge

Coulombs = amps x timeCoulombs = amps x time= (1.5 amps)(15.0 min)(60 s/min) = (1.5 amps)(15.0 min)(60 s/min) = 1350 C= 1350 C


120120


1.50 amps flow thru a Ag1.50 amps flow thru a Ag++(aq) solution for 15.0 (aq) solution for 15.0 min. What mass of Ag metal is deposited?min. What mass of Ag metal is deposited?

SolutionSolution

(a)(a) Charge = 1350 CCharge = 1350 C(b)(b) Calculate moles of eCalculate moles of e-- used used

1350 C • 1 mol e-

96,500 C 0.0140 mol e-

(c)(c) Calculate quantity of AgCalculate quantity of Ag0.0140 mol e- •

1 mol Ag1 mol e-

0.0140 mol Ag or 1.51 g Ag

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Quantitative Aspects of ElectrochemistryQuantitative Aspects of ElectrochemistryQuantitative Aspects of ElectrochemistryQuantitative Aspects of ElectrochemistryThe anode reaction in a lead storage battery isThe anode reaction in a lead storage battery is

Pb(s) + HSOPb(s) + HSO44--(aq) ---> PbSO(aq) ---> PbSO44(s) + H(s) + H++(aq) + 2e(aq) + 2e--

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

SolutionSolutiona)a) 454 g Pb = 2.19 mol Pb454 g Pb = 2.19 mol Pbb)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 e • 96,500 C/mol e-- = 423,000 C = 423,000 C

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SolutionSolutiona)a) 454 g Pb = 2.19 mol Pb454 g Pb = 2.19 mol Pbb)b) Mol of e- = 4.38 molMol of e- = 4.38 molc)c) Charge = 423,000 CCharge = 423,000 C

d)d) Calculate timeCalculate time

Time (s) = Charge (C)

I (amps)

Time (s) = 423,000 C1.50 amp

= 282,000 s

About 78 hoursAbout 78 hours

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• Remember:Remember:–The balanced half-reaction The balanced half-reaction provides the conversion provides the conversion from moles electrons to from moles electrons to moles of species sought.moles of species sought.

Now solve the problem:Now solve the problem:

• Calculate the grams of aluminum Calculate the grams of aluminum formed at the cathode if a current formed at the cathode if a current of 4.40 A flows for 1.50 hours.of 4.40 A flows for 1.50 hours.

COUNTING ELECTRONSCOUNTING ELECTRONS

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BALANCING REDOX BALANCING REDOX EQUATIONSEQUATIONS

The Half-Reaction MethodThe Half-Reaction Method• Separate the equation into half-Separate the equation into half-

reactions.reactions.

• Balance the half-reactions.Balance the half-reactions.

• Combine the half-reactions to form a Combine the half-reactions to form a balanced equation containing no balanced equation containing no electrons.electrons.

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Balancing Half-ReactionsBalancing Half-ReactionsBalancing Half-ReactionsBalancing Half-Reactions

• First balance the element changing First balance the element changing oxidation state.oxidation state.

• Balance the oxygen atoms with water.Balance the oxygen atoms with water.

• Balance the hydrogen atoms with HBalance the hydrogen atoms with H++..

• Balance theBalance the charge with electrons.charge with electrons.

After combining the half-reactions, After combining the half-reactions,

check for mass and charge balance.check for mass and charge balance.

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Practice ProblemsPractice ProblemsPractice ProblemsPractice ProblemsBalance the following equation:Balance the following equation:

MnOMnO44-- + H + H22SOSO3 3 ----> Mn ----> Mn+2+2 + SO + SO44

-2-2

MnOMnO44-- ----> Mn ----> Mn+2+2

HH22SOSO3 3 ----> SO ----> SO44-2-2

2. Balance atoms:

8 H8 H++ + MnO + MnO44-- ----> Mn ----> Mn+2+2 + 4 H + 4 H22O O

HH22O + HO + H22SOSO3 3 ----> SO ----> SO44-2-2 + 4 H + 4 H++

1. Separate into half reactions:

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Practice ProblemsPractice ProblemsPractice ProblemsPractice Problems3. Balance charges:

5 e5 e-- + 8 H + 8 H++ + MnO + MnO44------> Mn----> Mn+2+2 + 4 H + 4 H22O O

HH22O + HO + H22SOSO3 3 ----> SO ----> SO44-2-2 + 4 H + 4 H++ + 2 e + 2 e--

4. Equal electrons gained and lost:

2(2(5 e5 e-- + 8 H + 8 H++ + MnO + MnO44-- ----> Mn ----> Mn+2+2 + 4 H + 4 H22OO))

5(5(HH22O + HO + H22SOSO3 3 ----> SO ----> SO44-2-2+ 4 H+ 4 H+++ 2 e+ 2 e--))

10 16 2 2 8

5 5 5 20 10

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Practice ProblemsPractice ProblemsPractice ProblemsPractice Problems

5. Simplify and Add:

2(2(5 e5 e-- + 8 H + 8 H++ + MnO + MnO44-- ----> Mn ----> Mn+2+2 + 4 H + 4 H22OO))

5(5(HH22O + HO + H22SOSO3 3 ----> SO ----> SO44-2-2 + 4 H + 4 H+++2 e+2 e--))

10 16 2 2 8

5 5 5 20 10

2MnO2MnO44-- + 5 H + 5 H22SOSO3 3 ----> ---->

2 Mn2 Mn+2 +2 + 5 SO+ 5 SO44-2-2 + 4 H + 4 H++ + + 33 HH22O O

4

3

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Practice ProblemsPractice ProblemsPractice ProblemsPractice ProblemsBalance the following equation:Balance the following equation:

Al + NOAl + NO33-- ---> Al(OH) ---> Al(OH)44

-- + NH + NH33

Al ----> Al(OH)Al ----> Al(OH)44--

NONO33-- ----> NH ----> NH33

2. Balance atoms:

4 H4 H22O + Al ----> Al(OH)O + Al ----> Al(OH)44-- + 4 H + 4 H++

9 H9 H+ + + NO+ NO33-- ----> NH ----> NH3 3 + 3 H+ 3 H22OO

1. Separate into half reactions:

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Practice ProblemsPractice ProblemsPractice ProblemsPractice Problems3. Balance charges:

4 H4 H22O + Al ----> Al(OH)O + Al ----> Al(OH)44-- + 4 + 4

HH++ + 3 e + 3 e--8 e8 e-- + 9 H + 9 H+ + + NO+ NO33-- ----> NH ----> NH3 3 + 3 + 3

HH22OO 4. Equal electrons gained and lost:

8(8(4 H4 H22O + Al ----> Al(OH)O + Al ----> Al(OH)44-- + 4 H + 4 H++

+ 3 e+ 3 e--)) 3(3(8 e8 e-- + 9 H + 9 H+ + + NO+ NO33

-- ----> NH ----> NH3 3 + 3 + 3

HH22OO))

32 8 8 32 24

24 27 3 3 9

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5. Simplify and Add:

8(8(4 H4 H22O + Al ----> Al(OH)O + Al ----> Al(OH)44-- + 4 H + 4 H++

+ 3 e+ 3 e--))3(3(8 e8 e-- + 9 H + 9 H+ + + NO+ NO33

-- ----> NH ----> NH3 3 + 3 + 3

HH22OO))

32 8 8 32 24

24 27 3 3 9

23 H23 H22O + 8 Al + 3 NOO + 8 Al + 3 NO33-- ----> ---->

8 8

Al(OH)Al(OH)44-- + 5 H + 5 H++ + 3 NH + 3 NH3 3

523

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6. Change to basic solution:

23 H23 H22O + 8 Al + 3 NOO + 8 Al + 3 NO33-- ----> ----> 8 Al(OH)8 Al(OH)44

--

+ 5 H+ 5 H++ + 3 NH + 3 NH3 3 + 5 OH- + 5 OH -

5 H2O

18

18 H18 H22O + 5 OHO + 5 OH-- + 8 Al + 3 NO + 8 Al + 3 NO33-- ---->8 ---->8

Al(OH)Al(OH)44-- + 3 NH + 3 NH3 3

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1. MnO2 + HBr --> Br2 + MnBr2

2e2e-- + 4 H + 4 H++ + MnO + MnO22 ----> Mn ----> Mn2+2+ + 2 + 2

HH22OO 2 Br2 Br-- ----> Br ----> Br2 2 + 2e+ 2e--

4 H4 H++ + MnO + MnO22 + 2 Br + 2 Br-- ----> Mn ----> Mn2+2+ + 2 + 2

HH22O + BrO + Br22

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2. Cl2 + NaBr --> NaCl + Br2

2e2e-- + Cl + Cl22 --> 2 Cl --> 2 Cl--

2 Br --> Br2 Br --> Br22 + 2e + 2e--

ClCl22 + 2 Br + 2 Br-- ----> 2 Cl ----> 2 Cl-- + +

BrBr2 2

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3. H2S + HNO3 --> S + NO

3(3(HH22S ----> S + 2 HS ----> S + 2 H++ + 2 e + 2 e--))

2(2(3 e3 e-- + 4 H + 4 H+ + + NO+ NO33-- ----> NO ----> NO + 2 + 2

HH22OO))

3 3 6 6

6 8 2 2 42

3 H2S + 2 H+ + 2 NO3- ----> 3 S + 2 NO + 4 H2O

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4. PbO2 + Sb --> PbO + NaSbO2

(base)

3(3(2 e- +2 e- + 2H2H++ + PbO + PbO22 ----> PbO + ----> PbO + HH22OO))

2(2(2 H2 H22O + Sb ----> SbOO + Sb ----> SbO22-- + 4 + 4

HH++ + 3 e + 3 e--))

6 6 3 3 3

4 2 2 8 6

3 PbO3 PbO22 + H + H22O + 2 Sb --->3 PbO + 2 O + 2 Sb --->3 PbO + 2

SbOSbO22-- + 2 H + 2 H++

21

+ 2 OH- + 2 OH-

2 H2O

3PbO3PbO22 + 2 OH- + 2Sb --->3PbO + + 2 OH- + 2Sb --->3PbO +

2SbO2SbO22-- + H + H22OO

1

1 CHAPTER 20 PRINCIPLES OF REACTIVITY: ELECTRON TRANSFER REACTIONS 1-11 + all bold numbered...

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Transcript of 1 CHAPTER 20 PRINCIPLES OF REACTIVITY: ELECTRON TRANSFER REACTIONS 1-11 + all bold numbered...