CHEM 160 General Chemistry II Lecture Presentation Electrochemistry December 1, 2004 Chapter 20.

CHEM 160 General Chemistry IICHEM 160 General Chemistry IILecture PresentationLecture Presentation

ElectrochemistryElectrochemistry

December 1, 2004

Chapter 20


Electrochemistry deals with interconversion between chemical and

electrical energy


Electrochemistry deals with the interconversion between chemical and

electrical energy involves redox reactions


Electrochemistry deals with interconversion between chemical and

electrical energy involves redox reactions

• electron transfer reactions

•Oh No! They’re back!

Redox reactions (quick review) Redox reactions (quick review)

Oxidation

Reduction

Reducing agent

Oxidizing agent

Redox reactions (quick review)Redox reactions (quick review)

Oxidation loss of electrons

Reduction

Reducing agent

Oxidizing agent



Reduction gain of electrons

Reducing agent

Oxidizing agent




Reducing agent donates the electrons and is oxidized

Oxidizing agent




Reducing agent donates the electrons and is oxidized

Oxidizing agent accepts electrons and is reduced

Redox ReactionsRedox Reactions

Direct redox reaction


Direct redox reaction Oxidizing and reducing agents are mixed together

CuSO4(aq) (Cu2+)

Zn rod

Direct Redox ReactionDirect Redox Reaction

CuSO4(aq) (Cu2+)

Zn rod

Deposit of Cu metal

forms

Direct Redox ReactionDirect Redox Reaction


Direct redox reaction Oxidizing and reducing agents are mixed together

Indirect redox reaction Oxidizing and reducing agents are separated but

connected electrically• Example

– Zn and Cu2+ can be reacted indirectly

Basis for electrochemistry– Electrochemical cell

Electrochemical CellsElectrochemical Cells


Voltaic Cell cell in which a spontaneous redox reaction generates

electricity chemical energy electrical energy


Voltaic Cell



Electrolytic Cell electrochemical cell in which an electric current

drives a nonspontaneous redox reaction electrical energy chemical energy

Cell PotentialCell Potential


Cell Potential (electromotive force), Ecell (V) electrical potential difference between the two

electrodes or half-cells• Depends on specific half-reactions, concentrations, and

temperature

• Under standard state conditions ([solutes] = 1 M, Psolutes = 1 atm), emf = standard cell potential, Ecell

• 1 V = 1 J/C

driving force of the redox reaction

high electrical high electrical potentialpotential

low electrical low electrical potentialpotential



Ecell = Ecathode - Eanode = Eredn - Eox

E°cell = E°cathode - E°anode = E°redn - E°ox

(Ecathode and Eanode are reduction potentials by definition.)


E°cell = E°cathode - E°anode = E°redn - E°ox Ecell can be measured

• Absolute Ecathode and Eanode values cannot

Reference electrode has arbitrarily assigned E used to measure relative Ecathode and Eanode for half-

cell reactionsStandard hydrogen electrode (S.H.E.)

conventional reference electrode

Standard Hydrogen ElectrodeStandard Hydrogen Electrode

E = 0 V (by definition; arbitrarily selected)

2H+ + 2e- H2

Example 1Example 1

A voltaic cell is made by connecting a standard Cu/Cu2+ electrode to a S.H.E. The cell potential is 0.34 V. The Cu electrode is the cathode. What is the standard reduction potential of the Cu/Cu2+ electrode?

Example 2Example 2

A voltaic cell is made by connecting a standard Zn/Zn2+ electrode to a S.H.E. The cell potential is 0.76 V. The Zn electrode is the anode of the cell. What is the standard reduction potential of the Zn/Zn2+ electrode?

Standard Electrode PotentialsStandard Electrode Potentials

Standard Reduction Potentials, E° E°cell measured relative to S.H.E. (0 V)

• electrode of interest = cathode

If E° < 0 V:• Oxidizing agent is harder to reduce than H+

If E° > 0 V:• Oxidizing agent is easier to reduce than H+

Standard Reduction PotentialsStandard Reduction PotentialsReduction Half-Reaction E(V)

F2(g) + 2e- 2F-(aq) 2.87

Au3+(aq) + 3e- Au(s) 1.50

Cl2(g) + 2 e- 2Cl-(aq) 1.36

Cr2O72-(aq) + 14H+(aq) + 6e- 2Cr3+(aq) + 7H2O 1.33

O2(g) + 4H+ + 4e- 2H2O(l) 1.23

Ag+(aq) + e- Ag(s) 0.80

Fe3+(aq) + e- Fe2+(aq) 0.77

Cu2+(aq) + 2e- Cu(s) 0.34

Sn4+(aq) + 2e- Sn2+(aq) 0.15

2H+(aq) + 2e- H2(g) 0.00

Sn2+(aq) + 2e- Sn(s) -0.14

Ni2+(aq) + 2e- Ni(s) -0.23

Fe2+(aq) + 2e- Fe(s) -0.44

Zn2+(aq) + 2e- Zn(s) -0.76

Al3+(aq) + 3e- Al(s) -1.66

Mg2+(aq) + 2e- Mg(s) -2.37

Li+(aq) + e- Li(s) -3.04

Uses of Standard Reduction Uses of Standard Reduction PotentialsPotentials

Compare strengths of reducing/oxidizing agents. the more - E°, stronger the red. agent the more + E°, stronger the ox. agent


F2(g) + 2e- 2F-(aq) 2.87

Au3+(aq) + 3e- Au(s) 1.50

Cl2(g) + 2 e- 2Cl-(aq) 1.36


O2(g) + 4H+ + 4e- 2H2O(l) 1.23

Ag+(aq) + e- Ag(s) 0.80

Fe3+(aq) + e- Fe2+(aq) 0.77

Cu2+(aq) + 2e- Cu(s) 0.34

Sn4+(aq) + 2e- Sn2+(aq) 0.15

2H+(aq) + 2e- H2(g) 0.00

Sn2+(aq) + 2e- Sn(s) -0.14

Ni2+(aq) + 2e- Ni(s) -0.23

Fe2+(aq) + 2e- Fe(s) -0.44

Zn2+(aq) + 2e- Zn(s) -0.76

Al3+(aq) + 3e- Al(s) -1.66

Mg2+(aq) + 2e- Mg(s) -2.37

Li+(aq) + e- Li(s) -3.04

Ox.

age

nt s

tren

gth

incr

ease

sR

ed. agent strength increases


Determine if oxidizing and reducing agent react spontaneously diagonal rule

ox. agent

red. agent

spontaneous

spontaneous


Determine if oxidizing and reducing agent react spontaneously

Cathode (reduction) E°redn (cathode)

more +

Anode (oxidation)

E° re

dn (

V)

E°redn (anode)

more -

Spontaneous rxn if Spontaneous rxn if EE°°cathodecathode > E > E°°anodeanode


F2(g) + 2e- 2F-(aq) 2.87

Au3+(aq) + 3e- Au(s) 1.50

Cl2(g) + 2 e- 2Cl-(aq) 1.36


O2(g) + 4H+ + 4e- 2H2O(l) 1.23

Ag+(aq) + e- Ag(s) 0.80

Fe3+(aq) + e- Fe2+(aq) 0.77

Cu2+(aq) + 2e- Cu(s) 0.34

Sn4+(aq) + 2e- Sn2+(aq) 0.15

2H+(aq) + 2e- H2(g) 0.00

Sn2+(aq) + 2e- Sn(s) -0.14

Ni2+(aq) + 2e- Ni(s) -0.23

Fe2+(aq) + 2e- Fe(s) -0.44

Zn2+(aq) + 2e- Zn(s) -0.76

Al3+(aq) + 3e- Al(s) -1.66

Mg2+(aq) + 2e- Mg(s) -2.37

Li+(aq) + e- Li(s) -3.04


Calculate E°cell

E°cell = E°cathode - E°anode

• Greater E°cell, greater the driving force

E°cell > 0 : spontaneous redox reactions

E°cell < 0 : nonspontaeous redox reactions

Example 3Example 3

A voltaic cell consists of a Ag electrode in 1.0 M AgNO3 and a Cu electrode in 1 M Cu(NO3)2. Calculate E°cell for the spontaneous cell reaction at 25°C.


F2(g) + 2e- 2F-(aq) 2.87

Au3+(aq) + 3e- Au(s) 1.50

Cl2(g) + 2 e- 2Cl-(aq) 1.36


O2(g) + 4H+ + 4e- 2H2O(l) 1.23

Ag+(aq) + e- Ag(s) 0.80

Fe3+(aq) + e- Fe2+(aq) 0.77

Cu2+(aq) + 2e- Cu(s) 0.34

Sn4+(aq) + 2e- Sn2+(aq) 0.15

2H+(aq) + 2e- H2(g) 0.00

Sn2+(aq) + 2e- Sn(s) -0.14

Ni2+(aq) + 2e- Ni(s) -0.23

Fe2+(aq) + 2e- Fe(s) -0.44

Zn2+(aq) + 2e- Zn(s) -0.76

Al3+(aq) + 3e- Al(s) -1.66

Mg2+(aq) + 2e- Mg(s) -2.37

Li+(aq) + e- Li(s) -3.04

Example 4Example 4

A voltaic cell consists of a Ni electrode in 1.0 M Ni(NO3)2 and an Fe electrode in 1 M Fe(NO3)2. Calculate E°cell for the spontaneous cell reaction at 25°C.


F2(g) + 2e- 2F-(aq) 2.87

Au3+(aq) + 3e- Au(s) 1.50

Cl2(g) + 2 e- 2Cl-(aq) 1.36


O2(g) + 4H+ + 4e- 2H2O(l) 1.23

Ag+(aq) + e- Ag(s) 0.80

Fe3+(aq) + e- Fe2+(aq) 0.77

Cu2+(aq) + 2e- Cu(s) 0.34

Sn4+(aq) + 2e- Sn2+(aq) 0.15

2H+(aq) + 2e- H2(g) 0.00

Sn2+(aq) + 2e- Sn(s) -0.14

Ni2+(aq) + 2e- Ni(s) -0.23

Fe2+(aq) + 2e- Fe(s) -0.44

Zn2+(aq) + 2e- Zn(s) -0.76

Al3+(aq) + 3e- Al(s) -1.66

Mg2+(aq) + 2e- Mg(s) -2.37

Li+(aq) + e- Li(s) -3.04


Is there a relationship between Ecell and G for a redox reaction?


Relationship between Ecell and G:

G = -nFEcell

• F = Faraday constant = 96500 C/mol e-’s, n = # e-’s transferred redox rxn.



G = -nFEcell

• F = Faraday constant = 96500 C/mol e-’s, n = # e-’s transferred redox rxn.

• 1 J = CVG < 0, Ecell > 0 = spontaneous

Equilibrium Constants from EEquilibrium Constants from Ecellcell


G = -nFEcell

• F = Faraday constant = 96500 C/mol e-’s, n = # e-’s transferred redox rxn


Under standard state conditions: G° = -nFE°cell



G = -nFEcell





Relationship between Ecell and G: G = -nFEcell


• 1 J = CV G < 0, Ecell > 0 = spontaneous


and G° = -RTlnK

so -nFE°cell = -RTlnK

H° S°

Calorimetric Data

G°Electrochemical

DataComposition

Data

E°cell

Equilibrium constants

K

Example 5Example 5

Calculate E°cell, G°, and K for the voltaic cell that uses the reaction between Ag and Cl2 under standard state conditions at 25°C.

The Nernst EquationThe Nernst EquationG depends on concentrations

G = G° + RTlnQ

andG = -nFEcell and G° = -nFE°cell

thus-nFEcell = -nFE°cell + RTlnQ

or Ecell = E°cell - (RT/nF)lnQ (Nernst eqn.)

The Nernst EquationThe Nernst Equation

Ecell = E°cell - (RT/nF)lnQ (Nernst eqn.) At 298 K (25°C), RT/F = 0.0257 V

soEcell = E°cell - (0.0257/n)lnQ

orEcell = E°cell - (0.0592/n)logQ

Example 7Example 7

Calculate the voltage produced by the galvanic cell which uses the reaction below if [Ag+] = 0.001 M and [Cu2+] = 1.3 M.

2Ag+(aq) + Cu(s) 2Ag(s) + Cu2+(aq)


F2(g) + 2e- 2F-(aq) 2.87

Au3+(aq) + 3e- Au(s) 1.50

Cl2(g) + 2 e- 2Cl-(aq) 1.36


O2(g) + 4H+ + 4e- 2H2O(l) 1.23

Ag+(aq) + e- Ag(s) 0.80

Fe3+(aq) + e- Fe2+(aq) 0.77

Cu2+(aq) + 2e- Cu(s) 0.34

Sn4+(aq) + 2e- Sn2+(aq) 0.15

2H+(aq) + 2e- H2(g) 0.00

Sn2+(aq) + 2e- Sn(s) -0.14

Ni2+(aq) + 2e- Ni(s) -0.23

Fe2+(aq) + 2e- Fe(s) -0.44

Zn2+(aq) + 2e- Zn(s) -0.76

Al3+(aq) + 3e- Al(s) -1.66

Mg2+(aq) + 2e- Mg(s) -2.37

Li+(aq) + e- Li(s) -3.04

Ox.

age

nt s

tren

gth

incr

ease

sR

ed. agent strength increases

Commercial Voltaic CellsCommercial Voltaic CellsBattery

commercial voltaic cell used as portable source of electrical energy

types primary cell

• Nonrechargeable

• Example: Alkaline battery

secondary cell• Rechargeable

• Example: Lead storage battery

How Does a Battery WorkHow Does a Battery Work

cathode (+)

anode (-)

Electrolyte Paste

Seal/cap

Assume a generalized battery

BatteryBattery

cathode (+): Reduction occurs

here

anode (-): oxidation

occurs here

e- flow

Electrolyte paste: ion migration occurs

here

Placing the battery into a flashlight, etc., and turning the power on completes the circuit and allows

electron flow to occur

How Does a Battery WorkHow Does a Battery WorkBattery reaction when producing electricity

(spontaneous): Cathode: O1 + e- R1

Anode: R2 O2 + e-

Overall: O1 + R2 R1 + O2

Recharging a secondary cell Redox reaction must be reversed, i.e., current is

reversed (nonspontaneous)

Recharge: O2 + R1 R2 + O1

Performed using electrical energy from an external power source

BatteriesBatteries

Read the textbook to fill in the details on specific batteries. Alkaline battery Lead storage battery Nicad battery Fuel cell

CorrosionCorrosionCorrosion

deterioration of metals by a spontaneous redox reaction

• Attacked by species in environment– Metal becomes a “voltaic” cell

• Metal is often lost to a solution as an ion

Rusting of Iron

Corrosion of IronCorrosion of Iron


Half-reactions

anode: Fe(s) Fe2+(aq) + 2e-

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

overall: 2Fe(s) + O2(g) + 4H+(aq) 2Fe2+(aq) +

2H2O(l)

Ecell > 0 (Ecell = 0.8 to 1.2 V), so process is spontaneous!


Rust formation:

4Fe2+(aq) + O2(g) + 4H+(aq) 4Fe3+(aq) + 2H2O(l)

2Fe3+(aq) + 4H2O(l) Fe2O3H2O(s) + 6H+(aq)

Prevention of CorrosionPrevention of Corrosion

Cover the Fe surface with a protective coating Paint Passivation

• surface atoms made inactive via oxidation

2Fe(s) + 2Na2CrO4(aq) + 2H2O(l) --> Fe2O3(s) + Cr2O3(s) + 4NaOH(aq)

Other metal• Tin

• Zn– Galvanized iron

Prevention of CorrosionPrevention of Corrosion

Cathodic Protection metal to be protected is brought into contact with a

more easily oxidized metal “sacrificial” metal becomes the anode

• “Corrodes” preferentially over the iron

• Iron serves only as the cathode

Standard Electrode PotentialsStandard Electrode Potentials

Half-reaction E°F2(g) + 2e- -> 2F-(aq) +2.87 V

Ag+(aq) + e- -> Ag(s) +0.80 V

Cu2+(aq) + 2e- -> Cu(s) +0.34 V

2H+(aq) + 2e- -> H2(g) 0 V

Ni2+(aq) + 2e- -> Ni(s) -0.25 V

Fe2+(aq) + 2e- -> Fe(s) -0.44 V

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

Al3+(aq) + 3e- -> Al(s) -1.66 V

Mg2+(aq) + 2e- ->Mg(s) -2.38 V

Metals more easily oxidized than Fe have

more negative E°’s

Cathodic ProtectionCathodic Protection

galvanized steel (Fe)

Cathodic ProtectionCathodic Protection

(cathode)

(electrolyte)

(anode)

ElectrolysisElectrolysis

Electrolysis process in which electrical energy drives a

nonspontaneous redox reaction• electrical energy is converted into chemical energy

Electrolytic cell electrochemical cell in which an electric current

drives a nonspontaneous redox reaction

ElectrolysisElectrolysis

Same principles apply to both electrolytic and voltaic cells oxidation occurs at the anode reduction occurs at the cathode electrons flow from anode to cathode in the external

circuit• In an electrolytic cell, an external power source pumps

the electrons through the external circuit

Electrolysis of Molten NaClElectrolysis of Molten NaCl

Quantitative Aspects of Electrochemical CellsQuantitative Aspects of Electrochemical Cells

For any half-reaction, the amount of a substance oxidized or reduced at an electrode is proportional to the number of electrons passed through the cell Faraday’s law of electrolysis Examples

• Na+ + 1e- Na

• Al3+ + 3e- Al

Number of electrons passing through cell is measured by determining the quantity of charge (coulombs) that has passed

• 1 C = 1 A x 1 s

• 1 F = 1 mole e- = 96500 C

Steps for Quantitative Electrolysis Steps for Quantitative Electrolysis CalculationsCalculations

current (A) and time (s), A x s

charge in coulombs

(C)

Number of moles of e-

moles of substance oxidized or reduced

mass of substance oxidized or reduced

Example 8Example 8

What mass of copper metal can be produced by a 3.00 A current flowing through a copper(II) sulfate (CuSO4) solution for 5.00 hours?

Example 9Example 9

An aqueous solution of an iron salt is electrolyzed by passing a current of 2.50 A for 3.50 hours. As a result, 6.1 g of iron metal are formed at the cathode. Calculate the charge on the iron ions in the solution.

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CHEM 160 General Chemistry II Lecture Presentation Electrochemistry December 1, 2004 Chapter 20.

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