ElectrochemistryElectrochemistry

Chapter 17

ElectrochemistryElectrochemistry

The study of the The study of the interchangeinterchange of of chemical and electrical energy.chemical and electrical energy.

Review of TermsReview of Terms

oxidation-reduction (redox) reactionoxidation-reduction (redox) reaction: : involves a transfer of electrons from the involves a transfer of electrons from the reducingreducing agent to the agent to the oxidizing agentoxidizing agent..

oxidationoxidation: loss of electrons: loss of electrons

reductionreduction: gain of electrons: gain of electrons

What are some examples of What are some examples of Redox Reactions?Redox Reactions?

What are some examples What are some examples of redox reactions?of redox reactions?

•forest fireforest fire

•rusting steelrusting steel

•combustion in auto enginecombustion in auto engine

•metabolism of food in the metabolism of food in the bodybody

What everday uses depend on What everday uses depend on redox reactions?redox reactions?

•starting a carstarting a car

•calculatorcalculator

•digital watchdigital watch

•portable radioportable radio

•portable CD playerportable CD player

Half-ReactionsHalf-Reactions

The overall reaction is split into two The overall reaction is split into two half-reactionshalf-reactions, , one involving one involving oxidationoxidation and one and one reductionreduction..

8H8H++ + MnO + MnO44 + 5Fe + 5Fe2+2+ Mn Mn2+2+ + 5Fe + 5Fe3+3+ + 4H + 4H22OO

Reduction:Reduction: 8H 8H++ + MnO + MnO44 + 5e + 5e Mn Mn2+2+ + 4H + 4H22OO

Oxidation:Oxidation: 5Fe 5Fe2+2+ 5Fe 5Fe3+3+ + 5e + 5e

Anode and CathodeAnode and Cathode

OXIDATIONOXIDATION occurs at the occurs at the ANODEANODE..

REDUCTIONREDUCTION occurs at the occurs at the CATHODECATHODE..

AN OXAN OX RED CATRED CAT

RedoxRedox

Oxidation is the loss of electrons--oxidation Oxidation is the loss of electrons--oxidation number becomes more positive.number becomes more positive.

Reduction is the gain of electrons--oxidation Reduction is the gain of electrons--oxidation number becomes more negative.number becomes more negative.

OIL RIGOIL RIG

Oxidation Is Loss.

Reduction Is Gain.

Redox ReactionsRedox Reactions

Loss and gain of electrons must be Loss and gain of electrons must be simultaneous.simultaneous.

Loss and gain of electrons must be equal.Loss and gain of electrons must be equal.

Why must the loss and gain of electrons be Why must the loss and gain of electrons be equal?equal?Law of Conservation of Matter

Redox ReactionsRedox Reactions

Redox reactions are reactions in which Redox reactions are reactions in which electrons are transferred. electrons are transferred.

Decomposition and synthesis reactions Decomposition and synthesis reactions maymay be be redox.redox.

Single replacement reactions are Single replacement reactions are alwaysalways redox. redox.

Double replacement reactions are Double replacement reactions are nevernever redox. redox.

Combustion reactions are Combustion reactions are alwaysalways redox. redox.

Identifying Oxidation & Reduction Identifying Oxidation & Reduction in a Reactionin a Reaction

Identify the element which is Identify the element which is oxidizedoxidized and the one and the one which is which is reducedreduced..

1. 2Mg1. 2Mg(s)(s) + O + O2(g)2(g) ---> 2MgO ---> 2MgO(s)(s)

2. 2Al2. 2Al(s)(s) + 3I + 3I2(s)2(s) ---> 2AlI ---> 2AlI3(s)3(s)

3. 2Cu3. 2Cu(s)(s) + O + O2(g)2(g) ---> 2CuO ---> 2CuO(s)(s)

4. 2Cs4. 2Cs(s)(s) + F + F2(g)2(g) ---> 2CsF ---> 2CsF(s)(s)

Rules for Assigning Oxidation StatesRules for Assigning Oxidation States1.1. Oxidation state of an atom in an element = 0Oxidation state of an atom in an element = 0

2.2. Oxidation state of monatomic element = chargeOxidation state of monatomic element = charge

3.3. Oxygen = Oxygen = 2 in covalent compounds (except in 2 in covalent compounds (except in peroxides where it = peroxides where it = 1)1)

4.4. H = +1 in covalent compoundsH = +1 in covalent compounds

5.5. Fluorine = Fluorine = 1 in compounds1 in compounds

6.6. Sum of oxidation states = 0 in compounds Sum of oxidation states = 0 in compounds Sum of oxidation states = charge of the ionSum of oxidation states = charge of the ion

Charges & Oxidation StatesCharges & Oxidation States

Oxidation states are written as +2.Oxidation states are written as +2.

Charges are written 2+.Charges are written 2+.

Ca2

2Ca

Determining Oxidation StatesDetermining Oxidation States

SFSF66 [NO[NO33]]--

+6+6 -6 = 0-6 = 0 +5 -6 = -1 +5 -6 = -1

(-1 for each F) (-1 for each F) (-2 for each (-2 for each O)O)

The most electronegative element is assigneda negative oxidation number--see electronegativity chart on page 354.

Determining Oxidation StatesDetermining Oxidation States

substancesubstance

NaNa(s)(s)

NaFNaF(s)(s)

SOSO2(g)2(g)

HH22OO22

NHNH44ClCl(s)(s)

oxidation number rule

0 1

+1, -1 2

+4, -2 3

+1, -1 3

-3, +1, -1 5, 4, 7, & 2

Redox Redox

metalsmetals nonmetalsnonmetals

oxidation reduction

reducing agents oxidizing agents

metal ions nonmetal ions

reduction oxidation

oxidizing agents reducing agents

RedoxRedoxOxidizing agent is the electron acceptor--Oxidizing agent is the electron acceptor--

usually a nonmetal.usually a nonmetal.

Reducing agent is the electron donor--usually Reducing agent is the electron donor--usually a metal. a metal.

CHCH4(g)4(g) + 2O + 2O2(g)2(g) ----> CO ----> CO2(g)2(g) + 2HOH + 2HOH(g)(g)

Carbon is oxidized.Carbon is oxidized.

Oxygen is reduced.Oxygen is reduced.

CH4 is the reducing agent.

O2 is the oxidizing agent.

-4 +1 0 +4 -2 +1-2+1

Redox ReactionsRedox ReactionsIdentify the substance oxidized and the substance reduced Identify the substance oxidized and the substance reduced

as well as the oxidizing and reducing agents.as well as the oxidizing and reducing agents.

PbOPbO(s)(s) + CO + CO(g)(g) ---> Pb ---> Pb(s)(s) + CO + CO2(g)2(g)

oxidizedoxidized

reducedreduced

oxidizing agentoxidizing agent

reducing agentreducing agent

+2 -2 +2 -2 0 +4 -2

Carbon

Lead

PbO

CO

Redox ReactionsRedox ReactionsIdentify the substance oxidized and the substance reduced Identify the substance oxidized and the substance reduced

as well as the oxidizing and reducing agents.as well as the oxidizing and reducing agents.

2PbS2PbS(s)(s) + + 3O3O2(g) 2(g) ---> 2PbO---> 2PbO(s)(s) + 2SO + 2SO2(g)2(g)

oxidizedoxidized

reducedreduced

oxidizing agentoxidizing agent

reducing agentreducing agent

+2 -2 0 +2 -2 +4 -2

sulfur

oxygen

O2

PbS

Balancing by Half-Reaction MethodBalancing by Half-Reaction Method

1.1. Write separate reduction, oxidation Write separate reduction, oxidation reactions.reactions.

2.2. For each half-reaction:For each half-reaction:

-- Balance elements (except H, O)Balance elements (except H, O)

-- Balance O using HBalance O using H22OO

-- Balance H using HBalance H using H++

-- Balance charge using electronsBalance charge using electrons

Balancing by Half-Reaction Balancing by Half-Reaction Method Method (continued)(continued)

3.3. If necessary, multiply by integer to If necessary, multiply by integer to equalize electron count.equalize electron count.

4.4. Add half-reactions & cancel identical Add half-reactions & cancel identical species.species.

5.5. Check that both Check that both elementselements and and chargescharges are are balanced.balanced.

Balancing By Half-Reaction Balancing By Half-Reaction Acidic SolutionAcidic Solution

HH++(aq)(aq) + Cr + Cr22OO77

2-2-(aq)(aq) + C + C22HH55OHOH(l)(l) ---> Cr ---> Cr3+3+

(aq)(aq) + CO + CO2(g)2(g) + HOH + HOH(l)(l)

Red CrRed Cr22OO772-2-

(aq)(aq) ---> Cr ---> Cr3+3+(aq) (aq)

OxOx CC22HH55OHOH(l)(l) ---> CO---> CO2(g)2(g)

Red 2(6eRed 2(6e-- + 14H + 14H++(aq) (aq) + Cr+ Cr22OO77

2-2-(aq)(aq) ---> 2Cr ---> 2Cr3+3+

(aq) (aq) + 7HOH+ 7HOH(l)(l)))

OxOx C C22HH55OHOH(l)(l) + 3HOH + 3HOH(l) (l) ---> 2CO---> 2CO2(g)2(g) + 12H + 12H++(aq)(aq) + 12e + 12e--

16H16H++(aq)(aq) + 2Cr + 2Cr22OO77

2-2-(aq)(aq) + C + C22HH55OHOH(l)(l) ---> 4Cr ---> 4Cr3+3+

(aq)(aq) + 11HOH + 11HOH(l)(l) + + 2CO2CO2(g)2(g)

12+ = 12+12+ = 12+

Redox ReactionsRedox Reactions

Always add electrons to the side of Always add electrons to the side of the half-reaction with excess the half-reaction with excess positive charge!positive charge!

Balancing By Half-Reaction Balancing By Half-Reaction Acidic SolutionAcidic Solution

CuCu(s)(s) + HNO + HNO3(aq)3(aq) ---> Cu(NO ---> Cu(NO33))2(aq)2(aq) + NO + NO(g)(g) + HOH + HOH(l)(l)

Ox CuOx Cu(s)(s) + HNO + HNO3(aq)3(aq) ---> Cu(NO ---> Cu(NO33))2(aq) 2(aq)

RedRed HNOHNO3(aq)3(aq) ---> NO---> NO(g)(g)

Ox 3(CuOx 3(Cu(s) (s) + 2HNO+ 2HNO3(aq)3(aq) ---> Cu(NO ---> Cu(NO33))2(aq) 2(aq) + 2H+ 2H++(aq) (aq) + 2e+ 2e--))

RedRed 2(3e 2(3e-- + 3H + 3H++(aq) (aq) ++ HNOHNO3(aq)3(aq) ---> NO---> NO(g)(g) + 2HOH + 2HOH(l)(l) ) )

3Cu3Cu(s)(s) + 8HNO + 8HNO3(aq)3(aq) ---> 3Cu(NO ---> 3Cu(NO33))2(aq)2(aq) + 4HOH + 4HOH(l)(l) + 2NO + 2NO(g)(g)

0 = 00 = 0

Half-Reaction Method - Half-Reaction Method - Balancing in BaseBalancing in Base

1.1. Balance as in acid.Balance as in acid.

2.2. Add OHAdd OH that equals H that equals H++ ions (both sides!) ions (both sides!)

3.3. Form water by combining HForm water by combining H++, OH, OH..

4.4. Check elements and charges for balance.Check elements and charges for balance.

Balancing By Half-Reaction Balancing By Half-Reaction Basic SolutionBasic Solution

AgAg(s)(s) + CN + CN--(aq)(aq) + O + O2(g)2(g) ---> Ag(CN) ---> Ag(CN)22

--(aq)(aq)(Basic)(Basic)

Ox CNOx CN--(aq) (aq) + Ag+ Ag(s)(s) ---> Ag(CN) ---> Ag(CN)22

--(aq)(aq)

RedRed O O2(g)2(g) ---> --->

Ox 4(2CNOx 4(2CN--(aq) (aq) + Ag+ Ag(s)(s) ---> Ag(CN) ---> Ag(CN)22

--(aq) (aq) + e+ e--))

RedRed O O2(g)2(g) + 4H + 4H++(aq) (aq) + 4e+ 4e-- ---> 2HOH ---> 2HOH(l)(l)

8CN8CN--(aq)(aq) + 4Ag + 4Ag(s) (s) + O+ O2(g)2(g) + 4H + 4H++

(aq) (aq) ---> 4Ag(CN)---> 4Ag(CN)22--(aq) (aq) + + 2HOH2HOH(l)(l)

Balancing By Half-Reaction Balancing By Half-Reaction Basic SolutionBasic Solution

8CN8CN--(aq)(aq) + 4Ag + 4Ag(s) (s) + O+ O2(g)2(g) + 4H + 4H++

(aq) (aq) + 4OH+ 4OH--(aq)(aq)

---> 4Ag(CN)---> 4Ag(CN)22--(aq) (aq) + 2HOH+ 2HOH(l) (l) + 4OH+ 4OH--

(aq)(aq)

8CN8CN--(aq)(aq) + 4Ag + 4Ag(s) (s) + O+ O2(g)2(g) + 4HOH + 4HOH(l)(l)

---> 4Ag(CN)---> 4Ag(CN)22--(aq) (aq) + 2HOH+ 2HOH(l) (l) + 4OH+ 4OH--

(aq)(aq)

88-- = 8 = 8--

+ 2HOH+ 2HOH(l)(l)

Galvanic CellGalvanic Cell

A device in which chemical A device in which chemical energy is changed to electrical energy is changed to electrical energy.energy.

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Porous disk

Reducingagent

Oxidizingagent

e –

e –

e – e –

e –

e –

CathodeAnode (b)(a)

Electrons are transferred at the interface between theelectrodes and the solution. Porous disk allows ion flow.

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e–

e – e–

e–

Zn 2+

SO4 2–

Zn( s)

1.0 M Zn 2+

solution

Anode

1.0 M Cu 2+

solution

Cathode

Cu 2+

SO4 2–

Cu( s)

A galvanic cell (Daniell Cell) involving Zn and Cu electrodes. This cell was the energy source for telegraphyduring the War Between the States.

Cu-Zn (Daniell Cell) on the microscopic level.

Zinc electrode compared to a Standard Hydrogen Electrode(SHE). The Zn has a potential of 0.76 V.

Standard Reduction PotentialsStandard Reduction Potentials

The The EE values corresponding to reduction half- values corresponding to reduction half-reactions with all solutes at 1M and all gases at 1 reactions with all solutes at 1M and all gases at 1 atm.atm.

CuCu2+2+ + 2e + 2e Cu Cu EE = 0.34 V vs. SHE = 0.34 V vs. SHE

SOSO4422 + 4H + 4H++ + 2e + 2e H H22SOSO33 + H + H22O O

EE = 0.20 V vs. SHE = 0.20 V vs. SHE

SHE = Standard Hydrogen ElectrodeSHE = Standard Hydrogen Electrode

0.54 Li+ + e Li 3.05

17_01TTable 17.1 Standard Reduction Potentials at 25°C (298 K) for Many Common Half-reactions Half-reaction ° (V) Half-reaction

2.87 O2 + 2H2O + 4e 4HO 0.401.99 Cu2+ + 2e Cu 0.341.82 Hg2Cl2 + 2e 2Hg + 2Cl 0.341.78 AgCl + e Ag + Cl 0.221.70 SO4

+ 4H+ 2e H2SO3 + H2SO3 + H2O 0.201.69 Cu2+ + e Cu+ 0.161.68 2H+ + 2e H2 0.001.60 Fe3+ + 3e Fe 0.0361.51 Pb2+ + 2e Pb 0.131.50 Sn2+ + 2e Sn 0.141.46 Ni2+ + 2e Ni 0.231.36 PbSO4 + 2e Pb + SO4

2 0.351.33 Cd2+ + 2e Cd 0.401.23 Fe2+ + 2e Fe 0.441.21 Cr3+ + e Cr2+ 0.501.20 Cr3+ + 3e Cr 0.731.09 Zn2+ + 2e Zn 0.761.00 2H2O + 2e H2 + 2OH 0.830.99 Mn2+ + 2eMn 1.180.96 Al3+ + 3e Al 1.660.954 H2 + 2e 2H 2.230.91 Mg2+ + 2eMg 2.370.80 La3+ + 3e La 2.370.80 Na+ + e Na 2.710.77 Ca2+ + 2e Ca 2.760.68 Ba2+ + 2e Ba 2.90

I2 + 2e 2I

F2 + 2e 2F

Ag2+ + e Ag+

Co3+ + e Co2+

H2O2 + 2H+ + 2e 2H2OCe4+ + e Ce3+

PbO2 + 4H+ + SO42 + 2ePbSO4 + 2H2O

MnO4 + 4H+ + 3eMnO2 + 2H2O

2e + 2H+ + IO4IO3

+ H2OMnO4

+ 8H+ + 5eMn2+ + 4H2OAu3+ + 3e AuPbO2 + 4H+ + 2ePb2+ + 2H2OCl2 + 2e 2Cl

Cr2O72 + 14H+ + 6e2Cr3+ + 7H2O

O2 + 4H+ + 4e 2H2OMnO2 + 4H+ + 2eMn2+ + 2H2OIO3

+ 6H+ + 5e½I2 + 3H2OBr2 + 2e 2Br

VO2 + 2H+ + e VO2+ + H2OAuCl4

+ 3e Au + 4Cl

NO3 + 4H+3e NO + 2H2O

ClO2 + e ClO2

2Hg2+ + 2e Hg22+

Ag+ + e AgHg2

2+ + 2e 2Hg+

Fe3+ + e Fe2+

O2 + 2H+ + 2e H2O2MnO4

+ eMnO42 0.56 K+ + e K 2.92

0.52Cu+ + e Cu

° (V)

Cell PotentialCell Potential

Cell Potential Cell Potential or or Electromotive Electromotive Force Force (emf): The “pull” or driving (emf): The “pull” or driving force on the electrons.force on the electrons.

Cell Potential CalculationsCell Potential CalculationsTo Calculate cell potential using Standard To Calculate cell potential using Standard

Reduction Potentials:Reduction Potentials:

1. One reaction and its cell potential’s sign 1. One reaction and its cell potential’s sign must be reversed--it must be chosen such must be reversed--it must be chosen such that the overall cell potential is positive.that the overall cell potential is positive.

2. The half-reactions must often be multiplied 2. The half-reactions must often be multiplied by an integer to balance electrons--this is by an integer to balance electrons--this is not done for the cell potentials.

Cell Potential Calculations Cell Potential Calculations ContinuedContinued

FeFe3+3+(aq)(aq) + Cu + Cu(s)(s) ----> Cu ----> Cu2+2+

(aq)(aq) + Fe + Fe2+2+(aq)(aq)

FeFe3+3+(aq)(aq) + e + e- - ----> Fe ----> Fe2+2+

(aq) (aq) EEoo = 0.77 V= 0.77 V

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

Reaction # 2 must be reversed.Reaction # 2 must be reversed.

Cell Potential Calculations Cell Potential Calculations ContinuedContinued

2 (Fe2 (Fe3+3+(aq)(aq) + e + e- - ----> Fe ----> Fe2+2+

(aq)(aq))) EEoo = 0.77 V= 0.77 V

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

2Fe2Fe3+3+(aq)(aq) + Cu + Cu(s)(s) ----> Cu ----> Cu2+2+

(aq)(aq) + 2Fe + 2Fe2+2+(aq)(aq)

EEoo = 0.43 V = 0.43 V

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Reference solution ofdilute hydrochloric acid

Silver wire coated withsilver chloride

Thin-walled membrane

Ion selective electrodes are glass electrodes that measures a change in potential when [H+] varies. Used to measure pH.

BatteriesBatteries

A A batterybattery is a galvanic cell or, is a galvanic cell or, more commonly, a more commonly, a groupgroup of of galvanic cells connected in series.galvanic cells connected in series.

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H2SO4

electrolytesolution

Anode (leadgrid filled withspongy lead) Cathode (lead

grid filled withspongy PbO2)

A lead storage battery consists of a lead anode, lead dioxide cathode, and an electrolyte of 38% sulfuric acid.

Lead Storage BatteryLead Storage Battery

Anode reaction:Anode reaction:

PbPb(s)(s) +H +H22SOSO4(aq)4(aq) ---> PbSO ---> PbSO4(aq)4(aq) + 2H + 2H++(aq) (aq) + 2e+ 2e--

Cathode reaction: Cathode reaction:

PbOPbO2(s)2(s) + H + H22SOSO4(aq) 4(aq) ++ 2e2e-- + 2H + 2H++(aq) (aq) ---> PbSO---> PbSO4(aq)4(aq) + +

2HOH 2HOH(l) (l)

Overall reaction:Overall reaction:

PbPb(s)(s)+ PbO+ PbO2(s) 2(s) ++ 2H2H22SOSO4(aq) 4(aq) ---> PbSO---> PbSO4(aq)4(aq) + + 2HOH 2HOH(l)(l)

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Cathode(graphite rod)

Anode(zinc inner case)

Paste of MnO2,NH4CL, andcarbon

Common dry cell and its components.

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Cathode (steel)

Anode (zinc container)

Solution of HgO (oxidizingagent) in a basic medium (KOHand Zn(OH)2)

Insulation

Mercury battery used in calculators.

Fuel CellsFuel Cells

. . .. . . galvanic cells for which the reactants are galvanic cells for which the reactants are continuously supplied.continuously supplied.

2H2H22((gg) + O) + O22((gg) ) 2H 2H22O(O(ll))

anodeanode: 2H: 2H22 + 4OH + 4OH 4H 4H22O + 4eO + 4e

cathodecathode: 4e: 4e + O + O22 + 2H + 2H22O O 4OH 4OH

emf and Workemf and Work

emf potential difference Vwork J

ch e C ( )

( )arg ( )

Free Energy and Cell Free Energy and Cell PotentialPotential

GG = = nFEnFE

nn = number of moles of electrons = number of moles of electrons

FF = Faraday = 96,485 coulombs per mole of = Faraday = 96,485 coulombs per mole of electronselectrons

Concentration CellConcentration Cell

. . .. . . a cell in which both compartments a cell in which both compartments have the have the same components same components but at but at different concentrationsdifferent concentrations..

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e–

e–e–

e–

Ag

1 M Ag+

1 M NO3–

Anode Cathode

Porousdisk

Ag

0.1 M Ag+

0.1 M NO3–

The Nernst EquationThe Nernst Equation

We can calculate the potential of a cell in We can calculate the potential of a cell in which some or all of the components are not which some or all of the components are not in their standard states.in their standard states.

E E 0 .0 5 9 1

n lo g Q

Calculation of Equilibrium Calculation of Equilibrium Constants for Redox ReactionsConstants for Redox Reactions

At equilibrium, At equilibrium, EEcellcell = 0 and = 0 and QQ = = KK..

lo g ( ).

KnE

0 0 5 9 1

a t 2 5 C

Stoichiometry of ElectrolysisStoichiometry of Electrolysis

- How much How much chemical change chemical change occurs with occurs with the flow of a given current for a specified the flow of a given current for a specified time?time?

current and time current and time quantity of charge quantity of charge moles of electrons moles of electrons moles of analyte moles of analyte grams of analytegrams of analyte

1 amp = 1 C/s1 amp = 1 C/s

Electrolytic CalculationsElectrolytic Calculations

How many grams of copper can be plated out when a How many grams of copper can be plated out when a current of 10.0 amps is passed through a current of 10.0 amps is passed through a CuCu2+2+solution for 30.0 minutes?solution for 30.0 minutes?

(10.0 C/s)(30.0 min)(60 s/1 min)(1 mol e(10.0 C/s)(30.0 min)(60 s/1 min)(1 mol e --/96,485 C)/96,485 C)

(1 mol Cu/2 mole e(1 mol Cu/2 mole e--)(63.5 g/1 mol) = 5.94 g Cu )(63.5 g/1 mol) = 5.94 g Cu

Electrolytic CalculationsElectrolytic Calculations

How long must a current of 5.00 A be applied to a How long must a current of 5.00 A be applied to a solution of Agsolution of Ag1+1+ to produce 10.5 g of silver? to produce 10.5 g of silver?

(10.5 g Ag)(1 mol/107.86 g)(1 mol e(10.5 g Ag)(1 mol/107.86 g)(1 mol e --/1 mol Ag)/1 mol Ag)

(96,485 C/1 mole e(96,485 C/1 mole e--)(1 s/5.00 C)(1 min/60.0s))(1 s/5.00 C)(1 min/60.0s)

= 31.3 min = 31.3 min

Electrochemical <---> ElectrolyticElectrochemical <---> Electrolytic

Electrochemical <---> Electrolytic

Spontaneous <---> Nonspontaneous

Energy released <---> Energy absorbed

Cu2+(aq) + Mg(s) <---> Cu(s) + Mg2+

(aq)

Electrochemical cell -- chemical energy to electrical energy.

Electrolytic cell -- electrical energy to chemical energy.

Galvanic CellGalvanic CellA device in which chemical energy A device in which chemical energy is changed to electrical energy. The is changed to electrical energy. The basic parts are:basic parts are:

anodeanode

cathodecathode

electrochemical solutionelectrochemical solution

porous disk or salt bridgeporous disk or salt bridge

CorrosionCorrosionSome metals, such as copper, gold, Some metals, such as copper, gold, silver and platinum, are relatively silver and platinum, are relatively difficult to oxidize. These are often difficult to oxidize. These are often called called noble metalsnoble metals..

About 1/5 of all iron and steel About 1/5 of all iron and steel produced each year is used to replace produced each year is used to replace rusted metal.rusted metal.

Self-protecting MetalsSelf-protecting Metals

Some metals such as aluminum, copper, and Some metals such as aluminum, copper, and silver form a silver form a protective coatingprotective coating that keeps that keeps them from corroding further. them from corroding further.

The protective coating for iron and steel The protective coating for iron and steel flakes away opening new layers of metal flakes away opening new layers of metal to corrosion.to corrosion.

Corrosion of IronCorrosion of Iron

Iron is oxidized at the anodic reaction and oxygen is reducedat the cathodic reaction. Dissolved ions are necessary totransfer electrons between the anodic and cathodic areas.

Prevention of CorrosionPrevention of CorrosionCoatingCoating--painting or applying oil to keep out --painting or applying oil to keep out

oxygen and moisture.oxygen and moisture.

GalvanizingGalvanizing--dipping a metal in a more active --dipping a metal in a more active metal -- galvanized steel bucket.metal -- galvanized steel bucket.

AlloyingAlloying -- mixing metals with iron to prevent -- mixing metals with iron to prevent corrosion -- stainless steel.corrosion -- stainless steel.

Cathodic protectionCathodic protection -- attaching a more active -- attaching a more active metal. Serves as sacrificial metal--used to metal. Serves as sacrificial metal--used to protect ships, gas lines, and gas tanks.protect ships, gas lines, and gas tanks.

ElectrolysisElectrolysis

. . .. . . forcingforcing a current through a cell to a current through a cell to produce a chemical change for which produce a chemical change for which the cell potential is negative.the cell potential is negative.

Electrolysis of WaterElectrolysis of Water

The electrolysis of water is:The electrolysis of water is:

DCDC

2HOH2HOH(l)(l) -----> 2H -----> 2H2(g)2(g) + + OO2(g)2(g)

Why would this process be important on a Why would this process be important on a

submarine?submarine?

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Moltenaluminum

Carbondioxideformed atthe anodes

Carbon-lined iron tank

Plug

Molten Al2O3/Na3AlF6mixture

Electrodes ofgraphite rods

To externalpower source

Schematic of Hall-Heroult process. Molten sinks andis tapped off at the bottom. Bauxite-cryolite mixturefloats on top and is electrolyzed by the carbon electrodes.

Why did Napoleon III of France served his most honored guests with aluminum silverware?