Chapter NineteenElectrochemistry

1

Electrochemistry

The study of chemical reactions through electrical circuits. Monitor redox reactions by controlling electron transfer

REDOX: Shorthand for “REDuction-OXidation”

Use redox reactions to experimentally measure:Reaction progress (kinetics)Composition (equilibrium constants)Energy changes (thermodynamics)

2

Oxidation-Reduction ReactionsOxidation-reduction reactions (REDOX reaction) occur when

electrons are transferred from one reactant to another during a chemical reaction. There is a change in oxidation number for

both substancesOxidation Number: Theoretical charge on an ion

Oxidation is the process where the oxidation number increases.Electrons are lost from the substance

Reduction is the process where the oxidation number decreases. Electrons are gained by the substance

Oxidation and reduction always accompany each other;Neither can occur alone

3

Redox Reaction: Magnesium Burning

4

Oxidation: Mg (s) Mg2+ + 2e-Reduction: 1/2O2(g) + 2e- O2-

Reaction: Mg (s) + 1/2O2(g) MgO(s)

LEO the lion says GER

5

LEOLoseElectrons Oxidation

GERGainElectronsReduction

Oxidation Number Rules: See Chapter 4The rule earlier in the list always takes

precedence.1) ON = 0 for a compound or ionic charge for an ion2)ON = +1 for IA elements and

ON = +2 2A elements3) ON= -2 for oxygen

4)ON= -1 for 7A elementsIf both elements in 7A, then the one higher in the list is -1

5)ON = -2 for 6A elements other than oxygen6) ON = -3 for 5A elements (very shaky!!!)

6

Common Oxidation NumbersMust be able to determine charges on each element

7

Balancing Redox Reactions: Balance Elements

1. Write equation for the ions in acid solution: Fe2+ + Cr2O7

2- Fe3+ + Cr3+

2. Divide equation into 2 half reactionsFe2+ Fe3+ Cr2O7

2- 2Cr3+

3. Balance all elements except H and OFe2+ Fe3+ Cr2O7

2- 2Cr3+

4. Balance O with H2OFe2+ Fe3+ Cr2O7

2- 2Cr3+ + 7H2O5. Balance H with H+

Fe2+ Fe3+ Cr2O72- + 14H+ 2Cr3+ + 7H2O

8

Balancing Redox Reactions: Charge6. Balance chargeFe2+ Fe3+ + 1e- 6e- + Cr2O7

2- + 14H+ 2Cr3+ + 7H2OOxidation: lose electron Reduction: gain electrons7. Multiply by an integer to equalize # electrons 6Fe2+ 6Fe3+ + 6e- 6e- + Cr2O7

2- + 14H+ 2Cr3+ + 7H2O

8. Add reactions together and cancel like species6Fe2+ + 6e- + Cr2O7

2- + 14H+ 2Cr3+ + 7H2O+ 6Fe3+ + 6e-

9. Check balance of final # of atoms and charge6Fe2+ + Cr2O7

2- + 14H+ 2Cr3+ + 7H2O+ 6Fe3+

(12+) + (2-) + (14+) = +24 (6+) + (0) + (18+) = +24

9

Balancing Redox Reactions: Basic Reactions

Need OH- instead of H+ in final equation

1. Balance as if the reaction is in acidic solution.6Fe2+ + Cr2O7

2- + 14H+ 2Cr3+ + 7H2O+ 6Fe3+

2. Add OH- to both sides to match H+

14OH- + 6Fe2+ + Cr2O72- + 14H+ 2Cr3+ + 7H2O+ 6Fe3++14OH-

3. Combine OH- and H+ to make water14H2O + 6Fe2+ + Cr2O7

2- 2Cr3+ + 7H2O+ 6Fe3++14OH-

4. Cancel water from both sides7H2O + 6Fe2+ + Cr2O7

2- 2Cr3+ + 6Fe3++14OH-

5. Check to see if balanced

10

11

Galvanic Cells

Parts of an Electrochemical CellIonic SolutionsProvide ions to transfer chargeSolution + Electrode= Half-cell

ElectrodesAnode: oxidation occurs Cathode: reduction occurs

Salt bridge.Keeps 2 half-cells connectedIons flow, but solution doesn’t

Metal wires Connect the electrodes to the terminals of the voltmeter.Provide means of transporting electrons between electrodes

Voltmeter Measures the electron flow in the system

12

Galvanic Cell: Daniell Cell

13

Types of Cells

Voltaic or Galvanic CellNet oxidation/reduction reaction is spontaneousConvert energy to useful work Batteries

Electrolytic CellNet redox reaction is non-spontaneousWork is done on the cell, energy must be supplied: Recharge a battery, electroplating

14

15

Standard Reduction Potentials

Cell DiagramReaction: Zn(s) + Cu2+ Zn2+ + Cu

Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s)anode solution salt bridge solution cathode

Single bar (|)Divides reduced from oxidizedShow phase differences

Double bars (||) Represent salt bridgeDivides redox half reactions

Goes left to right Anode written firstRemainder in order of actual cell

16

Cell DiagramZn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s)anode solution salt bridge solution cathode

Anode (oxidation) reaction: Zn(s)Zn2+(aq) + 2e–

Solid zinc (Zn(s)) oxidized to Zn2+(aq)Solid zinc is the physical electrode

Cathode (reduction) reaction: Cu2+(aq) + 2e–Cu(s) Cu2+(aq) is reduced to solid copper (Cu(s)) Solid copper is the physical electrode

Net Reaction: Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)

17

Standard Electrode PotentialsStandard Cell Potentials

Eocell = Eo

reduction - Eooxidation

Standard Hydrogen Electrode (SHE)2H+(aq) + 2e– H2(g)

Standard Electrode Potential, Eored

[H+] = 1.00 MPH2= 1.00atmT = 298KEo

red= 0.000VCalculate potential of oxidation half reactions(Eo

oxid)Eo

oxid= Eored - Eo

cell = 0.000V - Eocell = -Eo

cell

18

Spontaneous: Galvanic CellPt | H2(g) | H+(aq) || Cu2+(aq) | Cu(s)H2(g) + Cu2+(aq) 2H+(aq) + Cu(s)

Cell Potential:Eo

cell = Eored - Eo

oxidEo

cell = +.340VAnode Reaction:

H2(g) 2H+(aq) + 2e–

Eooxid = 0.000 V

Cathode Reaction: Cu2+(aq) + 2e– Cu(s)Eo

red = +.340V

+ Eocell; Reaction is spontaneous as written

(Galvanic)

19

Nonspontaneous: Electrolytic CellPt | H2(g) | H+(aq) || Zn2+(aq) | Zn(s)H2(g)+ Zn2+(aq)2H+(aq) + Zn(s)

Cell Potential:Eo

cell = Eored - Eo

oxid Eocell = -0.763V

Anode Reaction: H2(g) 2H+(aq) + 2e– Eo

oxid = 0.00 V

Cathode Reaction: Zn2+(aq) + 2e– Zn(s) Eo

red = -.763V

- Eocell: Reaction nonspontaneous (electrolytic)

20

Daniell cell: Cu2+(aq) + Zn(s)Cu(s) + Zn2+

(aq)

Cell ReactionsReduction: Cu2+(aq) + 2e–Cu(s) Oxidation: Zn(s) Zn2+(aq) + 2e–

From Reduction TablesCu2+ Eo

red = + 0.34 VZn2+ Eo

ox = – 0.76V

Cell potential, Eocell

Eocell= Eo

red - Eooxid = +0.34V - (-0.76 V)

Eocell= +1.10V

21

Calculating Cell Potentials

25 oC

22

For a Spontaneous ReactionReduction reaction is the more positive, higher in table

Oxidation reaction is lower in table and needs to be reversed

Redox Rules

23

1. E0 is for the reaction as written

2. The more positive E0 the greater the tendency for the substance to be reduced

3. Half-cell reactions are reversibleThe sign of E0 changes when the reaction is reversed

4. Changing stoichiometric coefficients of a half-cell reaction does not change E0

Intensive Property: Amount doesn’t matter

Will Br2(l) spontaneously oxidize Fe2+(aq)? If so, what are the net cell reaction and the

Eocell?

24

Figure out what is being asked: Oxidize: Cause to lose e-: Fe2+(aq) Fe3+(aq)

Write and balance chemical equation:2Fe2+(aq) + Br2(l) 2Fe3+(aq) + 2 Br– (aq)

Determine half-reactions and get Eo from table:Br2(l) + 2e– 2Br–(aq) reduction Eo = +1.07 VFe2+(aq) Fe3+(aq) + e– oxidation Fe3+(aq) + e– Fe2+(aq Eo = + 0.77 V from tableCell Potential

Eocell= Eo

reduction- Eooxidation= +1.07V - 0.77V = +0.30V

Eocell>0 Spontaneous reaction

Will I–(aq) reduce Cr3+(aq) to the free metal. If so, what are the net reaction and the Eo

cell?

25

Figure out what is being asked: Reduce: Cause something to gain e-: Cr3+(aq) + 3e- Cr(s)

Write and balance chemical equation:2Cr3+(aq) + 6I–(aq) 2Cr(s) + 3I2(s)

Cell ReactionsOxidation: 2I–(aq) I2(s) + 2e–

Table: I2(s) + 2e– 2I–(aq) Eo = +0.53 VReduction: Cr3+(aq) + 3e– Cr(s) Eo = -0.74 V

Cell PotentialEo

cell= Eoreduction - Eo

oxidation = (-0.74) - (+.53V) = -1.27VEo

cell <0 Non-spontaneous Reaction NR

26

Thermodynamics of Redox Reactions

Gibb’s Energy and Cell PotentialsElectrical work, w

w (joules)= Total cell charge (coulombs) x cell potential (V)Total cell charge = # mol electrons(n) x (coulombs/mol e-s)Faraday’s constant, F= 96485 Coulombs/1 mol e-s

Gibb's Energy:GG =wmax = –nFEcellAt standard conditions: Go = –nFEo

cell = -RTlnK(Voltaic cells are spontaneous, Ecell must be positive)

Cell potential, Ecell–nFEo

cell = -RTlnK so Eocell=( RT/nF)lnK

27

Summarizing the Important Relationships

28

ΔG° K E°cell Spontaneity Direction(-) >1 (+) Spontaneous More products

0 1 0 At equilibrium Products= reactants

(+) <1 (-) Nonspontaneous More reactants

Find Keq at 25 oC for the reaction; Sn(s) + 2Cu2+(aq) Sn2+(aq) + 2Cu+(aq)

29

Balance net reaction:Sn(s) + 2Cu2+(aq) Sn2+(aq) + 2Cu+(aq)

Cell ReactionsOxidation: Sn(s) Sn2+(aq) + 2e– Eo

ox= - 0.14VReduction : 2Cu2+(aq) + 2e- 2Cu+(aq) Eo

red= + 0.15 VLink Eo

cell and K and fill in constantsEo

cell = (RT/nF) lnKeq lnKeq= Eocell(nF/RT)

R = 8.314 J/mol K n = 2 Eocell= +0.29 V

T = 25°C = 298K F = 96485 coul/molSolve for Keq

ln Keq = 22.6 Keq = e22.6 = 7x109

30

Concentration and Cell Potential

The Nernst Equation:Cell Potentials at Nonstandard Conditions

31

From Thermodynamics:G = Go + RT ln Q

Linking Thermodynamics & ElectrochemistryG = –nFE and Go = –nFEo

Substitute into first equation to remove G :(–nF)E = (–nF)Eo + RT ln Q

Divide by –nF to get the Nernst Equation:E = Eo

cell - (RT/nF) ln QCan now change temperature and concentrations

Concentration Cells:Daniell Cell at Nonstandard Conditins

32

Electrochemical cell that does not have 1M solutions Find the EMF at the following conditions:

1. [Cu2+] = 1.00 M, [Zn2+] = 1.0×10–9 M 2. [Cu2+] = 0.10 M, [Zn2+] = 0.90 M

Daniell CellZn(s) + Cu2+(aq) Cu(s) + Zn2+(aq) Eocell = +1.10 V

Nernst EquationE = Eo

cell - (RT / nF) ln QQ = [Cu(s)] [Zn2+(aq)] / [Zn(s)] [Cu2+(aq)] = [Zn2+/Cu2+]

Plugging in from equationE = Eo

cell - (RT / nF) ln [Zn2+/Cu2+]

Concentration Cells: Daniell Cell at Nonstandard Conditions

33

Solve for E

Conditions of First Cell:[Cu2+] = 1.00 M, [Zn2+] =1.0×10–9 M

E = 1.37 VConditions of Second Cell

[Cu2+] = 0.10 M [Zn2+] = 0.90 ME = 1.07 V

]][ln

/96485*2298*/314.810.1 2

2

CuZn

molcoulmolKmolKJVE

Changing Zinc Concentration

34

After the initial drop, concentration has a minimal effect on voltage

35

Galvanic Cells:Batteries & Corrosion

The Dry Cell (Disposable Batteries)Portable electronic devices

Zn(s) | ZnCl2(aq), NH4Cl(aq) || Mn2O3(aq) | MnO2(s)|C(s)Oxidation: Zn Zn2+

(aq)+ 2e-Reduction: 2MnO2 + 2NH4

+ + 2e- Mn2O3 (s) + 2NH3 + H2OCell Reaction: Zn+2MnO2+ 2NH4

+ Zn2++Mn2O3+2NH3+ H2O

Cell VoltageCalculated:+ 1.36V Produced: + 1.5V

Irreversible ReactionCell is “dead” when reactants are used up

36

Button Batteries: Mercury BatteryPacemakers, hearing aids, watches

Zn | ZnO, OH- || HgO, OH- | Hg(s) | Fe(s)Oxidation: Zn + 2OH- ZnO + H2O +2e-Reduction: HgO + H2O + 2e- Hg + 2OH-Cell Reaction: Zn+ HgO + H2O ZnO + Hg

Cell VoltageConstant OH- composition More constant voltageReported voltage: + 1.5V

Irreversible ReactionCell is “dead” when reactants are used up

37

38

Lead Storage BatteryCar and boat batteries

Pb | H2SO4 (38%) || H2SO4 (38%) | PbO2Oxidation: Pb + SO4

2- PbSO4 +2e-Reduction: PbO2 + 4H++ SO4

2- + 2e- PbSO4 + 2H2OCell Reaction: Pb + PbO2 + 4H++ SO4

2- 2PbSO4 + 2H2OCell Voltage

Voltage: + 2.0VUsually 6 cells=12V

Reversible ReactionReaction reverses whenengine is running

Battery rechargesElectrolytic cell reaction

39

Fuel CellsGalvanic cells: Continually renew reactants

C-Ni (catalyst) | H2 | OH- (aq)|| OH- (aq)| O2 | C-Ni (catalyst)

Oxidation: 2H2 + 4OH- 4H2O +4e-Reduction: O2 + 2H2O + 4e- 4OH-

Cell Reaction: 2H2 + O2 2H2OCell Voltage

Voltage: + 1.23VNonreversible Reaction

Need fresh reactantsRemoval of productsNeed electrocatalysts

Corrosion: Deterioration of metal through an electrochemical process

Oxidation of metals: Rust, tarnish, “patina”Rust: Fe2O3•xH2O

2Fe + O2 + 4H+ 2Fe2+ + 2H2O E=1.67V4Fe2+ + O2+ (4+2x)H2O 2Fe2O3•xH2O + 8H+ E>0

Cell ReactionSpontaneous reactionsSalts increase rate

Cathodic protectionMore reactive metal protects (Zn)Reacts in place of protected metal (Fe)

40

41

Electrolysis and

Electrometallurgy

Electrolytic CellsNon-spontaneous processes driven by the application of an external power supply

Side reactions from solvent and dissolved ions

Determine reactions that are most spontaneous

Least Negative Ecell

42

A 1 M solution of potassium iodide is electrolyzed under acidic conditions.What are the products?

43

Half-Cell Potentials from tableK+(aq) + e–K(s) Eo = –2.92 V Reduction2I–(aq) I2(s) + 2e– Eo = - 0.54 V Oxidation 2H+(aq) + 2e– H2(g) Eo = 0.00 V Reduction2H2O(l) O2(g) + 4H+(aq) + 4e– Eo = -1.23 V Oxidation

Dissociation Reaction: KI(aq) K+(aq) + I–(aq)Resulting Products: K+, I-, H+, H2O

Cation Reduction: K+(aq) + e–K(s)Acid Reduction to H2 : 2H+(aq) + 2e– H2(g)Anion Oxidation: 2I–(aq) I2(s) + 2e–

H2O Oxidation to O2: 2H2O(l) O2(g) + 4H+(aq) + 4e–

A 1 M solution of potassium iodide is electrolyzed under acidic conditions.

What are the products?

44

Less Negative ReactionsReduction: 2H+(aq) + 2e– H2(g) Eo = 0.00 VOxidation: 2I–(aq) I2(s) + 2e– Eo = + 0.54 V

Standard PotentialEo = 0.00 - (+0.54) = – 0.54V

need at least .54V to start reaction

Net Reaction2H+(aq) + 2I–(aq) I2(s) + H2(g)

ProductsSolid Iodine:I2(s) and Hydrogen gas: H2(g)

Producing Products by ElectrolysisElectroplating

Coating one metal onto anotherSilver or gold over iron or steelCheaperProduct often more durable

Purification of copperImpure copper anodeMore reactive impurities oxidized (ions)Less reactive impurities drop to bottom

Gold, silver, etc. now separatedCopper built up on cathode 99.5% pure

45

Quantitative ElectrolysisUse Electrolytic Cells to find

stoichiometry Measure the charge passed through the cell

The current (amps, A) is the rate of charge flow1amp = 1 coulomb per second.

nF = Atn = number of moles of electrons F = Faraday's constant = 96485 C/molA = current in amps = 1 coulomb/sect = time in seconds

46

47

AxmA

AmAxA 2105.21000125 sec100.9

min1sec60min15 2xxt

Quantitative Electrolysis of Water: 2H2OO2 + 2H2

CxsCxAtnF 5.22sec)100.9)(/105.2( 22

2542 108.5103.2

41

2 Oelectronselectrons

OO molxmolxx

molmolemol

mLLKmolKLatmmolx

PnRTV O 4.10014.0)298)(0821.0)(108.5( 2

5

Water is electrolyzed in a cell at 25 mA for 15 minutes. How many mL of oxygen gas are produced at 1 atm & 25 °C?

electronsmolxmolC

CFCn 4103.2

/964855.22