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24 1 Electrochemistry
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Electrochemistry

Jan 02, 2016

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Electrochemistry. Dr. M. Sasvári. Chemistry Lectures. Electrochemistry Galvanic Cells. Electrode Potential. Half cell reactions (Electrode potential). electrons are lost oxidation (reducing agent). electrons are taken reduction (oxidizing agent). Anode (-). Cathode (+). - PowerPoint PPT Presentation
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Page 1: Electrochemistry

24 1

Electrochemistry

Page 2: Electrochemistry

24 2

ElectrochemistryGalvanic Cells

ElectrochemistryGalvanic Cells

Chemistry Lectures Chemistry Lectures

Dr. M. Sasvári

Page 3: Electrochemistry

24 3

Electrode Potential Electrode Potential

Half cell reactions (Electrode potential)Half cell reactions (Electrode potential)

electrons are lostoxidation

(reducing agent)

Anode (-)Anode (-)

e.g. Zn (s) / Zn 2+ 0 = - 0.76 V

e.g. Cu (s) / Cu 2+ 0 = + 0.34 V

electrons are takenreduction

(oxidizing agent)

Cathode (+)Cathode (+)

Page 4: Electrochemistry

24 4

Cation Electrodes

+

Cation electrodeswith

negativeelectrode potential

are ready to loose electrons

-Zn2+Zn + 2 e-

red. form ox. form

--e-

e-

Page 5: Electrochemistry

24 5

Cation Electrodes

+

Cation electrodeswith

positiveelectrode potential

are ready to take electrons

+Cu2+ Cu+ 2 e-

red. formox. form

++e-

e-

Page 6: Electrochemistry

24 6

The Hydrogen ElectrodeThe Hydrogen Electrode

H+ (aq) H2 PtH+ (aq) H2 Pt

0 = 00 = 0

Page 7: Electrochemistry

24 7

Voltaic (Galvanic) CellsThe Electromotive Force.Voltaic (Galvanic) Cells

The Electromotive Force.

Two electrodes are connected: Redox reaction occurs

Electromotive force (Emf):

Emf = +)--)

The difference of two electrode potentials

If separated in space: Generation of electric current

Page 8: Electrochemistry

24 8

Calculation of Emf?

Redox reactions

Oxidation Reduction

Oxidation potential(to loose e-)

Reduction potential(to take e-)

Def

Emf = red.pot.(+) + (- red. pot) (-)

Page 9: Electrochemistry

24 9

e-

Galvanic Cell

+

Cured. form

+Cu2+

ox. form

+

Znred. form

Zn2+

ox. form

e-

-

+

+1

e-

Page 10: Electrochemistry

24 10

Daniell Cell: A Zinc-Copper Voltaic CellDaniell Cell: A Zinc-Copper Voltaic Cell

Page 11: Electrochemistry

24 11

Why do we need salt bridge?

Daniell cell:

Cu +Cu2+ Zn2++ZnSO4

2- SO42-

•Transports counter ions•Closes the circuit

Page 12: Electrochemistry

24 12

Normal/Standard Electrode PotentialsNormal/Standard Electrode Potentials

Zn (s) / Zn2+ (1M)anode (-)

Zn

H2(g) / H+ (1M)cathode (+)

Measured Emf = 0.76Measured Emf = 0.76

0 = - 0.76 Voxidation

Zn Zn2+ + 2e-

0 = 0 Vreduction

H+ + e- 1/2 H2

Page 13: Electrochemistry

24 13

Normal/standard electrode potentialsNormal/standard electrode potentials

Cu (s) / Cu2+ (1M)

cathode (+)

Cu

H2(g) / H+ (1M)anode (-)

Measured Emf = 0.34Measured Emf = 0.34

0 = + 0.34 Vreduction

Cu2+ + 2e- Cu

0 = 0 Voxidation

1/2 H2 H+ + e-

Page 14: Electrochemistry

24 14

Measuring Normal/standard electrode potentials of Copper

Measuring Normal/standard electrode potentials of Copper

Page 15: Electrochemistry

24 15

• A comparison to Normal H electrode• Concentrations are 1 M (1 activity)• pH = 0, Temp= 0oC

Normal electrode potential:Normal electrode potential:

Standard electrode potential:Standard electrode potential:

• pH = 0, Temp= 25oC

Biochemistry: pH = 7 and 37oC

Page 16: Electrochemistry

24 16

Electromotive Force (Emf) and the Gibbs free energy change of a reaction

Electromotive Force (Emf) and the Gibbs free energy change of a reaction

G= max. useful work

where G: Gibbs free energy change

Where• Emf: Voltage difference (V)• Q: Electric charge (Cb)• n= number of electrons• F= Faraday number

W= Emf Q = Emf n F

G0 = - n F Emf 0G0 = - n F Emf 0

Page 17: Electrochemistry

24 17

Calculating G0 from EmfCalculating G0 from Emf

Calculating Keq from G0Calculating Keq from G0

G0 = - RT lnKeq = - 2.3RT log Keq

Calculating the Keq from EmfCalculating the Keq from Emf

- n F Emf0 = - 2.3RT log Keq

Emf0 =(2.3 RT/nF)log Keq

at 25 degree: Emf0 = (0.059/n)log Keq

e.g. Daniell cell: Emf = 1.1 V G0 = -2 x 96500 x 1.1 (J)

Page 18: Electrochemistry

24 18

The Nernst EquationThe Nernst Equation

Dependence of Emf on the concentrations:Dependence of Emf on the concentrations:

G = G0 + 2.3 RT log Q-nF Emf = -nF Emf0 + 2.3 RT log Q

Emf = Emf0 - (2.3 RT/nF) log Q

Nernst equation:Nernst equation:

where Q=cproducts/creactants

Page 19: Electrochemistry

24 19

Concentration dependence of the electrode potentialConcentration dependence of the electrode potential

Nernst equation for half cells:Nernst equation for half cells:

: Reduction potential:: Reduction potential:

ox. form + e-

reactantred. formproduct

0 - (2.3 RT/nF) log (cred/cox) 0 - (0.059/n) log (cred/cox) =0 + (0.059/n) log (cox/cred) 0 + (0.06/n) log (cox/cred)

e-

Page 20: Electrochemistry

24 20

Voltaic cells: ExamplesVoltaic cells: Examples

A solid-state lithium battery implanted within the chest to power heart pacemakersLasts about 10 yearsIf discharged: has come to equilibrium

Anode (-) : Li/Li+

Cathode (+) : I-/I2 complex

LiI crystals

Li Li+ + e - 1/2 I2 + e- I -

Page 21: Electrochemistry

24 21

Rechargeable batteries: Lead storage cellRechargeable batteries: Lead storage cell

Anode (-) : Pb/Pb2+

Cathode (+) : Pb4+/Pb2+

Pb Pb2+ + 2e- Pb4+ + 2e- Pb2+

Pb(s) + H2SO4 PbSO4(s) + 2H+ PbO2(s) + 4H+ PbSO4(s) + 2H2O

Page 22: Electrochemistry

24 22

Fuel cells (see: Ebbing)

e.g. supplying space shuttle orbiters by electricity

2H2 + O2 = 2H2OA Hydrogen-Oxygen fuel cell:

The galvanic cell:

2H2(g)+4 OH- 4H2O + 4e-

0= -0.42

Ox.Anode (-)

Chatode (+)

O2(g) + 2 H2O + 4e- 4 OH-

0= +1.23

Red.

Eme0= +1.23 - (-0.42)=+1.65 V (25 degree, 1 atm, pH 7)

Electromotive force:

200oC, 20-40 atm, alkalic pH : Emf is much higher

Page 23: Electrochemistry

24 23

Rusting of iron is an electrochemical processRusting of iron is an electrochemical process

Anode (-): Fe/Fe2+ Cathode (+) : O2/OH -

Fe Fe2+ + 2e- 1/2 O2+H2O + 2e- + 2OH-

A single drop of water on the ironforms a galvanic cell with the oxygen of the air

Page 24: Electrochemistry

24 24

Cathodic protection of a buried steel pipe

Mg2+ + 2e- Mg 0 = - 2,38 V

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

0 = + 1,23 V

-

+Redox reaction:

2Mg+ O2+2H2O 2Mg(OH)2

+2

Fe2+ + 2e- Fe 0 = - 0,41 V

(see: Ebbing)

Page 25: Electrochemistry

24 25

+e-