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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 (+).

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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 (+)

e.g. Cu (s) / Cu 2+

0 = + 0.34 V

e.g. Zn (s) / Zn 2+

0 = - 0.76 V


-

-

Cation Electrodes

e-

Cation electrodes

with

negative

electrode potential

are ready

to loose electrons

+

e-

-

Zn

Zn2+

+ 2 e-

red. form

ox. form


+

+

Cation Electrodes

e-

Cation electrodes

with

positive

electrode potential

are ready

to take electrons

+

e-

+

+ 2 e-

Cu2+

Cu

ox. form

red. form


The Hydrogen Electrode

H+ (aq)  H2 Pt

e0= 0


Voltaic (Galvanic) Cells

The Electromotive Force.

Two electrodes are connected: Redox reaction occurs

If separated in space: Generation of electric current

Electromotive force (Emf):

The difference of two electrode potentials

Emf = (+)- (-)


Def

Calculation of Emf?

Redox reactions

Oxidation

Reduction

Oxidation potential

(to loose e-)

Reduction potential

(to take e-)

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


Galvanic Cell

e-

+

+

e-

e-

-

+1

Zn

Cu2+

Cu

Zn2+

+

+

red. form

ox. form

red. form

ox. form



Why do we need salt bridge?

Daniell cell:

Zn

Cu2+

Cu

Zn2+

+

+

SO42-

SO42-

  • Transports counter ions

  • Closes the circuit


Zn

Normal/Standard Electrode Potentials

H2(g) / H+ (1M)

cathode (+)

Zn (s) / Zn2+ (1M)

anode (-)

Measured Emf = 0.76

e0 = - 0.76 V

oxidation

Zn Zn2+ + 2e-

e0 = 0 V

reduction

H+ + e- 1/2 H2


Cu

Normal/standard electrode potentials

H2(g) / H+ (1M)

anode (-)

Cu (s) / Cu2+ (1M)

cathode (+)

Measured Emf = 0.34

e0 = 0 V

oxidation

1/2 H2 H+ + e-

e0 = + 0.34 V

reduction

Cu2+ + 2e- Cu


Measuring Normal/standard

electrode potentials of Copper


Normal electrode potential:

  • A comparison to Normal H electrode

  • Concentrations are 1 M (1 activity)

  • pH = 0, Temp= 0oC

Standard electrode potential:

  • pH = 0, Temp= 25oC

Biochemistry: pH = 7 and 37oC


Electromotive Force (Emf)

and the Gibbs free energy change of a reaction

G= max. useful work

W= Emf Q = Emf n F

where

G: Gibbs free energy change

  • Where

  • Emf: Voltage difference (V)

  • Q: Electric charge (Cb)

  • n= number of electrons

  • F= Faraday number

G0 = - n F Emf 0


Calculating G0 from Emf

e.g. Daniell cell: Emf = 1.1 V

G0 =-2 x 96500 x 1.1 (J)

Calculating Keq from G0

G0 = - RT lnKeq = - 2.3RT log Keq

Calculating 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


The Nernst Equation

Dependence of Emf on the concentrations:

G = G0 + 2.3 RT log Q

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

Nernst equation:

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

where Q=cproducts/creactants


e-

Concentration dependence of the electrode potential

: Reduction potential:

ox. form + e-

reactant

red. form

product

Nernst equation for half cells:

=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)


Li Li+ + e -

1/2 I2 + e-I -

Voltaic cells: Examples

A solid-state lithium battery

implanted within the chest to power heart pacemakers

Lasts about 10 years

If discharged: has come to equilibrium

LiI crystals

Anode (-) :

Li/Li+

Cathode (+) :

I-/I2 complex


Pb4+ + 2e-Pb2+

Pb Pb2+ + 2e-

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

Pb(s) + H2SO4PbSO4(s) + 2H+

Rechargeable batteries: Lead storage cell

Anode (-) :

Pb/Pb2+

Cathode (+) :

Pb4+/Pb2+


Fuel cells (see: Ebbing)

e.g. supplying space shuttle orbiters by electricity

2H2 + O2 = 2H2O

A Hydrogen-Oxygen fuel cell:

The galvanic cell:

Anode (-)

Ox.

2H2(g)+4 OH- 4H2O + 4e-

e0= -0.42

Chatode (+)

Red.

O2(g) + 2 H2O + 4e- 4 OH-

e0= +1.23

Electromotive force:

Eme0= +1.23 - (-0.42)=+1.65 V

(25 degree, 1 atm, pH 7)

200oC, 20-40atm, alkalic pH :

Emf is much higher


Rusting of iron is an electrochemical process

Fe Fe2+ + 2e-

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

A single drop of water on the iron

forms a galvanic cell with the oxygen of the air

Anode (-): Fe/Fe2+

Cathode (+) : O2/OH -


Cathodic protection of a buried steel pipe

Mg2+ + 2e- Mg

e0 = - 2,38 V

Fe2+ + 2e- Fe

e0 = - 0,41 V

-

O2(g) + 2H2O(l)+ 4 e- 4OH-(l)

e0 = + 1,23 V

+

Redox reaction:

+2

2Mg+ O2+2H2O  2Mg(OH)2

(see: Ebbing)


+

e-

Thank you for your attention!


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