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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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Presentation Transcript
slide2
Dr. M. Sasvári

Chemistry Lectures

Electrochemistry

Galvanic Cells

slide3
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

slide4
-

-

Cation Electrodes

e-

Cation electrodes

with

negative

electrode potential

are ready

to loose electrons

+

e-

-

Zn

Zn2+

+ 2 e-

red. form

ox. form

slide5
+

+

Cation Electrodes

e-

Cation electrodes

with

positive

electrode potential

are ready

to take electrons

+

e-

+

+ 2 e-

Cu2+

Cu

ox. form

red. form

slide6
The Hydrogen Electrode

H+ (aq)  H2 Pt

e0= 0

slide7
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 = (+)- (-)

slide8
Def

Calculation of Emf?

Redox reactions

Oxidation

Reduction

Oxidation potential

(to loose e-)

Reduction potential

(to take e-)

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

slide9
Galvanic Cell

e-

+

+

e-

e-

-

+1

Zn

Cu2+

Cu

Zn2+

+

+

red. form

ox. form

red. form

ox. form

slide11
Why do we need salt bridge?

Daniell cell:

Zn

Cu2+

Cu

Zn2+

+

+

SO42-

SO42-

  • Transports counter ions
  • Closes the circuit
slide12
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

slide13
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

slide14
Measuring Normal/standard

electrode potentials of Copper

slide15
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

slide16
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

slide17
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

slide18
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

slide19
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)

slide20
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

slide21
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+

slide22
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

slide23
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 -

slide24
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)

slide25
+

e-

Thank you for your attention!

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