1 / 25

Electrochemistry

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 (+).

shaeleigh-aguirre
Télécharger la présentation

Electrochemistry

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Electrochemistry

  2. Dr. M. Sasvári Chemistry Lectures Electrochemistry Galvanic Cells

  3. 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

  4. - - Cation Electrodes e- Cation electrodes with negative electrode potential are ready to loose electrons + e- - Zn Zn2+ + 2 e- red. form ox. form

  5. + + Cation Electrodes e- Cation electrodes with positive electrode potential are ready to take electrons + e- + + 2 e- Cu2+ Cu ox. form red. form

  6. The Hydrogen Electrode H+ (aq)  H2 Pt e0= 0

  7. 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 = (+)- (-)

  8. Def Calculation of Emf? Redox reactions Oxidation Reduction Oxidation potential (to loose e-) Reduction potential (to take e-) Emf = red.pot.(+)+ (- red. pot) (-)

  9. Galvanic Cell e- + + e- e- - +1 Zn Cu2+ Cu Zn2+ + + red. form ox. form red. form ox. form

  10. Daniell Cell: A Zinc-Copper Voltaic Cell

  11. Why do we need salt bridge? Daniell cell: Zn Cu2+ Cu Zn2+ + + SO42- SO42- • Transports counter ions • Closes the circuit

  12. 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

  13. 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

  14. Measuring Normal/standard electrode potentials of Copper

  15. 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

  16. 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

  17. 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

  18. 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

  19. 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)

  20. 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

  21. 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+

  22. 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

  23. 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 -

  24. 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)

  25. + e- Thank you for your attention!

More Related