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Lecture 2. Basic Electrochemistry

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  1. Lecture 2. Basic Electrochemistry • Basic Concepts • Electric Work and Gibb's Free Energy • Standard Cell Potential • Nernst Equation

  2. Basic Concepts Oxidation number/state: number of electrons gained or lost by an atom of that element when it forms compounds Oxidation: algebraic increase in oxidation number and loss of electrons, 4Fe + 3O22Fe2O3;H22H+ + 2e- Reduction: algebraic decrease in oxidation number and gain of electrons, WO3 + 3H2W + 3H2O; O2 +2H+ + 2e- H2O Redox: Oxidation-reduction reactions which involve electron transfer from one species to another. WO3 + 3H2W + 3H2O

  3. Basic Concepts Electrode: surface upon which oxidation or reduction half-reactions occur Electrolytic Cell: electrical energy from external sources causes nonspontaneous chemical reactions to occur (electrolysis) Voltaic/Galvanic Cell: spontaneous chemical reactions produce electricity and supply it to an external circuit (fuel cell) Electrolytes: substances whose aqueous solutions conduct electricity by the motion of ions, also called ionic conductor.

  4. Useful Units in Fuel Cells Voltage: Volt (V); 1V=1000mV Current: Ampere (A); 1A=1000mA Power: P=V x I, Watt(W); 1W=1000mW Charge: Coulomb (C), Faraday (F) 1F=96485 C 1e=1.602x10-19C 1F=N·e=(6.022x1023)x(1.6022x10-19)C 96485C (1F: charge of 1 mol electrons) 1A= 1C/s Energy: Joule (J) = V·I·t (J) Power: J/s=W; 1.0hp (Horsepower) = 746 W 1J = (1V)·(1A) ·(1s) = V·(C/s)·s = V·C = charge ·voltage

  5. Electrolysis of Aqueous NaCl Anode reaction (oxidation): 2Cl-Cl2 + 2e- Cathode reaction (reduction): 2H2O + 2e-2OH- +H2 Overall cell reaction: 2H2O + 2Cl-2OH- +H2+Cl2 + 2Na+  + 2Na+ 2NaCl 2NaOH

  6. Faraday’s Law of Electrolysis The amount of substance that undergoes oxidation or reduction at each electrode during electrolysis is directly proportional to the amount of electricity that passes through the cell. 1 mole of electrons = 96485 coulombs = 1 faraday electricity 1 ampere = 1 coulomb/second

  7. Voltaic/Galvanic Cell Components in Voltaic Cells Two half-cells Electric circuit Ionic conductor For Znic-Copper Cell Zn (s)+Cu2+(aq.)  Zn2+(aq.)+Cu (s)

  8. Electric Work and Gibb's Free Energy In a redox reaction, the energy released in a reaction due to movement of charged particles give rise to a potential difference. The maximum potential difference is called the electromotive force, (EMF), E and the maximum electric work W is the product of charge q in Coulomb (C), and the potential E in Volt ( J / C) or EMF. W (J) = q*E [C *J/C] E is determined by the nature of the reactants and electrolytes.

  9. Electric Work and Gibb's Free Energy The Gibb's free energy G is the negative value of maximum electric work, G = - W= - q E A redox reaction equation represents definite amounts of reactants in the formation of also definite amounts of products. The number (z) of electrons in such a reaction equation, is related to the amount of charge transferred when the reaction is completed.

  10. Electric Work and Gibb's Free Energy Since each mole of electron has a charge of 96485 C (known as the Faraday's constant, F), q = z F G = - z F E At standard conditions (298.15 K, 1 bar), G° = - z F E° E° = -G°/ (z F ) G° is the Gibbs free energy change of the redox reaction at standard conditions.

  11. Standard Cell Potential Every oxidation must be accompanied by a reduction. It is impossible to determine experimentally the potential of any single electrode. We establish an arbitrary standard by choosing the standard hydrogen electrode (SHE) as the reference electrode. The standard electrode potential (E0) for both oxidation and reduction of hydrogen is set to be zero (1M H+ solution). H22H+ + 2e- E0= 0.0000 V as anode 2H+ + 2e-  H2 E0= 0.0000 V as cathode

  12. Standard Cell Potential

  13. The Nernst Equation E° = -G°/ (z F ) The general Nernst equation correlates the Gibb's Free Energy G and the EMF of a chemical system known as the galvanic cell. For the reaction a A + b B = c C + d D and It has been shown in thermodynamics that G = G° + R T ln Q

  14. The Nernst Equation G = - z FE. - z F E = - z F E° + R T ln Q where R, T, Q and F are the gas constant (8.314 J/(molK), temperature (in K), reaction quotient, and Faraday constant (96487 C) respectively. Thus, we have This is known as the Nernst equation which allows us to calculate the cell potential of any galvanic cell for any concentrations.

  15. The Nernst Equation It is interesting to note the relationship between equilibrium and the Gibb's free energy at this point. When a system is at equilibrium, E = 0, and Qeq = K. Therefore, we have, Thus, the equilibrium constant and E° are related.

  16. Application of The Nernst Equation Example 1. Calculate the EMF of the cell Zn(s) | Zn2+(0.024 M) || Zn2+(2.4 M) | Zn(s) Solution: Zn2+ (2.4 M) + 2e- = Zn Reduction Zn = Zn 2+ (0.024 M) + 2e- Oxidation Zn 2+ (2.4 M) = Zn 2+ (0.024 M), E° = 0.00 - - Net reaction

  17. Application of The Nernst Equation

  18. Open Cell Voltage for Other FCs Example 2. Alkali Battery Anode Reaction: Zn + 2OH- ZnO + H2O + 2e- Cathode Reaction: 2MnO2 + H2O+ 2e- Mn2O3 + 2OH- 1J = (1V)·(1A) ·(1s) = V·(C/s)·s = V·C = charge ·voltage z=2, ΔE0 = 277000 (J/mol)/(2x96485 C/mol)=1.44 V

  19. Open Cell Voltage for Other FCs Example 2. Direct Methanol Fuel Cell Anode Reaction: CH3OH + H2O  6H+ + CO2 + 6e- Cathode Reaction: 1.5O2 + 6H++ 6e- 3H2O z=6, E = 698200/(6x96485)=1.21 V

  20. Open Cell Voltage for Other FCs Example 3. Alkaline Methanol Fuel Cell Anode Reaction: CH3OH + 6OH- 5H2O + CO2 + 6e- Cathode Reaction: 1.5O2 + 3H2O + 6e- 6OH- z=6, E = 698200/(6x96485)=1.21 V

  21. Open Cell Voltage for Other FCs Example 4. Alkaline Electrolyte Fuel Cells H2 + 0.5 O2 H2O Anode Reaction: H2 + 2OH- 2H2O + 2e- Cathode Reaction: 0.5O2 + H2O + 2e- 2OH- z=2, E = 220400/(2x96485)=1.14 V

  22. Open Cell Voltage for Other FCs Example 5. Molten Carbon Fuel Cells (400 C) H2 + 0.5 O2 +CO2 (C) H2O + CO2(A) Anode Reaction: H2 + CO3= H2O + CO2 + 2e- Cathode Reaction: 0.5O2 + CO2 + 2e- CO3= z=2, E = 210300/(2x96485)=1.11 V

  23. Open Cell Voltage for Other FCs Example 6. Solid Oxide Fuel Cells (650C) CO + 0.5 O2  CO2 Anode Reaction: CO +O= CO2 + 2e- Cathode Reaction: 0.5O2 + 2e- O= z=2, E = 19700/(2x96485)=1.02 V