1. Tutorial schedule (3:30 – 4:50 PM) No. 1 (Chapter 7: Chemical Equilibrium) January 31 at Biology building, room 113 February 1 at Dillion Hall, room 254 No. 2 (Chapter 22: The rates of chemical reactions) March 6 at Biology building, room 113 March 7 at Dillion Hall, room 254 No. 3 (Chapter 24: Molecular Reaction Dynamics) March 27 at Biology building, room 113 March 28 at Dillion Hall, room 361

2. Extended Debye-Hückel law B is an adjustable empirical parameter. It is different for each electrolyte.

3. Calculating parameter B Example : The mean activity coefficient of NaCl in a diluted aqueous solution at 25 o C is 0.907 (at 10.0 mmol kg -1 ). Estimate the value of B in the extended Debye-Huckel law. Solution: First calculate the ionic strength I = ½[1 2 *0.01 + (-1) 2 *0.01] = 0.01 From equation log(0.907) = - (0.509|1*(-1)|*0.01 1/2 )/(1+ B*0.01 1/2 ) B = - 1.67

4. Half-reactions and electrodes Two types of electrochemical cells: 1. Galvanic cell: is an electrochemical cell which produces electricity as a result of the spontaneous reactions occurring inside it. 2. Electrolytic cell: is an electrochemical cell in which a non-spontaneous reaction is driven by an external source of current.

5. Other important concepts include: Oxidation: the removal of electrons from a species. Reduction: the addition of electrons to a species. Redox reaction: a reaction in which there is a transfer of electrons from one species to another. Reducing agent: an electron donor in a redox reaction. Oxidizing agent: an electron acceptor in a redox reaction. Two type of electrodes: Anode: the electrode at which oxidation occurs. Cathode: the electrode at which reduction occurs

6. Typical Electrodes

7. Electrochemical cells Liquid junction potential: due to the difference in the concentrations of electrolytes. The right-hand side electrochemical cell is often expressed as follows: Zn(s)|ZnSO 4 (aq)||CuSO 4 (aq)|Cu(s) The cathode reaction (copper ions being reduced to copper metal) is shown on the right. The double bar (||) represents the salt bridge that separates the two beakers, and the anode reaction is shown on the left: zinc metal is oxidized into zinc ions

8. In the above cell, we can trace the movement of charge. –Electrons are produced at the anode as the zinc is oxidized –The electrons flow though a wire, which we can use for electrical energy –The electrons move to the cathode, where copper ions are reduced. –The right side beaker builds up negative charge. Cl - ions flow from the salt bridge into the zinc solution and K + ions flow into the copper solution to keep charge balanced. To write the half reaction for the above cell, Right-hand electrode: Cu 2+ (aq) + 2e - → Cu(s) Left-hand electrode: Zn 2+ (aq) + 2e - → Zn(s) The overall cell reaction can be obtained by subtracting left-hand reaction from the right-hand reaction: Cu 2+ (aq) + Zn(s) → Cu(s) + Zn 2+ (aq)

9. Expressing a reaction in terms of half-reactions Example : Express the formation of H 2 O from H 2 and O 2 in acidic solution as the difference of two reduction half- reactions. (in class discussion) Redox couple: the reduced and oxidized species in a half- reaction such as Cu 2+ /Cu, Zn 2+ /Zn…. Ox + v e - → Red The quotient is defined as: Q = a Red /a Ox Example: Write the half-reaction and the reaction quotient for a chlorine gas electrode. (in class discussion)

10. Varieties of cells

11. Notation of an electrochemical cell Phase boundaries are denoted by a vertical bar. A double vertical line, ||, denotes the interface that the junction potential has been eliminated. Start from the anode. A general format: Solid | gas phase | aqueous phase || aqueous phase | gas phase | solid

12. Concentration Cells M | M + (aq, L) || M + (aq, R) | M Cell reaction: M + (aq, R) → M + (aq, L) since Δ r G θ = 0 (why?)

13. Cell Potential Cell potential: the potential difference between two electrodes of a galvanic cell (measured in volts V). Maximum electrical work : w e,max = ΔG Electromotive force, E, Relationship between E and Δ r G: Δ r G = - ν FE where ν is the number of electrons that are exchanged during the balanced redox reaction and F is the Faraday constant, F = eN A. At standard conditions, this equation can be written as Δ r G θ = -νFE θ

14. The cell emf The emf of a cell can be calculated by the difference of the potentials of the two electrodes, E cell = E right - E left The potential of an electrode can be calculated from its standard potential. For example Fe +3 (aq) + e - → Fe +2 (aq) Cu +2 (aq) + 2e - → Cu(s) Consider: Ag(s)|Ag + (aq) || Cl - (aq) |AgCl(s)| Ag(s) E cell = E(AgCl/Ag, Cl - ) – E(Ag + /Ag) = E θ - ??

15. In the lead storage battery (used in automobiles), Pb | PbSO 4 | H 2 SO 4 | PbSO 4 |PbO 2 | Pb would the voltage change if you changed the concentration of H 2 SO 4 ? (yes/no) Answer... Yes, because The net cell reaction is Pb + PbO 2 + 2HSO 4 - + 2H + → 2 PbSO 4 + 2 H 2 O The Nernst equation E = E° - (0.0594/2)log{1/{[HSO 4 - ] 2 [H + ] 2 }}.

16. Previous two examples involve the same number of electron transfer at the cathode and anode, how to calculate emf if the number of electrons transferred at the two electrodes are different? Example: What is the emf for the cell : Mn(s)|Mn +2 ||Fe +3 |Fe +2 |Pt(s) Solution: The two reduction half reactions Right: Fe +3 (aq) + e - → Fe +2 (aq) Left: Mn +2 (aq) + 2e - → Mn(s) It shows that the above two half reactions have different number of electrons being transferred! The cell reaction is obtained via 2*R – L, 2Fe +3 (aq) + Mn(s) → 2Fe +2 (aq) + Mn +2 (aq) should the standard cell potential be calculated as 2*E ө (R) - E ө (L) ? Answer: NO! it is still calculated with E cell = E right - E left Consider: Δ r G θ = 2Δ r G θ (R) - Δ r G θ (L) 2FE θ = 2(1*F* E θ (R) – 2*F* E θ (L) it leads to E ө cell = E ө (R) - E ө (L) = 0.769 - (- 1.182) = 1.951 V

17. Standard Cell emf E θ cell = E θ (right) – E θ (Left) Calculating equilibrium constant from the standard emf : Evaluate the solubility constant of silver chloride, AgCl, from cell potential data at 298.15K. Solution: AgCl(s) → Ag + (aq) + Cl - (aq) Establish the electrode combination: Right: AgCl + e - → Ag(s) + Cl - (aq) E θ = 0.22V Left: Ag + (aq) + e - → Ag(s) E θ = 0.80V The standard cell emf is : E θ (right) – E θ (Left) = - 0.58V K = 1.6x10 -10 The above example demonstrates the usefulness of using two half reactions to represent a non redox process. What would be the two half reactions for the autoprotolysis of H 2 O?

18. The measurement of standard potentials The potential of standard hydrogen electrode: Pt(s)|H 2 (g)|H + (aq) is defined as 0 at all temperatures. The standard potential of other electrodes can be obtained by constructing an electrochemical cell, in which hydrogen electrode is employed as the left-hand electrode (i.e. anode) Example: the standard potential of the AgCl/Ag couple is the standard emf of the following cell: Pt(s)|H 2 (g)|H + (aq), Cl - (aq)|AgCl(s)|Ag(s) or Pt(s)|H 2 (g)|H + (aq) || Cl - (aq)|AgCl(s)|Ag(s) with the cell reaction is: ½ H 2 (g) + AgCl(s) → H + (aq) + Cl - (aq) + Ag(s)