1. Midterm Exam 1: Feb. 2, 1:00- 2:10 PM at Toldo building, Room 100

2. Example: Calculate emf of the cell : Mn(s)|Mn +2 ||Fe +3, Fe +2 |Pt(s) Solution: The two reduction half reactions Right (cathode): Fe +3 (aq) + e - → Fe +2 (aq) Left (anode): Mn +2 (aq) + 2e - → Mn(s) It shows that the above two half reactions have different numbers of electrons transferred The cell reaction is obtained through 2*R – L, 2Fe +3 (aq) + Mn(s) → 2Fe +2 (aq) + Mn +2 (aq) should the cell potential be calculated as 2E right - E left ? Answer: ??? Δ r G = 2Δ r G (R) - Δ r G (L) 2FE = 2*1*F* E (R) – 2*F* E (L) it leads to E cell = E (R) - E (L) For the standard cell potential: E ө cell = 0.769 - (- 1.182) = 1.951 V

3. 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: (Assume this one) Right: AgCl + e - → Ag(s) + Cl - (aq) E θ = 0.22V (Obtain this one through C – R) 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?

4. 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)

5. 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)

6. The Nernst equation of the above cell reaction is Using the molality and the activity coefficient to represent the activity: E = E θ – (RT/vF)ln(b 2 ) - (RT/vF)ln(γ ± 2 ) Reorganized the above equation: E + (2RT/vF)ln(b) = E θ - (2RT/vF)ln(γ ± ) Since ln(γ ± ) is proportional to b 1/2, one gets E + (2RT/vF)ln(b) = E θ - C* b 1/2, C is a constant Therefore the plot of E + (2RT/vF)ln(b) against b 1/2 will yield a straight line with the interception that corresponds to E θ

7. Example plot from the text book (the interception corresponds to E θ )

8. Example: Devise a cell in which the cell reaction is Mg(s) + Cl 2 (g) → MgCl 2 (aq) Give the half-reactions for the electrodes and from the standard cell emf of 3.00V deduce the standard potential of the Mg 2+ /Mg couple. Solution:the above reaction indicates that Cl 2 gas is reduced and Mg is oxidized. Therefore, R: Cl 2 (g) + 2e - → 2Cl - (aq) (E ө = + 1.36 from Table 10.7) L: Mg 2+ (aq) + 2e - → Mg(s) (E ө = ? ) The cell which corresponds to the above two half-reactions is : Mg(s)|MgCl 2 (aq)|Cl 2 (g)|Pt E ө cell = E ө (R) - E ө (L) = 1.36 – E ө (Mg 2+ /Mg) E ө (Mg 2+ /Mg) = 1.36V – 3.00V = - 1.64V

9. Example: Consider a hydrogen electrode in aqueous HCl solution at 25 o C operating at 105kPa. Calculate the change in the electrode potential when the molality of the acid is changed from 5.0 mmol kg -1 to 50 mmol kg -1. Activity coefficient can be found from Atkin’s textbook (Table 10.5 ). Solution: first, write down the half reaction equation: H + (aq) + e - → ½ H 2 (g) Based on Nernst equation So E 2 – E 1 = - ln( ) = - 25.7(mV)x ln( ) = 56.3 mV

10. Choose the correct Nernst equation for the cell Zn(s) | Zn 2+ || Cu 2+ | Cu(s). A: Δ E = Δ E° - 0.032 log([Zn 2+ ]/[Cu 2+ ]) B: Δ E = Δ E° + 0.024 log([Cu 2+ ] / [Zn 2+ ]) C: Δ E = Δ E° - 0.021 log(Zn / Cu) D: Δ E = Δ E° - 0.018 log(Cu / Zn) Answer... Hint... The cell as written has Reduction on the Right: Cu 2+ + 2e = Cu oxidation on the left: Zn = Zn 2+ + 2e Net reaction of cell is Zn(s) + Cu 2+ = Cu(s) + Zn 2+

11. Applications of standard potentials

12. The electrochemical series For two redox couples Ox 1 /Red 1 and Ox 2 /Red 2, Red 1, Ox 1 || Red 2, Ox 2 E θ = E θ 2 – E θ 1 The cell reaction: Red 1 + Ox 2 → Ox 1 + Red 2 If E θ > 0 then Δ r G θ E θ 1, the Ox2 has the thermodynamic tendency to oxidize Red1.

13. Which metal is more Suitable for anode? Cathode?

14. The determination of activity coefficients Once the standard potential of an electrode (E θ ) is known, one can use the following equation E + (2RT/vF)ln(b) = E θ - (2RT/vF)ln(γ ± ) to determine the mean activity coefficient of the ions at the concentration of interest via measuring the cell emf (E). For example: H 2 (g) + Cl 2 (g) → 2HCl(aq)

15. The determination of equilibrium constants Self-test 7.11 Calculate the solubility constant (the equilibrium constant for reaction Hg 2 Cl 2 (s) ↔ Hg 2 2+ (aq) + 2Cl - (aq)) and the solubility of mercury(I) chloride at 298.15K. The mercury(I) ion is the diatomic species Hg 2 2+. Answer: This chemical process does not involve electron transfer, i.e. is not a redox reaction. Choosing cathode reaction as: Hg 2 Cl 2 (s) + 2e → 2Hg( l ) + 2Cl - (aq) from reference table 7.2, E θ = 0.27 V the anode reaction can be obtained through R – Cell Hg 2 2+ (aq) + 2e → 2Hg( l ) from reference table 7.2, E θ = 0.79 V Therefore the standard cell potential = 0.27 – 0.79 = -0.52 V