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

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 2Eright - Eleft ? Answer: ??? ΔrG = 2ΔrG(R) - ΔrG(L) 2FE = 2*1*F* E(R) – 2*F* E(L) it leads to Ecell = E(R) - E(L) For the standard cell potential: Eөcell = 0.769 - (- 1.182) = 1.951 V

3. Standard Cell emf • Eθcell = Eθ (right) – Eθ(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) Eθ = 0.22V (Obtain this one through C – R) Left: Ag+(aq) + e- → Ag(s) Eθ = 0.80V The standard cell emf is : Eθ (right) – Eθ(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 H2O?

4. The measurement of standard potentials • The potential of standard hydrogen electrode: Pt(s)|H2(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)|H2(g)|H+(aq), Cl-(aq)|AgCl(s)|Ag(s) or Pt(s)|H2(g)|H+(aq) || Cl-(aq)|AgCl(s)|Ag(s) with the cell reaction is: ½ H2(g) + AgCl(s) → H+(aq) + Cl-(aq) + Ag(s)

5. The measurement of standard potentials • The potential of standard hydrogen electrode: Pt(s)|H2(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)|H2(g)|H+(aq), Cl-(aq)|AgCl(s)|Ag(s) or Pt(s)|H2(g)|H+(aq) || Cl-(aq)|AgCl(s)|Ag(s) with the cell reaction is: ½ H2(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(b2) - (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 b1/2, one gets E + (2RT/vF)ln(b) = E θ - C* b1/2, C is a constant Therefore the plot of E + (2RT/vF)ln(b) against b1/2 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) + Cl2(g) → MgCl2(aq) Give the half-reactions for the electrodes and from the standard cell emf of 3.00V deduce the standard potential of the Mg2+/Mg couple. Solution: the above reaction indicates that Cl2 gas is reduced and Mg is oxidized. Therefore, R: Cl2(g) + 2e- → 2Cl-(aq) (Eө = + 1.36 from Table 10.7) L: Mg2+(aq) + 2e- → Mg(s) (Eө = ? ) The cell which corresponds to the above two half-reactions is : Mg(s)|MgCl2(aq)|Cl2(g)|Pt Eөcell = Eө(R) - Eө(L) = 1.36 – Eө(Mg2+/Mg) Eө(Mg2+/Mg) = 1.36V – 3.00V = - 1.64V

9. Example: Consider a hydrogen electrode in aqueous HCl solution at 25oC 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 H2(g) Based on Nernst equation So E2 – E1 = - ln( ) = - 25.7(mV)x ln( ) = 56.3 mV

10. Choose the correct Nernst equation for the cell Zn(s) | Zn2+ || Cu2+ | Cu(s). A: ΔE = ΔE° - 0.032 log([Zn2+ ]/[Cu2+]) B: ΔE = ΔE° + 0.024 log([Cu2+] / [Zn2+]) C: ΔE = ΔE° - 0.021 log(Zn / Cu) D: ΔE = ΔE° - 0.018 log(Cu / Zn) • Answer ... Hint... • The cell as written hasReduction on the Right: Cu2+ + 2e = Cuoxidation on the left: Zn = Zn2+ + 2e • Net reaction of cell is Zn(s) + Cu2+ = Cu(s) + Zn2+

11. Applications of standard potentials

12. The electrochemical series • For two redox couples Ox1/Red1 and Ox2/Red2, Red1, Ox1 || Red2, Ox2 Eθ = Eθ2 – Eθ1 The cell reaction: Red1 + Ox2→ Ox1 + Red2 If Eθ > 0 then ΔrGθ< 0 (Nernst equation), the reaction will take place spontaneously (??). In other words, if Eθ2 > Eθ1, 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 (Eθ) 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: H2(g) + Cl2(g) → 2HCl(aq)

15. The determination of equilibrium constants • Self-test 7.11 Calculate the solubility constant (the equilibrium constant for reaction Hg2Cl2(s) ↔ Hg22+(aq) + 2Cl-(aq)) and the solubility of mercury(I) chloride at 298.15K. The mercury(I) ion is the diatomic species Hg22+. • Answer: This chemical process does not involve electron transfer, i.e. is not a redox reaction. Choosing cathode reaction as: Hg2Cl2(s) + 2e → 2Hg(l) + 2Cl-(aq) from reference table 7.2, Eθ = 0.27 V the anode reaction can be obtained through R – Cell Hg22+(aq) + 2e → 2Hg(l) from reference table 7.2, Eθ = 0.79 V Therefore the standard cell potential = 0.27 – 0.79 = -0.52 V