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

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 = ( ) = V

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 K. 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) = V K = 1.6x 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?

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)

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)

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

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

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 = 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 = V

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( ) = (mV)x ln( ) = 56.3 mV

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

Applications of standard potentials

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.

Which metal is more Suitable for anode? Cathode?

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)

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 K. The mercury(I) ion is the diatomic species Hg 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 = V