1. http://mutuslab.cs.uwindsor.ca/Wang/59- 241/index.htmhttp://mutuslab.cs.uwindsor.ca/Wang/59- 241/index.htm On CLEW website, click Syllabus Midterm Exam: Students whose family name starts with letter A, B, C, or D will write exam in DH354.

2. The mean activity coefficient γ  is calculated as (γ + γ - ) 1/2 The chemical potential for individual ions M + and X - then becomes: μ + = μ ideal + RTln(γ  ) μ - = μ ideal + RTln(γ  ) For a general compound of the following form: M p X q, the mean activity coefficient is expressed as: γ  = [(γ + ) p (γ - ) q ] 1/s with s = p+q Debye-Hückel limiting law is employed to calculate the mean activity coefficient: log(γ  ) = -|z + z - | A I 1/2 A = 0.509 for aqueous solution at 25 o C. I is the ionic strength, which is calculated as the following: I = ½  z i 2 (b i /b ө ) where z i is the charge number of the ion j and b i is the molarlity of ion j.

3. Example: Relate the ionic strength of (a) MgCl 2, (b) Fe 2 (SO 4 ) 3 solutions to their molalities, b. Solution: Assuming the molality of the solution is b (a) MgCl 2 : From its molecular formula, we can get b(Mg 2+ ) = b; b(Cl - ) = 2*b; Z(Mg 2+ ) = +2; Z(Cl - ) = -1; So I = ½((2) 2 *b +(-1) 2 *(2b)) = ½(4b + 2b) = 3b; (b) Fe 2 (SO 4 ) 3 : From the molecular formula, we can get b(Fe 3+ ) = 2*b; b(SO 4 2- ) = 3*b; Z(Fe 3+ ) = +3; Z(SO 4 2- ) = -2; So I = ½((3) 2 *(2b) +(-2) 2 *(3b)) = ½(18b + 12b) = 15b

4. The ionic strength of the solution equals the sum of the ionic strength of each individual compound. Example: Calculate the ionic strength of a solution that contains 0.050 mol kg -1 K 3 [Fe(CN) 6 ](aq), 0.040 mol kg -1 NaCl(aq), and 0.03 mol kg -1 Ce(SO 4 ) 2 (aq). Solution: I(K 3 [Fe(CN) 6 ]) = ½( 1 2 *(0.05*3) + (3) 2 *0.05 + (-1) 2 *(0.05*6)) = 0.45; I(NaCl) = ½(1 2 *0.04 + (-1) 2 *0.04) = 0.04; I(Ce(SO 4 ) 2 ) = ½(4 2 *0.03 + (-2) 2 *(2*0.03)) =0.36; So, I = I(K 3 (Fe(CN) 6 ]) + I(NaCl) + I(Ce(SO 4 ) 2 ) = 0.45 + 0.04 + 0.36 = 0.85 Solution that contains more than one types of electrolytes

5. Calculating the mean activity coefficient Example: Calculate the ionic strength and the mean activity coefficient of 2.0 mmol kg -1 Ca(NO 3 ) 2 at 25 o C. Solution: ionic strength I I = ½(2 2 *0.002 + (-1) 2 *(2*0.002)) = 3*0.002 = 0.006; plug I into the Debye-Huckel limiting equation log(γ ± ) = - |2*1|*A*(0.006) 1/2 ; = - 2*0.509*0.0775; = -0.0789; γ ± = 0.834;

6. Experimental test of the Debye-Hückel limiting law

7. Accuracy of the Debye-Hückel limiting law Example: The mean activity coefficient in a 0.100 mol kg -1 MnCl 2 (aq) solution is 0.47 at 25 o C. What is the percentage error in the value predicted by the Debye-Huckel limiting law? Solution: First, calculate the ionic strength I = ½(2 2 *0.1 + 1 2 *(2*0.1)) = 0.3 to calculate the mean activity coefficient. log(γ) = -|2*1|A*(0.3) 1/2 ; = - 2*0.509*0.5477 = - 0.5576 so γ = 0.277 Error = (0.47-0.277)/0.47 * 100% = 41%

8. Extended Debye-Hückel law B is an adjustable empirical parameter.

9. Calculating parameter B Example : The mean activity coefficient of NaCl in a diluted aqueous solution at 25 o C is 0.907 (at 10.0mmol 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

10. 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.

11. 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

12. Typical Electrodes

13. 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

14. 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)

15. 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. 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.

16. Varieties of cells

17. 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.

18. 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 = -νFE 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 concentrations at 25 o C, this equation can be written as Δ r G θ = -νFE θ