Midterm exam location Odette Building, room 104 Feb. 3rd from 1 - 2:20 p.m. March 10th from 1 - 2:20 p.m

1. Midterm exam location Odette Building, room 104 Feb. 3rd from 1 - 2:20 p.m. March 10th from 1 - 2:20 p.m

2. Solutions contain more than one types of electrolytes • 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 K3[Fe(CN)6](aq), 0.040 mol kg-1 NaCl(aq), and 0.03 mol kg-1 Ce(SO4)2 (aq). Solution: I (K3[Fe(CN)6]) = ½( 12*(0.05*3) + (-3)2*0.05) = ½ (0.15 + 0.45) = 0.3; I (NaCl) = ½(12*0.04 + (-1)2*0.04) = 0.04; I (Ce(SO4)2) = ½(42*0.03 + (-2)2*(2*0.03)) =0.36; So, I = I(K3(Fe(CN)6]) + I(NaCl) + I(Ce(SO4)2) = 0.3 + 0.04 + 0.36 = 0.7

3. Calculating the mean activity coefficient Example: Calculate the ionic strength and the mean activity coefficient of 2.0m mol kg-1 Ca(NO3)2 at 25 oC. Solution: In order to calculate the mean activity coefficient with the eq. 10.3, one needs to know the ionic strength of the solution. Thus, the right approach is first to get I and then plug I into the equation 10.3. I = ½(22*0.002 + (-1)2*(2*0.002)) = ½ *6*0.002 = 0.006; From Debye-Huckel limiting equation, log(γ±) = - |2*1|*A*(0.006)1/2; = - 2*0.509*0.0775; = -0.0789; γ± = 0.834;

4. Experimental test of the Debye-Hückel limiting law

5. Accuracy of the Debye-Hückel limiting law Example: The mean activity coefficient in a 0.100 mol kg-1 MnCl2(aq) solution is 0.47 at 25oC. What is the percentage error in the value predicted by the Debye-Huckel limiting law? Solution: First, calculate the ionic strength I = ½[22*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%

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

7. Calculating parameter B Example : The mean activity coefficient of NaCl in a diluted aqueous solution at 25oC 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[12*0.01 + (-1)2*0.01] = 0.01 From equation log(0.907) = - (0.509|1*(-1)|*0.011/2)/(1+ B*0.011/2) B = - 1.67

8. 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.

9. 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

10. Typical Electrodes

11. 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)|ZnSO4(aq)||CuSO4(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

12. 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: Cu2+(aq) + 2e- → Cu(s) Left-hand electrode: Zn2+(aq) + 2e- → Zn(s) The overall cell reaction can be obtained by subtracting left-hand reaction from the right-hand reaction: Cu2+(aq) + Zn(s) → Cu(s) + Zn2+(aq)

13. Expressing a reaction in terms of half-reactions Example : Express the formation of H2O from H2 and O2 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 Cu2+/Cu, Zn2+/Zn…. Ox + v e-→ Red The quotient is defined as: Q = aRed/aOx Example: Write the half-reaction and the reaction quotient for a chlorine gas electrode. (in class discussion)

14. Varieties of cells

15. 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

16. Cell Potential • Cell potential: the potential difference between two electrodes of a galvanic cell (measured in volts V). • Maximum electrical work : we,max = ΔG • Electromotive force, E, • Relationship between E and ΔrG: ΔrG = -νFE where ν is the number of electrons that are exchanged during the balanced redox reaction and F is the Faraday constant, F = eNA. • At standard conditions, this equation can be written as ΔrGθ = -νFEθ