2Example 0 AgCl(s) Ag+(aq) + Cl-(aq) calculate [Ag+] or [Cl-] Solubility of AgCl(s) in water at 25oC is x 10-5 mol kg-1.Calculate the solubility of AgCl(s) in mol kg-1 Na2SO4(aq).In the presence of Na2SO4 the solution is no longer ideal calculate activity coeff’sI < 0.05 Use Debye-Hückel lawCalculate Ksp for the ideal soln and assume it be the same for the non-ideal solnKsp = aAg+ aCl-For AgCl dissolving in H2O assume = 1 since m 0Ksp = (1.27410-5)2 = 1.62310-10For I = mol kg-1Ignore Ag+ and Cl- in solution as conc’s v. lowKsp = aAg+ aCl- = mAg+ mCl-1.62310-10 = (0.666)2 m2m = 1.9110-5 mol kg-1 = solubility
3Example 1 Units: w = V q Consider the Daniell cell: Using the standard reduction potentials, calculate the equilibrium constant at 25 C.Standard reduction potentials at 25 C:Cu2+ + 2e- Cu(s) Eo = VZn2+ + 2e- Zn(s) Eo = −0.76 VReduction rxnOxidation rxnOverall reaction:Cu2+(aq) + Zn(s) Cu(s) + Zn2+(aq)The cell potential at standard conditions:E°cell > 0 cell reaction isspontaneous expect K > 1= VAt equilibrium: Ecell = 0 VCu2+(aq) + Zn(s) Cu(s) + Zn2+(aq)Since K is large reactiongoes to completion K = 1.6 1037Units:But J = C Vw = V q
4Example 2The mean activity coefficients of HBr in 5.0 and 20.0 mmol kg–1 are and 0.879, respectively. Consider a hydrogen electrode in HBr(aq) solution at 25 °C operating at 1.15 atm.Calculate the change in the electrode potential when the molality of the acid solution is changed from 5.0 and 20.0 mmol kg–1.Electrochemical reaction involved: 2H+(aq) + 2e H2(g)Let E1 and E2 be potentials for the initial and final states, i.e. for 5 and 20 mmol kg–1and
5E°cell = E°cathode – E°anode = 1.36 V – x = 2.54 V Example 3Devise a cell in which the cell reaction is:Mn(s) + Cl2(g) MnCl2(aq)Give the half reactions at the electrodes and from the standard cell potential of 2.54 V deduce the standard potential for the Mn2+/Mn(s) redox couple.Given: E°(Cl2/Cl-) = VEcell > 0 Galvanic cell spontaneous reaction written as aboveHalf reactions:Reduction: Cl2(g) + 2e− 2Cl–(aq) E° = VSuggest a Pt electrode as cathode.Oxidation: Mn(s) Mn2+(aq) + 2e−Mn2+(aq) + 2e− Mn(s) E° = ? VSuggest a Mn electrode as anode.Cell:Mn(s) | Mn2+(aq), Cl–(aq) | Cl2(g) | Pt(s)Standard potential for the Mn2+/Mn(s) redox couple:E°cell = E°cathode – E°anode = 1.36 V – x = 2.54 V E°anode = V or E° for the redox couple Mn2+ / Mn(s) = V
6Example 4 Estimate the cell potential at 25°C for Ag(s)|AgBr(s)|KBr(aq, mol kg–1)||Cd(NO3)2(aq, mol kg–1)|Cd(s)E°(R-H) = –0.40 V E°(L-H) = V (assume non-ideal solutions)Write the spontaneous electrochemical reaction.(R-H): Cd2+(aq) + 2e Cd(s) E° = –0.40 V(L-H): 2AgBr(s) + 2e 2Ag(s) + 2Br–(aq) E° = VCell reaction: Cd2+(aq) + 2Ag(s) + 2Br–(aq) Cd(s) + 2AgBr(s)E°cell = E°cathode – E°anode = E°R-H – E°L-H = – 0.47 VNeed to calculate activities:ICd(NO3)2 = and IKBr = 0.050hence we can use the Debye-Hückel limiting law
8Example 5 The standard potential of the cell below at 25 °C is 0.95 V. Ag(s) |AgI(s) | AgI(aq) | Ag(s)Calculate: a) its solubility constant and b) the solubility of AgI .Note: the cell is considered at standard conditions and Ecell > 0(R-H) Ag+(aq) + e- Ag(s) E° = V(L-H) AgI(s) + e- Ag(s) + I–(aq) E° = –0.15 VSpontaneous electrochemical reaction:Ag+(aq) + I–(aq) AgI(s) E°cell = VK = 1.11016Therefore Ksp = K–1 = 8.710–17AgI(s) Ag+(aq) + I–(aq)Ksp = [Ag+] [I-] = [Ag+]2 = 8.7 10–17Solubility = [AgI(aq)] = 9.410–9 mol kg–1
9Example 6 Degree of ionisation, : Calculate the degree of ionization and the acid dissociation constant at 298 K for a M acetic acid solution that has a resistance of 2220 . The resistance of a M potassium chloride solution was also found to be .m(0.1 M KCl) = 129 S cm2 mol-1o(H+) = S cm2 mol-1o(CH3COO-) = 40.9 S cm2 mol-1Degree of ionisation, :whereTo find the cell constant (C), we can use the data for the KCl solution.
10Degree of ionisation, : Using KCl data to find the cell constant (C):c = M = mol dm-3 = 1.0010-4 mol cm-3Finding m:or %
12Example 7The molar conductivity of a strong electrolyte in water at 25 °C was found to be S cm2 mol-1 for a concentration of 6.2 10-3 mol L-1 and S cm2 mol-1 for a concentration of 1.5 10-3 mol L-1.Estimate the limiting molar conductivity of the electrolyte.Note: Strong electrolyte Kohlrausch law2 equations, 2 unknowns!Subtract two equations, solve for K.Note: is the same in for both eqn’s= S cm-2Using K, calculate limiting molar conductivity from Kohlrausch law.
13Example 8The limiting molar conductivities of KCl, KNO3, and AgNO3 are 149,9, 145.0, and S cm2 mol-1, respectively (all at 25 °C).What is the limiting molar conductivity of AgCl at this temperature?We can apply the Kohlrausch law of independent migration of ions.To solve: manipulate the 3 equations above to obtain the one for AgClRecall:
14Example 9The mobility of the NO3- ion in aqueous solution at 25 °C is 7.4010-8 m2 s-1 V-1. (Viscosity of water is 0.89110-3 kg m-1 s-1).Calculate its diffusion coefficient and the effective radius at this temperature.Use the Einstein relation between the mobility and diffusion coefficient:Calculate the hydrodynamic radius:Or use the equation:Having “a” and the radius of a simple ion (without coordinated water) plus the dimension of a single water molecule, you would be able to predict the number of molecules in the hydrated share of the ion.Remember: in the calculations you have to show the work on units.Without that, the work might be considered as not done at all.J = kg m2 s-2J = V C
15Considering units:BUT w = qV J = C VBUT J = kg m2 s-2Also:V = IR V = Aandq = It C = As
16Example 10The electron transfer coefficient of a certain electrode in contact with the redox couple M3+ / M4+ in aqueous solution at 25 °C is The current density is found to be 55.0 mA cm–2 when the overvoltage is 125 mV.What is the overvoltage required for a current density 75.0 mA cm–2?What is the exchange current density?Note that we deal with large and positive overpotential.Hence the Tafel equation for anodic current can be applied.M3+ M4+ + e- = Vj0 = 2.82 mA cm–2j1 = 55.0 mA cm– 1 = V j2 = 75.0 mA cm–2 2 = ?F = C mol-1 R = J K-1 mol T = 298 K n = = 0.39
17Example 11The exchange current density and the electron transfer coefficient for the reaction 2H+ + 2e H2(g) on nickel at 25 °C are 6.3 10–6 A cm–2 and 0.58, respectively.Determine what current density would be required to obtain an overpotential of 0.20 V as calculated from the Butler-Volmer equation and the Tafel equation.Butler-Volmer:for a 2 electron processTafel:The relatively high positive overpotential applied results in very little reduction taking place.Positive potential Anodic currentF = C mol-1 R = J K-1 mol T = 298 K n = 2 = = 0.20 V Jo = 6.3 10–6 A cm–2