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SOLUTIONS TO EXAMPLES

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Solubility of AgCl(s) in water at 25 o C is 1.274 x 10 -5 mol kg -1. Calculate the solubility of AgCl(s) in 0.010 mol kg -1 Na 2 SO 4 (aq). I < 0.05 Use Debye-Hückel law K sp = a Ag+ a Cl- For AgCl dissolving in H 2 O assume = 1 since m 0 K sp = (1.274 10 -5 ) 2 = 1.623 10 -10 For I = 0.030 mol kg -1 Ignore Ag + and Cl - in solution as conc’s v. low K sp = a Ag+ a Cl- = m Ag+ m Cl- 1.623 10 -10 = (0.666) 2 m 2 m = 1.91 10 -5 mol kg -1 = solubility In the presence of Na 2 SO 4 the solution is no longer ideal calculate activity coeff’s AgCl(s) Ag + (aq) + Cl - (aq) calculate [Ag+] or [Cl - ] Calculate K sp for the ideal soln and assume it be the same for the non-ideal soln Example 0

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Consider the Daniell cell: Using the standard reduction potentials, calculate the equilibrium constant at 25 C. Standard reduction potentials at 25 C: Cu 2+ + 2e - Cu(s) E o = +0.34 V Zn 2+ + 2e - Zn(s)E o = −0.76 V Overall reaction: Cu 2+ (aq) + Zn(s) Cu(s) + Zn 2+ (aq) The cell potential at standard conditions: Since K is large reaction goes to completion = +1.10 V At equilibrium:E cell = 0 V K = 1.6 10 37 Example 1 Reduction rxn Oxidation rxn Units: But J = C V w = V q E° cell > 0 cell reaction is spontaneous expect K > 1 Cu 2+ (aq) + Zn(s) Cu(s) + Zn 2+ (aq)

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Electrochemical reaction involved: 2H + (aq) + 2e H 2 (g) Let E 1 and E 2 be potentials for the initial and final states, i.e. for 5 and 20 mmol kg –1 The mean activity coefficients of HBr in 5.0 and 20.0 mmol kg –1 are 0.930 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. and Example 2

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E cell > 0 Galvanic cell spontaneous reaction written as above Reduction: Cl 2 (g) + 2e − 2Cl – (aq) E° = +1.36 V Suggest a Pt electrode as cathode. Oxidation:Mn(s) Mn 2+ (aq) + 2e − Mn 2+ (aq) + 2e − Mn(s) E° = ? V Suggest a Mn electrode as anode. E° cell = E° cathode – E° anode = 1.36 V – x = 2.54 V Mn(s) | Mn 2+ (aq), Cl – (aq) | Cl 2 (g) | Pt(s) Devise a cell in which the cell reaction is: Mn(s) + Cl 2 (g) MnCl 2 (aq) Give the half reactions at the electrodes and from the standard cell potential of 2.54 V deduce the standard potential for the Mn 2+ /Mn(s) redox couple. Given: E°(Cl 2 /Cl - ) = +1.36 V E° anode = -1.18 V or E° for the redox couple Mn 2+ / Mn(s) = -1.18 V Example 3 Half reactions: Cell: Standard potential for the Mn 2+ /Mn(s) redox couple:

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(R-H):Cd 2+ (aq) + 2e Cd(s)E° = –0.40 V (L-H):2AgBr(s) + 2e 2Ag(s) + 2Br – (aq) E° = +0.07 V Cell reaction: Cd 2+ (aq) + 2Ag(s) + 2Br – (aq) Cd(s) + 2AgBr(s) E° cell = E° cathode – E° anode = E° R-H – E° L-H = – 0.47 V Need to calculate activities: I Cd(NO3)2 = 0.010 and I KBr = 0.050 hence we can use the Debye-Hückel limiting law Estimate the cell potential at 25°C for Ag(s)|AgBr(s)|KBr(aq, 0.050 mol kg –1 )||Cd(NO 3 ) 2 (aq,0.0034 mol kg –1 )|Cd(s) E°(R-H) = –0.40 V E°(L-H) = +0.07 V (assume non-ideal solutions) Write the spontaneous electrochemical reaction. Example 4

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E cell < 0 non-spontaneous electrochemical reaction Cd 2+ (aq) + 2Ag(s) + 2Br – (aq) Cd(s) + 2AgBr(s) The spontaneous electrochemical reaction: Cd(s) + 2AgBr(s) Cd 2+ (aq) + 2Ag(s) + 2Br – (aq) -0.076 -0.084 E cell = –0.63 V = -0.076 = -0.084 -0.47

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Note: the cell is considered at standard conditions and E cell > 0 (R-H)Ag + (aq) + e - Ag(s) E° = +0.80 V (L-H)AgI(s) + e - Ag(s) + I – (aq) E° = –0.15 V Spontaneous electrochemical reaction: Ag + (aq) + I – (aq) AgI(s)E° cell = +0.95 V K = 1.1 10 16 Therefore K sp = K –1 = 8.7 10 –17 K sp = [Ag + ] [I - ] = [Ag + ] 2 = 8.7 10 –17 Solubility = [AgI(aq)] = 9.4 10 –9 mol kg –1 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. AgI(s) Ag + (aq) + I – (aq) Example 5

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Calculate the degree of ionization and the acid dissociation constant at 298 K for a 0.010 M acetic acid solution that has a resistance of 2220 . The resistance of a 0.100 M potassium chloride solution was also found to be 28.44 . m (0.1 M KCl) = 129 S cm 2 mol -1 o (H + ) = 349.6 S cm 2 mol -1 o (CH 3 COO - ) = 40.9 S cm 2 mol -1 To find the cell constant (C), we can use the data for the KCl solution. Degree of ionisation, : where Example 6

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Degree of ionisation, : Using KCl data to find the cell constant (C): c = 0.100 M = 0.100 mol dm -3 = 1.00 10 -4 mol cm -3 Finding m : or 4.23%

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Acid dissociation constant, K a : CH 3 COOH + H 2 O CH 3 COO - + H 3 O + Also pK a = -log(1.87 10 -5 ) = 4.73

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The molar conductivity of a strong electrolyte in water at 25 °C was found to be 109.9 S cm 2 mol -1 for a concentration of 6.2 10 -3 mol L -1 and 106.1 S cm 2 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 law Using K, calculate limiting molar conductivity from Kohlrausch law. Subtract two equations, solve for K. Note: is the same in for both eqn’s = -0.095 S cm -2 2 equations, 2 unknowns! Example 7

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The limiting molar conductivities of KCl, KNO 3, and AgNO 3 are 149,9, 145.0, and 133.4 S cm 2 mol -1, respectively (all at 25 °C). What is the limiting molar conductivity of AgCl at this temperature? To solve: manipulate the 3 equations above to obtain the one for AgCl We can apply the Kohlrausch law of independent migration of ions. Recall: Example 8

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The mobility of the NO 3 - ion in aqueous solution at 25 °C is 7.40 10 -8 m 2 s -1 V -1. (Viscosity of water is 0.891 10 -3 kg m -1 s -1 ). Calculate its diffusion coefficient and the effective radius at this temperature. Calculate the hydrodynamic radius: Use the Einstein relation between the mobility and diffusion coefficient: 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. Or use the equation: J = kg m 2 s -2 J = V C Example 9

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Considering units: BUT w = qV J = C V BUT J = kg m 2 s -2 Also: V = IR V = A and q = It C = As

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Note that we deal with large and positive overpotential. Hence the Tafel equation for anodic current can be applied. = 0.138 V j 0 = 2.82 mA cm –2 The electron transfer coefficient of a certain electrode in contact with the redox couple M 3+ / M 4+ in aqueous solution at 25 °C is 0.39. 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? M 3+ M 4+ + e - j 1 = 55.0 mA cm –2 1 = 0.125 V j 2 = 75.0 mA cm –2 2 = ? F = 96485 C mol-1 R = 8.315 J K-1 mol-1 T = 298 K n = 1 = 0.39 Example 10

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The relatively high positive overpotential applied results in very little reduction taking place. for a 2 electron process The exchange current density and the electron transfer coefficient for the reaction 2H + + 2e H 2 (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: Tafel: Positive potential Anodic current F = 96485 C mol -1 R = 8.315 J K -1 mol -1 T = 298 K n = 2 = 0.58 = 0.20 V J o = 6.3 10 –6 A cm –2 Example 11

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