Model Solutions to 3.53 Problem Set 5

Model Solutions to 3.53 Problem Set 5 Problem 8.3 In solving this problem, the Sand equation (8.2.14) is used. Co 2 = constant = k (8.2.14) For the known Pb2+ solution: (1) 273 mA x 25.98 1.00 mM = 1.389 As¹/2/mM For the known Cd²+ solution: kc²+ 136 mA x 42.0s 0.69 mM = 1.277 As¹/2/mM (2) The Er/4values are Er/4(Pb2+)=-0.38 V vs SCE and Er/4(Cd2+) = -0.56 V vs SCE. As these differ by approximately 180 mV, it is clear that the Pb2+ process finishes before the Cd2+ process begins. Thus, for the Pb2+/Cd2+ mixture, Сра+ = 0.0565 A × √7.088 1.389 As¹/2/m.M 0.108 mM (3) For Cd2+, the transition occurs at t = 71+72, when the concentration of Cd2+ drops to zero at the electrode surface. For the Cd2+ transition time alone, the Sand equation takes the following form. Cc+= i √T+T-√T 0.0565 A x [√7.08 s +7.00 -7.08 s kcd+ 1.277 A2/2/m.M 0.0483 mM (4) Problem 8.5 Figure 8.4.1 shows an experimental chronopotentiogram with current reversal for the oxidation of diphenylpicrylhydrazyl (DPPH) followed by reaction of the radical cation. From the figure, t₁ = 115 s and 72≈ 5 s. Since t₁ = 372 (equation (8.4.9)), the product, DPPH+, is stable and undergoes ann 1 reduction (DPPH++e DPPH) on current reversal with Er1/4 E≈ 0.73 V vs SCE. From the figure legend, CDPPH = 1.04 mM; A(Pt electrode) = 1.2 cm²; and i = 100 µA. Using the Sand equation (8.2.14) with i in mA, 7 in s, Co in mM and A in cm², allows one to calculate the diffusion coefficient Do in cm²/s. Do = 0.100 × √15 1.04 x 85.5 x 1.2 = 1.32 x 10-5cm²/s (1) Problem 8.6 From equation (8.7.1), the charge on Cinj1nF set by a 10 V battery is found. Aq=CinjxV10 Fx 10 V108 C When Aq is distributed over Cinj1nF and Ca=1 µF, the charge is conserved such that Aq = Qing + 2a = 10-8 C Also, the voltage drop across the two capacitors must be equal. Thus, = Linj Id Canj Ca 10-F10-6F (1) (2) (3) Solution of two equations in two unknowns yields q = 9.99×10-9 C and ging = 9.99 × 10-12 C. Thus, all of the charge is delivered from Cinj to Ca. The total capacitance in the Figure 8.9.1 circuit is found as follows. 1 1 +10° +10° 10° F-1 Cr Cinj Ca CT10°F (4) (5) The time constant ROCT 100 2 x 10-9 F≈ 107 s. From equation (1.2.6), the current for charging Ca drops to 5% of its initial value att 3 and 1% of its initial value at t=5r. Thus, Ca is 95% charged in 3 x 107 sor> 99% charged in ≈ 5 x 10-7 s. Problem 9.3 This problem is based on the data in Figure (9.3.8). From the Figure legend, f = 2500 rpm = 2500 rev/min x 1 min/60 8 = 41.67 του/ε w = 2x = 2 × 41.67 s1 = 262/s Au electrode, A = 0.196 cm² O2+H2O+2eHO₂ + OH- 1/2 - 16.28-1/2 Co, 1.00 mM (saturated) 1.0×10-3 mol/cm³ 712 (a). The Do, in 0.1 M NaOH is found from the i E curve in Figure (9.3.8a), where ite 6.5 x 10 A. From equation (9.3.22), 2/3 D = 0.62nF Aw/2-1/6C (6.5 x 10-4A) (0.01cm³/s)1/6 (0.62) (2) (96485C/mol) (0.196cm²) (16.28-1/2) (10-mol/cm³) (1) Do 7.94 x 10-4 cm/373/2 = 2.2 x 105 cm²/s (2) (b). The Koutecky-Levich equation (9.3.39) shows that a graph of 1/ite versus w-1/2 leads to an intercept of 1/ik. From Figure 9.3.8, the intercept at 0.75 V is i¹ yields ix 8.3 x 10 A. From equation (9.3.38), ix 1.2 mA-¹. This -1 K ky(E) = FACO (3) 8.3 x 10-4 A (2) (96485C/mol) (0.196cm²) (10-5mol/cm³) 2.2 x 102 cm/s at 0.75V Problem 9.4 (a). The first reduction step is characterized by the first current step for both the ring and the disk. Both waves yield E1/2~0.1 V, as the potential at half the plateau current. The ring current is shown for in = 0, such that equation (9.4.8) applies. İR iD --[()() 3 T 1100 μ.Α 300 μ.Α = 3.67 312/3 (9.4.8) Ог, в = 7.02. To determine the diffusion coefficient for Cu2+, the Levich equation for the mass transport limited current is used. iD,lim=0.62n.FAC 2/3-1/6 1/2 (9.3.22) Data provided in the correct units are idisk 3x 104 A, n1, A 0.962 cm², C = 5 x 10-6 mol/cm³, and w = 201 s-1. The kinematic viscosity, the ratio of viscosity to density, is provided in the caption, v = 0.011 cm²/s. Rearranging the above equation yields D レー1/6 73/2 1/2 3 x 10-4 А х (0.011 cm²/s) 1/6 0.62 x 1 x 96485 C/mol x 0.0962 cm² x 5 x 10-6 mol/cm³√201 s 6.48 x 106 cm²/s = (1) 3/2 = (b). If Ep-0.1 V, the disk is on the mass transport limited plateau for reduction of Cu2+, so the system is set up for a shielding experiment because Cu2+ will be reduced at both electrodes. Because in ip,lim, equation (9.4.20) applies. in,lim = iklim (1-N3-2/3) 0 (9.4.20) From the figure, Rim = 32/3ip,lim = 1100 μΑ ας the value recorded when in 0. It is 0 given that N = 0.53. From part (a), 32/3 1100 μ.Α (1 – 0.53/3.67) = 940 μ.Α. 3.67. Then, IR,lim = 1.Jim (1-N3-2/3) = (c). If ER+0.4 V and Ep-0.10 V, then the experiment is a collection experiment with mass transport limited current at both electrodes. From the collection efficiency (equation (9.4.15)), -in Nx ip = -0.53 × 300 μ.Α = -160 μ.Α.