t?dep- 11802105 (e) If the potentiometer is replaced by a 12.5 kn resistor and F
ID: 920715 • Letter: T
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t?dep- 11802105 (e) If the potentiometer is replaced by a 12.5 kn resistor and F the heat dissipated by the resistor is 1.00x10 s, at what rate (orams per second) N(G) dissoving? (n t 8 31e9 a/s Suppose that the concentrations of NaF and Ka were each 0.80 M in the following cel·(The Ko's of pb, and Apd are 3.5x10 0.80 M in the following cell. (The K's of Pbr, and Agcd are 3.6x10 and 1.8x10 respectivey.) (a) Using the following half-reactions, calculate the cell voltage. 2Agads) + 2 e. 2 Ag(s) + 2a. PDFa(s) + 2 e-=P(s) + 2 F- (b) By the reasoning in this figure, in which direction do electrons flow? Left to right. O Right to left. (c) Now calculate the cell voltage by using the following reactions 2 Ag +2 2 Ag(s) will ned th soluebity roducts for pod, and Aoo. For this part, you will need the solubility products for Pbf, and Aga. eBlook a te) 6. -16 points r 10 "in Mn'., o 3 Mmo. and 3.31 M HOO . (The Farada. constant is 9 64961 o co 10 M Ce 3.31 A solution contains 0.331 M ce". 3.31 substance Te (volts)Explanation / Answer
a) In such problems we have to calculate Ecell for each Half cell reaction by using the formula
E= E0 - 0.05916/ n log K
For Right side half cell, 2AgCl + 2e- = 2 Ag + 2 Cl-
Eright = E0 - 0.05916/ n log K = 0.222 - 0.05916/2 log [Cl-]^2 = 0.222 + 0.059162 = 0.281 V
For left side half cell, 2PbF2 + 2e- = Pb + 2 F-
Eright = E0 - 0.05916/ n log K = -0.350 - 0.05916/2 log [F-]^2 = -0.350 + 0.05916 = -0.291 V
The overall cell poterial E cell = Eright - Eleft = 0.281 - (-0.291) = 0.572 V
b) Given equations and solubility products data we have to calculate the solubility
2Ag+ + 2e- = 2Ag,
Pb2+ + 2e- = Pb
AgCl = Ag+ + Cl-, Ksp = 1.8 x 10-10 = [Ag+][Cl-]/AgCl
[Ag+] = Ksp / [Cl-] = 1.8 x 10-10 / 0.1 = 1.8 x 10-9 M
Similarly for [Pb2+] = 3.6 x 10-8 / [0.1]^2 = 3.6 x 10-6 M
Now calculate the half cellpotential by using the above equation
EAg = 0.799V - 0.05916/2 log (1/[1.8 x 10-9 ]) = 0.282 V
EPb = -0.126 - 0.05916/2 log(1/[3.6 x 10-6]) = -0.287 V
Ecell = Ered - Eox = 0.282 - (-0.287) = 0.569 V
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