E ε? rt2flnpb 2 012m 0150v 0150v 0 001284630lnpb 2

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Chemistry & Chemical Reactivity
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Chapter 14 / Exercise 84
Chemistry & Chemical Reactivity
Kotz/Treichel
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∆E = ∆Εο− (RT/2F)ln([Pb2+]/0.12M) = 0.150V.0.150V = 0 - (0.01284630)ln([Pb2+]/0.12M)ln([Pb2+]/0.12M) = -11.6765[Pb2+]/0.12M=8.4909x10-6M[Pb2+]=1.0189x10-6MKsp= [C2O42-][Pb2+] = 6.52 x 10-11 = 7 x 10-111 sf from[Pb2+]x 2 sf fromKa2 gives 1 sf in final answer. However, I would also accept 2 sf because if you did this problem with log instead of ln, there would be 2 digits after the mantissa. Worksheet:1. (1 pt) Part A: (Be sure to use the correct formula from the manual or from the bottles in the balance area to determine the molar mass of the compounds.)a. How much copper sulfate (in grams) do you need to make 100.ml of a 0.050M solution?
b. How much lead nitrate (in grams) do you need to make 20.ml of a 0.050M solution?
2. (1 pt) How does the salt bridge maintain electrical neutrality in the solutions?
3. (1 pt) Part C: You have two Cu2+ halfcells with different concentrations.a. Does the halfcell with the lower concentration of Cu2+ have a smaller or larger reduction potential than the one with the higher concentration?
b. Does the halfcell with the lower concentration of Cu2+ contain the anode or the cathode?
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Chemistry & Chemical Reactivity
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Chapter 14 / Exercise 84
Chemistry & Chemical Reactivity
Kotz/Treichel
Expert Verified
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