Exercise 18.78 A voltaic cell consists of a Pb/Pb half-cell and a Cu/Cu half. ce
ID: 1064494 • Letter: E
Question
Exercise 18.78 A voltaic cell consists of a Pb/Pb half-cell and a Cu/Cu half. cell at 25 °C. The initial concentrations of Pb and Cu are 5.30x10-2 M and 1.60 M respectively. Part A What is the initial cell potential? Express your answer using two significant figures. 0.51 V Subm My Answers Give Up Correct Part B what is the cell potential when the concentration of Cu has fallen to 0.220 M Express your answer using two significant figures. 0.45 V Submi My Answers Give Up Correct Part C What are the concentrations of Pb2+ and Cu when the cell potential falls to 0.370V? 2 Enter your answers numerically separated by a comma. Express your answer using two significant figures. o Ga Pb Cu 5.22.10 2, 5.22.10 2 Submit My Answers Give Up incorrect; Try Again: 2 attempts remainingExplanation / Answer
E = Eo - 0.059/2 * log [Pb2+]/[Cu2+]
let change in concentration be x. Since E < Eo, [Pb2+] has increased
E = Eo - 0.059/2 * log [Pb2+]/[Cu2+]
0.370 = 0.51 - 0.059/2 * log {(5.30*10^-2 + x)/(5.30*10^-2 - x)}
log {(5.30*10^-2 + x)/(5.30*10^-2 - x)} = 4.746
(5.30*10^-2 + x)/(5.30*10^-2 - x) = 55688
5.30*10^-2 + x = 2951.5 - 55688*x
x = 0.053 M
[Pb2+] = 5.30*10^-2 + 0.053 = 0.11 M
[Cu2+] = 5.30*10^-2 + 0.053 = 0
Answer: 0.11,0
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.