What is the initial cell potential? Express your answer using two decimal places. HA ? Ecell = 0.4465 V Submit Previous Answers Request Answer X Incorrect; Try Again

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Chapter1: Chemical Foundations
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A voltaic cell consists of a Pb/Pb?+ half-cell and a Cu/Cu?+ half-
cell at 25 °C. The initial concentrations of Pb2+ and Cu?+ are
0.0530 mol L-1 and 1.60 mol L1, respectively.
Part A
Pb2+ (aq) + 2 e → Pb(s)
Cu?+ (ag) + 2 e → Cu(s) E° = 0.34 V
E° = -0.13 V
What is the initial cell potential?
Express your answer using two decimal places.
Ecell = 0.4465
V
Submit
Previous Answers Request Answer
X Incorrect; Try Again
Part B
What is the cell potential when the concentration of Cu+ has fallen to 0.240 mol L-1?
Express your answer using two decimal places.
Ecell = 1.6431
m
Transcribed Image Text:A voltaic cell consists of a Pb/Pb?+ half-cell and a Cu/Cu?+ half- cell at 25 °C. The initial concentrations of Pb2+ and Cu?+ are 0.0530 mol L-1 and 1.60 mol L1, respectively. Part A Pb2+ (aq) + 2 e → Pb(s) Cu?+ (ag) + 2 e → Cu(s) E° = 0.34 V E° = -0.13 V What is the initial cell potential? Express your answer using two decimal places. Ecell = 0.4465 V Submit Previous Answers Request Answer X Incorrect; Try Again Part B What is the cell potential when the concentration of Cu+ has fallen to 0.240 mol L-1? Express your answer using two decimal places. Ecell = 1.6431 m
A voltaic cell consists of a Pb/Pb?+ half-cell and a Cu/Cu?+ half-
cell at 25 °C. The initial concentrations of Pb2+ and Cu?+ are
O 0530 mol L-1 and 1.60 mol L-1, respectively.
Express your answer using two decimal places.
Рb2+ (ag) + 2 е РЫ(s) E - -0.13 V
Cu²+ (ag) + 2 e → Cu(s) E° = 0.34 V
Ecell = 1.6431
Submit
Previous Answers Request Answer
X Incorrect; Try Again
Part C
What are the concentrations of Pb+ and Cu+ when the cell potential falls to 0.36 V?
Express your answers using two significant figures separated by a comma.
Pb*), [Cu**] =
mol L1
Transcribed Image Text:A voltaic cell consists of a Pb/Pb?+ half-cell and a Cu/Cu?+ half- cell at 25 °C. The initial concentrations of Pb2+ and Cu?+ are O 0530 mol L-1 and 1.60 mol L-1, respectively. Express your answer using two decimal places. Рb2+ (ag) + 2 е РЫ(s) E - -0.13 V Cu²+ (ag) + 2 e → Cu(s) E° = 0.34 V Ecell = 1.6431 Submit Previous Answers Request Answer X Incorrect; Try Again Part C What are the concentrations of Pb+ and Cu+ when the cell potential falls to 0.36 V? Express your answers using two significant figures separated by a comma. Pb*), [Cu**] = mol L1
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