voltaic cell consists of a Pb/Pb^2+ half-cell and a Cu/Cu^2+ half-cell at 25 ∘C. The initial concentrations of Pb^2+ and Cu^2+ are 0.0500 mol/L and 1.60 mol/L, respectively. Pb2+(aq)+2e−→Pb(s)E∘=−0.13V Cu2+(aq)+2e−→Cu(s)E∘=0.34V a) What is the initial cell potential? Express your answer using two decimal places. b) What is the cell potential when the concentration of Cu2+ has fallen to 0.240 mol/L? Express your answer using two decimal places.
voltaic cell consists of a Pb/Pb^2+ half-cell and a Cu/Cu^2+ half-cell at 25 ∘C. The initial concentrations of Pb^2+ and Cu^2+ are 0.0500 mol/L and 1.60 mol/L, respectively. Pb2+(aq)+2e−→Pb(s)E∘=−0.13V Cu2+(aq)+2e−→Cu(s)E∘=0.34V a) What is the initial cell potential? Express your answer using two decimal places. b) What is the cell potential when the concentration of Cu2+ has fallen to 0.240 mol/L? Express your answer using two decimal places.
Chemistry
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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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A voltaic cell consists of a Pb/Pb^2+ half-cell and a Cu/Cu^2+ half-cell at 25 ∘C. The initial concentrations of Pb^2+ and Cu^2+ are 0.0500 mol/L and 1.60 mol/L, respectively.
Pb2+(aq)+2e−→Pb(s)E∘=−0.13V
Cu2+(aq)+2e−→Cu(s)E∘=0.34V
a) What is the initial cell potential?
Express your answer using two decimal places.
b) What is the cell potential when the concentration of Cu2+ has fallen to 0.240 mol/L?
Express your answer using two decimal places.
c) What are the concentrations of Pb^2+ and Cu^2+ when the cell potential falls to 0.36 V?
Express your answers using two significant figures separated by a comma.
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