Question

In: Chemistry

21. What is the cell potential for the following cells at 25°C? a. Pb(s) | Pb2+...

21. What is the cell potential for the following cells at 25°C?

a. Pb(s) | Pb2+ (aq) (1.0 M) || Cu2+ (aq) (1.0x10-4 M) | Cu(s)

b. Pt(s) | Sn2+ (aq) (0.001 M), Sn4+(aq) (0.005 M) || Ag+(aq) (0.500 M) | Ag(s)

c. In part a, what will the concentration of Pb2+ be at equilibrium?

22. The following cell has a measured potential of 1.42 V at 25°C: Al (s) | Al3+ (aq) (0.01 M) || Co2+ (aq) (? M) | Co (s)

What is the concentration of the Co2+?

Solutions

Expert Solution


Related Solutions

A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 ∘C. The...
A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 ∘C. The initial concentrations of Pb2+ and Cu2+ are 5.30×10−2 M and 1.60 M , respectively. Part A What is the initial cell potential? Express your answer using two significant figures. Ecell= ? V Part B What is the cell potential when the concentration of Cu2+ has fallen to 0.230 M ? Express your answer using two significant figures. Part B What is the cell potential...
A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 ∘C. The...
A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 ∘C. The initial concentrations of Pb2+ and Cu2+ are 5.00×10−2 M and 1.70 M, respectively. What is the cell potential when the concentration of Cu2+ has fallen to 0.250 M ? What are the concentrations of Pb2+ and Cu2+ when the cell potential falls to 0.37 V?
A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 ∘C. The...
A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 ∘C. The initial concentrations of Pb2+ and Cu2+ are 0.0500 M and 1.50 M , respectively. Standard Reduction Half-Cell Potentials at 25 ∘C Half-Reaction E∘(V) Cu+(aq)+e−→Cu(s) 0.52 Cu2+(aq)+2e−→Cu(s) 0.34 Cu2+(aq)+e−→Cu+(aq) 0.16 Pb2+(aq)+2e−→Pb(s) −0.13 1. What is the initial cell potential? Express your answer using two decimal places. 2. What is the cell potential when the concentration of Cu2+ falls to 0.200 M ? Express your answer...
A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 ∘C. The...
A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 ∘C. The initial concentrations of Pb2+ and Cu2+ are 5.00×10−2 M and 1.50 M , respectively What is the initial cell potential? 0.51V What is the cell potential when the concentration of Cu2+ has fallen to 0.200 M ? 0.45V Part C What are the concentrations of Pb2+ and Cu2+ when the cell potential falls to 0.360 V ? I need help with Part C
A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 ∘C. The...
A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 ∘C. The initial concentrations of Pb2+ and Cu2+ are 5.20×10−2M and 1.50 M, respectively. A) What is the cell potential when the concentration of Cu2+ has fallen to 0.250 M ? B) What are the concentrations of Pb2+ and Cu2+ when the cell potential falls to 0.36 V?
A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 ?C. The...
A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 ?C. The initial concentrations of Pb2+ and Cu2+ are 5.30
1)1. You are given the following cell at 25°C: Pb(s) | Pb2+(aq, 0.60M) || Sn2+(aq, 0.1...
1)1. You are given the following cell at 25°C: Pb(s) | Pb2+(aq, 0.60M) || Sn2+(aq, 0.1 M), Sn4+(aq, 0.15 M) |Pt(s) Write the half reactions for the cell and using Appendix D in your text book, calculate the E°cell and the Ecell oxidation half reaction = Pb(s) ------------> Pb+2(aq) + 2e- reduction half reaction = Sn+4(aq) + 2e- --------> Sn+2(aq) E0cell = E0Sn+4/sn+2   E0Pb+2/Pb E0cell = 0.15 - (-0.126) E0cell = 0.276 V ( i'm not sure if it correct...
Calculate the Kc for the following reaction at 25 °C: Mg(s) + Pb2+(aq)?Mg2+(aq) + Pb(s) Kc...
Calculate the Kc for the following reaction at 25 °C: Mg(s) + Pb2+(aq)?Mg2+(aq) + Pb(s) Kc = × 10 Enter your answer in scientific notation.
An electrochemical cell is based on the following two half-reactions: Ox: Pb(s)→Pb2+(aq,Pb(s)→Pb2+(aq, 0.15 MM )+2e−)+2e− Red:...
An electrochemical cell is based on the following two half-reactions: Ox: Pb(s)→Pb2+(aq,Pb(s)→Pb2+(aq, 0.15 MM )+2e−)+2e− Red: MnO−4(aq,MnO4−(aq, 1.80 MM )+4H+(aq,)+4H+(aq, 1.9 MM )+3e−→)+3e−→ MnO2(s)+2H2O(l) Compute the cell potential at 25 ∘C∘C.
Comprehensive thermochemical problem: Consider an electrochemical cell with the following half- cells: Pb2+(aq)(0.01 M)/Pb(s) and Sn2+(aq)(2M)/Sn(s)...
Comprehensive thermochemical problem: Consider an electrochemical cell with the following half- cells: Pb2+(aq)(0.01 M)/Pb(s) and Sn2+(aq)(2M)/Sn(s) at 25 °C. a. Find the potential of the cell. b.Determine the oxidizing agent and reducing agent. c. Calculate the standard free energy change for the reaction. d. Find the free energy change for the cell under the current conditions. e. Determine the equilibrium concentrations of the solutions. f. Use ΔH°f data to determine ΔH°rxn. ΔH°f (Sn2+ (aq)) = -8.8 kJ/mol ΔH°f (Pb2+ (aq))...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT