Calculate Ecell for Pb(s) | PbSO4(s) | Pb2+(aq) || Cu2+(aq) | Cu(s) the way it is written here
given Cu2+(aq) + 2 e- → Cu(s) E° = +0.337 V
PbSO4(s) + 2 e- → Pb(s) + SO42-(aq) E° = -0.356 V
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Calculate Ecell for Pb(s) | PbSO4(s) | Pb2+(aq) || Cu2+(aq) | Cu(s) the way it is...
Question 7 (1 point) Consider the following cell: Pb(s) | PbSO4(s) S042-(aq) || Pb2+(aq) | Pb(s) The reaction utilized by this cell is O Pb(s) + 2H+(aq) --> Pb2+(aq) + H2(g) O s042-(aq) + H+(aq) --> HSO4-(aq) O PbSO4(s) --> Pb2+(aq) + SO42-(aq) O Pb2+(aq) + SO42- (aq) --> PbSO4(s) O s042-(aq) + H20(1) --> HS04"(aq) + OH(aq)
Candidate l: Zn(s) | Zn2+(aq,0.500 M) I Cu2+(aq, 1.00 M) Cu(s) Candidate 2: Pb(s) | Pb2+(aq, 0.500 M) || Cu2+(aq, 1.00 M) Cu(s) Candidate 3: Mg(s) | Mg2+(aq, 0.500 M) | Pb2+(aq, 1.00 M)| Pb(s) (a) 6 pts) Choose one of the candidate voltaic cells #1, #2, or #3. Draw a schematic cell diagram for the candidate voltaic cell of choice. Clearly label anode, cathode, electrodes, ions and their concentrations, salt bridge, and the flow of electrons. (b) (5 pts)...
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. What is the initial cell potential? Hint: the standard potential of Pb2+ + 2e → Pb(s) is -0.130 and the standard potential of Cu2+ + 2e → Cu(s) is +0.340. Hint #2: Use [Cu2+] as the product and [Pb2+] as the reactant. Ecell = 0.044 Ecell = 0.383 Ecell = 0.426 Ecell...
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. What are the concentrations of Pb2+ and Cu2+ when the cell potential falls to 0.370 V? Cu2+ (aq) + Pb(s) + Cu(s) + Pb2+ (aq) Hint: [Pb2+] + [Cu2+] = (1.5M + 0.05M) = 1.55 M total Hint: the standard potential of Pb2+ + 2e → Pb(s) is -0.130 and the...
Ecell = Pb (s) | Pb2+ [1.0 M] || Cu+ [1.0M] | Cu (s) Deterime the Ecell for the galvanic cell above if [Pb2+] = 0.025 M and [Cu'] = 0.130 M. Deterime the Ecell for the galvanic cell above if [Pb2+] = 0.025 M and [Cu'] = 0.130 M.
Question 8 0.5 pts 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. What is the initial cell potential? Hint: the standard potential of Pb2*+2e » Pb(s) is -0.130 and the standard potential of Cu2 + 2e -» Cu(s) is +0.340. Hint #2: Use [Cu2*] as the product and [Pb2'] as the reactant. Ecell 0.044 Ecell 0.383 Ecell 0.426 Ecell...
help 2. (4) Given: PbSO4(e)Pb2 (aq) + SO42 (aq) Calculate the equilibrium constant, if AGt for PbSO4(s) is-811 Kj/mole; for Pb2+ (aq) is 24.3 Kj/mole and for SO42 (ag) is -742 Kj/mole. Calculate the solubility of PbSO4, if the pH of the solution is 4.32 and the [NaHSO4] is 0.10 M. Ka for HSO4 1.02 x 10-2 1. (6) From this week's experiment, what would you expect the signs on ΔΗΡ and So for the observed reaction, Na2B O(OH)8 H2o...
A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 oC. The initial concentrations of Pb2+ and Cu2+ are 0.0500 M and 1.50 M, respectively. What are the concentrations of Pb2+ and Cu2+ when the cell potential falls to 0.370 V? Cu2+(ag) + Pb(s) → Cu(s) + Pb2+(ag) Hint: [Pb2+] + [Cu2+] = (1.5M + 0.05M) = 1.55 M total Hint: the standard potential of Pb2+ + 2e- → Pb(s) is -0.130 and the standard potential...
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Consider the cell. Cu ∣ Cu2+ (0.00534 M)∣∣ Pb2+ (0.00735 M) ∣∣ Pb Calculate the half‑cell reduction potential at 298 K at the cathode. ? cu2+/cu=+0.34V ?cathode=......V Calculate the half‑cell reduction potential at 298 K at the anode. ? pb2+/pb=-0.13V ?anode=....V What is the initial potential needed to provide a current of 0.0800A if the resistance of the cell is 4.13Ω? Assume that ?=298 K. ?applied=......V