Ecell = Pb (s) | Pb2+ [1.0 M] || Cu+ [1.0M] | Cu (s) Deterime 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. 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...
Determine the Eºcell for the galvanic cell depicted below. Pb(s) | Pb2+ (1.0 M) || Cut (1.0 M) Cu(s)
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...
data collected Cu(NO3)2 | Zn(NO3)2 = 0.999 V Pb | 1.0 M Pb(NO3)2 || 1.0 M Zn(NO3)2 | Zn = 0.396 V PART B: REDUCTION POTENTIALS 1. Report the measured cell potential for each galvanic cell and state which electrode corresponds to the cathode and which to the anode. 2. Given E = -0.76 V for the Zn/Zn half-cell, and your measured Ecell, calculate the reduction potential at the Cu and Pb electrodes and write the redox half- 6 reactions...
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...
Table view List view Ecell (calculated) Ecell (calculated) 1.115 1.124 Table 1. Voltaic cells data table Ecell (measured) Reaction Quotient Q 1. Zn | Zn2(1.0M) || Cu2+(1.0 M) | Cu 2. Zn | Zn2+(1.0 M) || Cu2+(0.1 M) Cu 1.067 3. ZnZn2+(0.1 M) || Cu2+(1.0 M) | Cu 4. Zn | Zn2+(1.0 M) || Pb2+(1.0 M) | Pb 5. ZnZn2(1.0M) || Pb2+(0.1 M) | Pb 0.600 6. Zn | Zn2+(0.1 M) || Pb2+(1.0 M) I Pb 0.644 7. ZnZn2+(1.0 M)...
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
Switch Voitmeter Salt bridge Cu(s) Pb(s) 1.0 M Pb."(aq) 1.0 M Cu(aq) [ Pb(s)+Cu(aq) Pb(aq)+Culs) a) When the switch is closed, which electrode will become heavier? Check the correct answerls) Neither Pb nor Cu Both Pb and Cu Cu Pb b) Which direction will the electrons flow? From Pb through the voltmeter to Cu From Cu through the voltmeter to Pb From Pb through the salt bridge to Cu From Cu through the salt bridge to Pb c) Which metal...
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...
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 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. The equation we are supposed to use is E= Eo-...