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Determine the Eºcell for the galvanic cell depicted below. Pb(s) | Pb2+ (1.0 M) || Cut...
Ecell = Pb (s) | Pb2+ [1.0 M] || Cu+ [1.0M] | Cu (s) Deterime 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.
Calculate e cell for the electrochemical cell below, Pb(s) |Pb2+(aq, 1.0 M) || Fe2+(aq, 1.0 M)...
Calculate e cell for the electrochemical cell below, Pb(s) |Pb2+(aq, 1.0 M) || Fe2+(aq, 1.0 M) | Fe(s) given the following reduction half-reactions. Pb2+(aq) + 2 e– ® Pb(s) E° = –0.126 V Fe2+(aq) + e– ® Fe(s) E° = –0.44 V
11. If the galvanic cell below is constructed with a Pb/Pb2-electrode and a Fe/Fe2+ electrode. determine...
11. If the galvanic cell below is constructed with a Pb/Pb2-electrode and a Fe/Fe2+ electrode. determine which electrode will be the cathode and which electrode will be the anode. Write each half cell reaction and the overall cell reaction. Oxidation: Reduction: Overall: Label the cell below with the following: a. the location of each substance (Pb, Pb2+, Fe, Fe2+) b. the cathode and anode C. the direction of electron flow d. If the salt bridge contains KCI, label the direction...
For which of the following voltaic cells will Ecell be greater than Eºcell? Pb(s) | Pb2+(0.0200...
For which of the following voltaic cells will Ecell be greater than Eºcell? Pb(s) | Pb2+(0.0200 M) || Ag+(0.0200 M)| Ag(s) Pb(s) | Pb2+(0.00200 M) || Ag+(0.200 M)| Ag(s) Pb(s) | Pb2+(0.200 M) || Ag+(0.200 M)| Ag(s) Pb(s) | Pb2+(0.00200 M) || Ag+(0.0200 M)| Ag(s) Pb(s) | Pb2+(0.200 M) || Ag+(0.0200 M)| Ag(s)
A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 °C. The initial concentrations of Pb2+...
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...
9) A galvanic cell Cr(s)|Cr2+(aq) || Pb2+(aq)|Pb(s) is constructed using a completely immersed Cr electrode that...
9) A galvanic cell Cr(s)|Cr2+(aq) || Pb2+(aq)|Pb(s) is constructed using a completely immersed Cr electrode that weighs 26.5 g and a Pb electrode immersed in 665 mL of 1.00 M Pb2+(aq) solution. A steady current of 0.0598 A is drawn from the cell as the electrons move from the Cr electrode to the Pb electrode. (a) Which reactant is the limiting reactant in this cell? ___ (b) How long does it take for the cell to be completely discharged? ___...
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)...
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...
Which is the correct line notation for the galvanic cell you constructed using a Pb2+ solution,...
Which is the correct line notation for the galvanic cell you constructed using a Pb2+ solution, a Pb electrode and a Saturated Calomel Electrode (SCE) as reference? Hg Ι Hg2Cl2(s) Ι KCl(sat) ΙΙ Pb2+(aq) Ι Pb Pb2+(aq) Ι Pb ΙΙ KCl(sat) Ι Hg Ι Hg2Cl2(s) Pb Ι Pb2+(aq) ΙΙ KCl(sat) Ι Hg2Cl2(s) Ι Hg Hg Ι Pb2+(aq) ΙΙ KCl(sat) Ι Hg2Cl2(s) Ι Pb Which is the correct full oxidation/reduction equation for the copper/SCE galvanic cell that you investigated in Part C...
A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 oC. The...
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...
data collected Cu(NO3)2 | Zn(NO3)2 = 0.999 V Pb | 1.0 M Pb(NO3)2 || 1.0 M...
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...
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.
11. If the galvanic cell below is constructed with a Pb/Pb2-electrode and a Fe/Fe2+ electrode. determine which electrode will be the cathode and which electrode will be the anode. Write each half cell reaction and the overall cell reaction. Oxidation: Reduction: Overall: Label the cell below with the following: a. the location of each substance (Pb, Pb2+, Fe, Fe2+) b. the cathode and anode C. the direction of electron flow d. If the salt bridge contains KCI, label the direction...
For which of the following voltaic cells will Ecell be greater than Eºcell? Pb(s) | Pb2+(0.0200 M) || Ag+(0.0200 M)| Ag(s) Pb(s) | Pb2+(0.00200 M) || Ag+(0.200 M)| Ag(s) Pb(s) | Pb2+(0.200 M) || Ag+(0.200 M)| Ag(s) Pb(s) | Pb2+(0.00200 M) || Ag+(0.0200 M)| Ag(s) Pb(s) | Pb2+(0.200 M) || Ag+(0.0200 M)| Ag(s)
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...
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...
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...
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