Homework Answers
The Correct Option will be the last option :
- H+(aq)/H2(g) is the Cathode in Cell 1 and Fe2+(aq)/Fe(s) is the Cathode in Cell 2.
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Separate galvanic cells are made from the following half-cells: cell 1: H+(aq)/H2(g) and Pb2+(aq)/Pb(s) cell 2:...
Candidate l: Zn(s) | Zn2+(aq,0.500 M) I Cu2+(aq, 1.00 M) Cu(s) Candidate 2: Pb(s) | Pb2+(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)...
Use the half-reactions below to produce a voltaic cell with the given standard cell potential. Standard...
Use the half-reactions below to produce a voltaic cell with the given standard cell potential. Standard Cell Potential Co- (aq) + e-Cot (aq) E = +1.82 V 1.53 V 2H(aq) + 2e-H2(g) E = +0.00 V Pb2+ (aq) + 2e-Pb(s) E = -0.13 V Fe (aq) + e-Fel+ (aq) E = +0.77 V Ag (aq) + e-Ag(s) E = +0.80 V Sn* (aq) + 2e Sne (aq) 20.13 V Cu- (aq) + e- Cu(aq) E = +0.15 V Zn²+ (aq)...
For each set of materials below: i.Write the cell notation (line notation)for a galvanic cell and...
For each set of materials below:
i.Write the cell notation (line notation)for a galvanic cell
and
ii.Calculate the standard cell potential
a. Pb(s), Zn(s), Pb2+(aq),Zn2+(aq)
b. Ag(s), Fe(s), Ag+(aq),Fe2+(aq)
c. Mg(s), Zn(s), Mg2+(aq),Zn2+(aq)
d. Fe(s), Mg(s), Fe2+(aq), Mg2+(aq
e. Ag(s), Pb(s), Pb2+(aq), Ag+(aq)
2. For each set of materials below: Write the cell notation (line notation) for a galvanic cell and Calculate the standard cell potential ii. a. Pb (). Zn (3), Pb2+(aq), Zn2+ (aq) 6. Ag (3), Fe (3),...
Problem Set #34: Galvanic Cells & Standard Reduction Potential Practice Sheet Directions: (a) Sketch the galvanic...
Problem Set #34: Galvanic Cells & Standard Reduction Potential Practice Sheet Directions: (a) Sketch the galvanic cell based on the following overall reaction OR half- reactions. o Show the direction of electron flow o Identify the cathode and anode (b) Give the overall balanced reaction (c) Determine the Evalues for each of the galvanic cells. (d) Give the standard line notation for each cell. (Assume that all concentrations are 1.0M and that all partial pressures are 1.0 atm) 1. 103(aq)...
A chemist designs a galvanic cell that uses these two half-reactions: standard reduction potential half-reaction +...
A chemist designs a galvanic cell that uses these two half-reactions: standard reduction potential half-reaction + O2(9)+4 H (aq)+4e' 2H20) = 1.23 V red Ered Fe+. (аq) Fe3(aq)+e = +0.771 V Answer the following questions about thiss cell Write a balanced equation for the half-reaction that happens at the cathode Write a balanced equation for the half-reaction that happens at the anode Write a balanced equation for the overall reaction that powers the cell. Be sure the reaction is spontaneous...
A certain half-reaction has a standard reduction potential +0.14 V. An engineer proposes using this half-reactio...
A certain half-reaction has a standard reduction potential +0.14 V. An engineer proposes using this half-reaction at the anode of a galvanic cell that must provide at least 0.80 V of electrical power. The cell will operate under standard conditions Note for advanced students: assume the engineer requires this half-reaction to happen at the anode of the cell. Is there a minimum standard reduction potential that the hall reaction used at the cathode of this cell can have? ves, there...
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...
Use the tabulated electrode potentials to calculate K for the oxidation of nickel by H+: Ni(s)+2H+(aq)→Ni2+(aq)+H2(g)...
Use the tabulated electrode potentials to calculate K for the oxidation of nickel by H+: Ni(s)+2H+(aq)→Ni2+(aq)+H2(g) Express your answer using two significant figures. Standard reduction half-cell potentials at 25∘C Half-reaction E∘ (V) Half-reaction E∘ (V) Au3+(aq)+3e−→Au(s) 1.50 Fe2+(aq)+2e−→Fe(s) −0.45 Ag+(aq)+e−→Ag(s) 0.80 Cr3+(aq)+e−→Cr2+(aq) −0.50 Fe3+(aq)+3e−→Fe2+(aq) 0.77 Cr3+(aq)+3e−→Cr(s) −0.73 Cu+(aq)+e−→Cu(s) 0.52 Zn2+(aq)+2e−→Zn(s) −0.76 Cu2+(aq)+2e−→Cu(s) 0.34 Mn2+(aq)+2e−→Mn(s) −1.18 2H+(aq)+2e−→H2(g) 0.00 Al3+(aq)+3e−→Al(s) −1.66 Fe3+(aq)+3e−→Fe(s) −0.036 Mg2+(aq)+2e−→Mg(s) −2.37 Pb2+(aq)+2e−→Pb(s) −0.13 Na+(aq)+e−→Na(s) −2.71 Sn2+(aq)+2e−→Sn(s) −0.14 Ca2+(aq)+2e−→Ca(s) −2.76 Ni2+(aq)+2e−→Ni(s) −0.23 Ba2+(aq)+2e−→Ba(s) −2.90 Co2+(aq)+2e−→Co(s) −0.28 K+(aq)+e−→K(s) −2.92 Cd2+(aq)+2e−→Cd(s)...
Using the standard reduction potentials given below, choose the reaction than can only be achieved through...
Using the standard reduction potentials given below, choose the reaction than can only be achieved through electrolysis. Cu2+(aq) + 2e → Cu(s) E° = 0.34 V Pb2+(aq) + 2e + Pb(s) E° = -0.13 V Fe2+(aq) + 2e Fe(s) E° = -0.44 V Zn2+(aq) + 2e + Zn(s) E° = -0.77 V Zn2+(aq) + Pb(s) → Zn(s) + Pb2+(aq) o Fe2+(aq) + Zn(s) → Fe(s) + Zn2+(aq) Pb2+(aq) + Fe(s) → Pb(s) + Fe2+(aq) Cu2+(aq) + Fe(s) → Cu(s) +...
A) Use tabulated electrode potentials to calculate ΔG∘ for the reaction. 2K(s)+2H2O(l)→H2(g)+2OH−(aq)+2K+(aq) B) (Refer to the...
A) Use tabulated electrode potentials to calculate ΔG∘ for the reaction. 2K(s)+2H2O(l)→H2(g)+2OH−(aq)+2K+(aq) B) (Refer to the following standard reduction half-cell potentials at 25∘C: VO2+(aq)+Ni2+(aq)2H+(aq)++2e−e−→ →Ni(s)VO2+(aq) +H2O(l)E∘=−0.23V E∘=0.99V) An electrochemical cell is based on these two half-reactions: Oxidation:Reduction:Ni(s)VO2+(aq,0.024M)+2H+(aq,1.4M)+e−→→Ni2+(aq,1.8M)+2e−VO2+(aq,1.8M)+H2O(l) Calculate the cell potential under these nonstandard concentrations. C) Standard reduction half-cell potentials at 25∘C Half-reaction E∘ (V ) Half-reaction E∘ (V ) Au3+(aq)+3e−→Au(s) 1.50 Fe2+(aq)+2e−→Fe(s) − 0.45 Ag+(aq)+e−→Ag(s) 0.80 Cr3+(aq)+e−→Cr2+(aq) − 0.50 Fe3+(aq)+3e−→Fe2+(aq) 0.77 Cr3+(aq)+3e−→Cr(s) − 0.73 Cu+(aq)+e−→Cu(s) 0.52 Zn2+(aq)+2e−→Zn(s) − 0.76...
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)...
Use the half-reactions below to produce a voltaic cell with the given standard cell potential. Standard Cell Potential Co- (aq) + e-Cot (aq) E = +1.82 V 1.53 V 2H(aq) + 2e-H2(g) E = +0.00 V Pb2+ (aq) + 2e-Pb(s) E = -0.13 V Fe (aq) + e-Fel+ (aq) E = +0.77 V Ag (aq) + e-Ag(s) E = +0.80 V Sn* (aq) + 2e Sne (aq) 20.13 V Cu- (aq) + e- Cu(aq) E = +0.15 V Zn²+ (aq)...
For each set of materials below:
i.Write the cell notation (line notation)for a galvanic cell
and
ii.Calculate the standard cell potential
a. Pb(s), Zn(s), Pb2+(aq),Zn2+(aq)
b. Ag(s), Fe(s), Ag+(aq),Fe2+(aq)
c. Mg(s), Zn(s), Mg2+(aq),Zn2+(aq)
d. Fe(s), Mg(s), Fe2+(aq), Mg2+(aq
e. Ag(s), Pb(s), Pb2+(aq), Ag+(aq)
2. For each set of materials below: Write the cell notation (line notation) for a galvanic cell and Calculate the standard cell potential ii. a. Pb (). Zn (3), Pb2+(aq), Zn2+ (aq) 6. Ag (3), Fe (3),...
Problem Set #34: Galvanic Cells & Standard Reduction Potential Practice Sheet Directions: (a) Sketch the galvanic cell based on the following overall reaction OR half- reactions. o Show the direction of electron flow o Identify the cathode and anode (b) Give the overall balanced reaction (c) Determine the Evalues for each of the galvanic cells. (d) Give the standard line notation for each cell. (Assume that all concentrations are 1.0M and that all partial pressures are 1.0 atm) 1. 103(aq)...
A chemist designs a galvanic cell that uses these two half-reactions: standard reduction potential half-reaction + O2(9)+4 H (aq)+4e' 2H20) = 1.23 V red Ered Fe+. (аq) Fe3(aq)+e = +0.771 V Answer the following questions about thiss cell Write a balanced equation for the half-reaction that happens at the cathode Write a balanced equation for the half-reaction that happens at the anode Write a balanced equation for the overall reaction that powers the cell. Be sure the reaction is spontaneous...
A certain half-reaction has a standard reduction potential +0.14 V. An engineer proposes using this half-reaction at the anode of a galvanic cell that must provide at least 0.80 V of electrical power. The cell will operate under standard conditions Note for advanced students: assume the engineer requires this half-reaction to happen at the anode of the cell. Is there a minimum standard reduction potential that the hall reaction used at the cathode of this cell can have? ves, there...
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...
Using the standard reduction potentials given below, choose the reaction than can only be achieved through electrolysis. Cu2+(aq) + 2e → Cu(s) E° = 0.34 V Pb2+(aq) + 2e + Pb(s) E° = -0.13 V Fe2+(aq) + 2e Fe(s) E° = -0.44 V Zn2+(aq) + 2e + Zn(s) E° = -0.77 V Zn2+(aq) + Pb(s) → Zn(s) + Pb2+(aq) o Fe2+(aq) + Zn(s) → Fe(s) + Zn2+(aq) Pb2+(aq) + Fe(s) → Pb(s) + Fe2+(aq) Cu2+(aq) + Fe(s) → Cu(s) +...
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