Homework Answers
#5:
The Nernst equation for the given Galvanic cell is:
Ecell = Eocell - (0.0591/n)*log([Pb2+(aq)] / [Cu2+(aq)])
In order to increase cell potential, Ecell, the value of log([Pb2+(aq)] / [Cu2+(aq)]) must decrease.
log([Pb2+(aq)] / [Cu2+(aq)]) will decrease when the ratio: [Pb2+(aq)] / [Cu2+(aq)] will decrease.
[Pb2+(aq)] / [Cu2+(aq)] will decrease when either [Cu2+(aq)] concentration is increased or [Pb2+(aq)] concentration is decreased
(Answer)
#6:
Since Pd2+(aq) has higher reduction potential, it undergoes reduction and Fe(s) undergoes oxidation. Hence
EoOxi = - (-0.45V) = 0.45 V
EoRed = 0.95 V
Eocell = EoOxi + EoRed = 0.45V + 0.95V = 1.40 V
The Nernst equation for the given Galvanic cell is:
Ecell = Eocell - (0.0591/n)*log([Fe2+(aq)] / [Pd2+(aq)])
Since Pd2+(aq) is reduced to Pd(s), 2 electrons are transferred. Hence n = 2
=> Ecell = 1.40V - (0.0591/2)*log(0.100 / 1.0*10-5)
=> Ecell = 1.2818 V (Answer)
Add Answer to:
5. Consider the galvanic cell, Pb(s) Pb(aq) || Cu(aq) Cu(s) What should be done to increase...
Candidate l: Zn(s) | Zn2+(aq,0.500 M) I Cu2+(aq, 1.00 M) Cu(s) Candidate 2: Pb(s) | Pb2+(aq,...
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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:...
Separate galvanic cells are made from the following half-cells: cell 1: H+(aq)/H2(g) and Pb2+(aq)/Pb(s) cell 2: Fe2+(aq)/Fe(s) and Zn2+(aq)/Zn(s) Which of the following is correct for the working cells? Standard reduction potentials, 298 K, Aqueous Solution (pH = 0): Cl2(g) + 2e --> 2C1-(aq); E° = +1.36 V Fe3+(aq) + e --> Fe2+(aq); E° = +0.77 V Cu2+(aq) + 2e --> Cu(s); E° = +0.34 V 2H+(aq) + 2e --> H2(g); E° = 0.00 V Pb2+(aq) + 2e --> Pb(s);...
need help with 8,9,10, and 11 Chem 103A Electrochemistry 8. Consider a voltaic cell that uses the reaction Cu(s) +...
need help with 8,9,10, and 11
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Consider the following cell reaction at 18°C: Ca(e)+Cu+ (aq) + Ca2+ (aq) + Cu() Calculate the...
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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...
5. The overall cell reaction in a galvanic cell is Sn(s) + Cu* (aq)>Sn (aq)+ Cu...
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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...
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An electrochemical cell is based on these two half-reactions: Ox: Pb(s) Pb?" (aq, 0.21 mol L ')+2 e Erode = -0.13 (V) Red: Mno. (aq, 1.40 mol L-)+4 H+ (aq, 1.5 mol L-') +3 e MnO2 (s) + 2 H2O (1) Ecathode = 1.68 (V) Part A Compute the cell potential at 25°C. Express your answer to two decimal places and include the appropriate units. Ecell = Value Units Submit Request Answer
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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: Fe2+(aq)/Fe(s) and Zn2+(aq)/Zn(s) Which of the following is correct for the working cells? Standard reduction potentials, 298 K, Aqueous Solution (pH = 0): Cl2(g) + 2e --> 2C1-(aq); E° = +1.36 V Fe3+(aq) + e --> Fe2+(aq); E° = +0.77 V Cu2+(aq) + 2e --> Cu(s); E° = +0.34 V 2H+(aq) + 2e --> H2(g); E° = 0.00 V Pb2+(aq) + 2e --> Pb(s);...
need help with 8,9,10, and 11
Chem 103A Electrochemistry 8. Consider a voltaic cell that uses the reaction Cu(s) + 2Fe'(aq) - Cr"(aq) + 2Fe""(aq) What is the potential of a cell at 25°C that has the following concentrations? [Fe'') - 1.0x 10M (Cu)-0.25 M [Fe] =0.20 M 9. Lead can displace silver from solution, and as a consequence, silver is a valuable by-product in the industrial extraction of lead from its ore: Pb(s) + 2 Ag" (aq) → Pb...
Consider the following cell reaction at 18°C: Ca(e)+Cu+ (aq) + Ca2+ (aq) + Cu() Calculate the standard cell potential of this cell from the standard electrode potentials, and from this, obtain AG" for the cell reaction. Calculate AF. Use these values of AN and AG to obtain AS for the cell reaction. Ca²+ (aq) +20 + Ca() --2.76 V Cu? (g) +20 + Cu(s) - 0.84 V AH;(O.*()) -- 542.8 kJ/mol AH;(Out (as)) - 64.8 kJ/mol V AG- AH- kJ...
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...
5. The overall cell reaction in a galvanic cell is Sn(s) + Cu* (aq)>Sn (aq)+ Cu (s). (a) Draw a diagram of this cell (similar to Figure 2) (b) Use standard reduction potentials (Table 1) to calculate the theoretical cell voltage, el
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