Part 2 Redox reaction Experimental Theoretical Cell potential Cell potential a. Zn(0.1M)-Cu(0.1M) 0.92 b. Zn(1M)-Cu(0.1M) 0.9 c. Zn(0.1M)-Cu(1M) 0.95 d. Zn(0.1M)-Zn(0.1M) 0.002 e. Zn(0.1M)-Zn(1M) 0.08 f. Cu(0.1M)-Cu(1M) 0.019 Theoretical cell potential E = E° cell –RT/nF InQ Q = [ product]/[reactant] R= 8.314 T= 298.15 F=96485 n=2 electron E° cell = E redox – E° oxidation
Part 2 Redox reaction Experimental Theoretical Cell potential Cell potential a. Zn(0.1M)-Cu(0.1M) 0.92 b. Zn(1M)-Cu(0.1M) 0.9 c. Zn(0.1M)-Cu(1M) 0.95 d. Zn(0.1M)-Zn(0.1M) 0.002 e. Zn(0.1M)-Zn(1M) 0.08 f. Cu(0.1M)-Cu(1M) 0.019 Theoretical cell potential E = E° cell –RT/nF InQ Q = [ product]/[reactant] R= 8.314 T= 298.15 F=96485 n=2 electron E° cell = E redox – E° oxidation
Chapter22: Bulk Electrolysis: Electrogravimetry And Coulometry
Section: Chapter Questions
Problem 22.30QAP
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Question
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![reactions appears on the list as it occurs. The other half-reaction occurs as oxidation and its
reverse is on the list. The oxidation potential is the negative of the listed reduction
potential. The standard cell potential is the sum of the reduction potential of the half-
reaction that occurs in the cathode compartment and the oxidation potential of the half-
reaction that occurs in the anode compartment. For the Zn/Cu Cell, the standard cell
potential, E°cell = E°red + E°ox = 0.34 + 0.76 = 1.10 V. For galvanic cells, E°cell is always
positive.
galvantc cell
Half reaction
E° (V)
Cu2+ + 2e → Cu
0.34
Fe2+ + 2e → Fe
-0.44
Zn2+ + 2e → Zn
-0.76
Table 16.1: Standard reduction potentials at 25 °C
Nernst equation
The Nernst equation allows us to determine the cell potential at non-standard conditions.
lues
RT
Ecell = E°cell -InQ,
where E°cell is the standard cell potential,
the gas constant (8.314 J/mol K), T the
temperature in K, F the Faraday constant (96485 C/mol), and Q the reaction quotient.
the iron and coper
In this laboratory experiment, you will study galvanic cells. You will construct different
cells, measure their cell potentials, and study the effect of concentration and temperature
on it. To study the temperature dependence, you need the following two equations:
AG° = AH° – TAS° and AG° = –NFE°
Combine these two equations to derive a relationship between E° and AH° and AS°.
Inotsb
Element/ion AH;° (kJ/mol) AG (kJ/mol) S° (J/mol K)oy
Cu(s)
Cu2"(aq)
Fe(s)
Fe2 (aq)
Zn(s)
33.2
-99.6
phise adhitN
64.8
65.5
27.3
0.
-78.9
-137.7
-89.1
26ulsv s
41.6
0.
-112.1
Zn*(aq)
-153.9
-147.2
Table 16.2: Thermodynamic Data for 25 °C
Experiment 16](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4f0bc5bd-8fc9-4f27-98a3-417216a571cb%2Ffe135108-efd4-41b8-ac4b-7363d3805c27%2Fvq1f3e_processed.jpeg&w=3840&q=75)
Transcribed Image Text:reactions appears on the list as it occurs. The other half-reaction occurs as oxidation and its
reverse is on the list. The oxidation potential is the negative of the listed reduction
potential. The standard cell potential is the sum of the reduction potential of the half-
reaction that occurs in the cathode compartment and the oxidation potential of the half-
reaction that occurs in the anode compartment. For the Zn/Cu Cell, the standard cell
potential, E°cell = E°red + E°ox = 0.34 + 0.76 = 1.10 V. For galvanic cells, E°cell is always
positive.
galvantc cell
Half reaction
E° (V)
Cu2+ + 2e → Cu
0.34
Fe2+ + 2e → Fe
-0.44
Zn2+ + 2e → Zn
-0.76
Table 16.1: Standard reduction potentials at 25 °C
Nernst equation
The Nernst equation allows us to determine the cell potential at non-standard conditions.
lues
RT
Ecell = E°cell -InQ,
where E°cell is the standard cell potential,
the gas constant (8.314 J/mol K), T the
temperature in K, F the Faraday constant (96485 C/mol), and Q the reaction quotient.
the iron and coper
In this laboratory experiment, you will study galvanic cells. You will construct different
cells, measure their cell potentials, and study the effect of concentration and temperature
on it. To study the temperature dependence, you need the following two equations:
AG° = AH° – TAS° and AG° = –NFE°
Combine these two equations to derive a relationship between E° and AH° and AS°.
Inotsb
Element/ion AH;° (kJ/mol) AG (kJ/mol) S° (J/mol K)oy
Cu(s)
Cu2"(aq)
Fe(s)
Fe2 (aq)
Zn(s)
33.2
-99.6
phise adhitN
64.8
65.5
27.3
0.
-78.9
-137.7
-89.1
26ulsv s
41.6
0.
-112.1
Zn*(aq)
-153.9
-147.2
Table 16.2: Thermodynamic Data for 25 °C
Experiment 16
![Part 2
Redox reaction
Experimental
Theoretical
Cell potential
Cell potential
a. Zn(0.1M)-Cu(0.1M)
0.92
b. Zn(1M)-Cu(0.1M)
0.9
c. Zn(0.1M)-Cu(1M)
0.95
d. Zn(0.1M)-Zn(0.1M)
0.002
e. Zn(0.1M)-Zn(1M)
0.08
f. Cu(0.1M)-Cu(1M)
0.019
Theoretical cell potential E = E° cell –RT/nF InQ
Q = [ product]/[reactant]
R= 8.314
T= 298.15
F=96485
n=2 electron
E° cell = E° redox – E° oxidation](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4f0bc5bd-8fc9-4f27-98a3-417216a571cb%2Ffe135108-efd4-41b8-ac4b-7363d3805c27%2F5qlxqje_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Part 2
Redox reaction
Experimental
Theoretical
Cell potential
Cell potential
a. Zn(0.1M)-Cu(0.1M)
0.92
b. Zn(1M)-Cu(0.1M)
0.9
c. Zn(0.1M)-Cu(1M)
0.95
d. Zn(0.1M)-Zn(0.1M)
0.002
e. Zn(0.1M)-Zn(1M)
0.08
f. Cu(0.1M)-Cu(1M)
0.019
Theoretical cell potential E = E° cell –RT/nF InQ
Q = [ product]/[reactant]
R= 8.314
T= 298.15
F=96485
n=2 electron
E° cell = E° redox – E° oxidation
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