Experiment 9
Results and Questions (Discussion)
A.
Reduction Potentials of Several Redox Couples
Galvanic Measured
Cell
Ecell (V)
Cu-Zn
0.891
Cu-Mg
1.704
Cu-Fe
0.645
Zn-Mg
0.766
Fe-Mg
1.040
Zn-Fe
0.291
Anode
Zn
Mg
Fe
Mg
Mg
Zn
Equation for
Anode Reaction
Zn → Zn2+ + 2e
Mg → Mg2+ + 2e
Fe → Fe2+ + 2e
Mg → Mg2+ + 2e
Mg → Mg2+ + 2e
Zn → Zn2+ + 2e
Cathode
Cu
Cu
Cu
Zn
Fe
Fe
Equation for
Cathode Reaction
Cu2+ + 2e → Cu
Cu2+ + 2e → Cu
Cu2+ + 2e → Cu
Zn2+ + 2e → Zn
Fe2+ + 2e → Fe
Fe2+ + 2e → Fe
1. Write the balanced net reaction for each of the galvanic cells.
Zn (s) + Cu2+ (aq) → Zn2+ (aq) + Cu (s)
Example: Cu-Zn
2.
0.766V
Mg
Zn
Cu
1.704 V
Based on Zn-Mg = 0.766 V, Mg-Cu = 1.704 V
Zu-Cu cell potential = 1.704 – 0.766 V = 0.938 V
Zn-Cu cell potential ( experiment) = 0.891 V Value nearly the same.
3.
0.291 V
Mg
Zn
Fe
0.766 V
Sum of Zn-Fe and Zn-Mg cell potentials
= 0.291 + 0.766 V
= 1.057 V
Fe-Mg cell potential (experiment) 1.040 V Value nearly the same.
4.
Redox Couple
Cu2+/ Cu
Fe2+/ Fe
Zn2+/Zn
Mg2+/Mg
Reduction Potential Reduction Potential
(Measured) (V)
(Calculated) (V)
0.101
0.310
-0.499
-0.440
-0.790
-0.76
-1.556
-0.2410
% Error
Sample Calculation for Reduction Potential (Measured) (V)
(a)
Ecell = Ecathode - Eanode
ECu2+/Cu – EZn2+/Zn = Ecell
ECu2+/Cu – (-0.79) = 0.891 V
ECu2+/Cu = 0.107 V
Sample Calculation for Reduction Potential (Calculated) (V)
0.0592
[ Zn 2 ]
(a)
Ecell = Eocell log
2
[Cu 2 ]
ECu2+/Cu – EZn2+/Zn = EoCu2+/Cu – EoZn2+/Zn -
ECu2+/Cu – EZn2+/Zn = [0.34- (-0.76)] -
0.0592
[ Zn 2 ]
log
2
[Cu 2 ]
0.0592
[0.1 ]
log
2
[0.1 ]
ECu2+/Cu – (-0.79) = 1.1
ECu2+/Cu = 0.31 V
These 2 values have to find from
Standard Reduction Potential Table from you text book.
% Error
Cu2+/Cu =
B.
0.31  0.101
 100%
0.31
= 67.21 %
Effect of Concentration Changes On Cell Potential
1. Cell Potential of “Concentrated Cell” = 0.051 V
Anode reaction : 1 M CuSO4
Cathode reaction : 0.001 M CuSO4
A potential is recorded because of the difference in the solutions
concentration.
2. Cell Potential from complex formation: 0.445 V
Anode:
Cathode:
Potential changed with the addition of NH3 (aq) because there are more
Cu2+ ions flow from cathode and the concentration of Cu2+ ions decrease.
3. Cell Potential from precipitate formation: 0.5 V
Observation of solution in half cell:
Potential changed with the addition of Na2S because of the formation of
CuS released electrons. As there are more electrons flowing from anode to
cathode, thus potential increased.
4. The cell potential would have decreased if NH3(aq) and/ or the Na2S(aq)
had been added to 1 M CuSO4 instead of 0.001 M CuSO4 solution of the
cell because the mole concentration of 1M CuSO4 has more copper than
0.001 M CuSO4
C.
The Nerst Equation and An Unknown Concentration
Solution
No.
1
2
3
4
[Cu(NO3)2]
(M)
0.1
0.001
0.00001
0.0000001
Log [Cu2+]
-1
-3
-5
-7
Ecell
(Measured)
0.962 V
0.874V
Sample Calculation
Cu(NO3)2 0.1 M
(b)
Ecell = Eocell -
0.0592
[ Zn 2 ]
log
2
[Cu 2 ]
ECu2+/Cu – EZn2+/Zn = EoCu2+/Cu – EoZn2+/Zn = 0.34 – (-0.76)
= 1.1 V.
0.0592
[0.1 ]
log
2
[0.1]
Ecell
(Calculated)
1.100 V
Experiment 10
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
Trial 1
0.388
0.412
900
0.15
0.345
0.456
0.044
6.925 × 10-4
1.385 × 10-3
135
8.4375 × 1020
6.092 × 1023
Trial 2
0.375
0.456
900
0.14
0.328
0.503
0.047
7.397 × 10-4
1.479 × 10-3
126
7.875 × 1020
5.323 × 1023
5.708 × 1023
6.022 × 1023
- 5.22 %
9.748 × 104
8.517 × 104
9.132 × 104
9.65 × 104
- 5.36 %
All Calculations follow the formulas in your lab manual.