FINAL LAB PRACTICAL–General Chemistry II Lab
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Part I: Equilibrium Constant Determination:
The equilibrium for the formation of FeSCN^{2+} has been achieved by combining 0.002 M solutions of Fe(NO_{3})_{3} and NaSCN.
Reaction studied: Fe^{3+} + SCN^{1} <=> FeSCN^{2+}, K_{f} = [FeSCN^{2+}]/[Fe^{3+}][ SCN^{1}]
Based on Beer’s Law concept, we achieved a linear plot by various dilutions of reactants (these data are not given). The linear equation is shown below.
A = 2622.52 x [FeSCN^{2+}] 2.691E4 R = 0.99965
Remember to use the formula C_{i}V_{i} = C_{f}V_{f} to determine the initial concentrations right before they react. (Note: V_{f} is 6.00 mL every time.)
Table 1: Initial concentration of Fe^{3+} and SCN^{–} solutions. (10 points)
Soln #  Volume of 2.00E3 M Fe^{3+} (mL)  Volume of 2.00E3 M SCN^{1}
(mL) 
Volume of water (mL)  Total Volume (mL)  [Fe^{3+}]o
M 
[SCN^{1}]o
M 
Blank  3.00  0.00  3.00  6.00  1.00E3  0 
1B  1.00  3.00  2.00  6.00  3.33E4  1.00E3 
2B  2.00  3.00  1.00  6.00  6.67E4  1.00E3 
3B  3.00  3.00  0.00  6.00  1.00E3  1.00E3 
4B  3.00  2.00  1.00  6.00  1.00E3  6.67E4 
5B  3.00  1.00  2.00  6.00  1.00E3  3.33E4 
Use Beer’s Law to determine the concentration from the absorbance provided and use ICE charts to determine the equilibrium concentrations and the K_{f}.
(15 points)
Soln #  Absorbance measured  [FeSCN^{2+}]eq
M 
[Fe^{3+}]eq
M 
[SCN^{1}]eq
M 
K_{f} 
Blank  —  —  —  —  — 
1B  0.105  0.0000399  0.0002901  0.0009601  14.33 
2B  .225  0.0000857  0.0005813  0.0009143  161.25 
3B  .345  0.000131  0.000869  0.000869  173.47 
4B  .451  0.000172  0.000828  0.000495  419.63 
5B  .564  0.000215  0.000785  0.000118  2321.06 
Calculate the average K_{f} and its standard deviation and perform the 4D test, if necessary, i.e., based on the discarding of n unwanted value. (5 points)
STANDARD DEVIATION
Mean = 14.33+161.25+173.47+419.63+2321.06 =3089.74/5 =617.95
x  xmean  (xmean)^{2} 
14.33  603.62  364357.1044 
161.25  456  207936 
173.47  444.48  197562.47 
419.63  198.32  39330.8224 
2321.06  1703.11  2900583.672 
Sum=3089.74  0  3709770.069 
Variance = 3709770.069/ 51 = 741954.0138
Standard deviation =
= 861.4
(a) Show calculations for the initial diluted concentrations of Fe^{3+} and SCN^{– }
(5 points)
[Fe^{3+}]o
M 
[SCN^{1}]o
M 
_{ }
1B = C_{i}V_{i} = C_{f}V_{f } 2.00E3 M X 1.00 = 6.00 X C_{f} _{ }C_{f }= 3.333 E4

1B = C_{i}V_{i} = C_{f}V_{f }
2.00E3 M X 3.00 = 6.00 X C_{f} _{ }C_{f }= 1.00 E3

2B = C_{i}V_{i} = C_{f}V_{f }
2.00E3 M X 2.00 = 6.00 X C_{f} _{ }C_{f }= 6.67E4

2B = C_{i}V_{i} = C_{f}V_{f }
2.00E3 M X 3.00 = 6.00 X C_{f} _{ }C_{f }= 1.00 E3

3B = C_{i}V_{i} = C_{f}V_{f }
2.00E3 M X 3.00 = 6.00 X C_{f} _{ }C_{f }= 1.00 E3

3B = C_{i}V_{i} = C_{f}V_{f }
2.00E3 M X 3.00 = 6.00 X C_{f} _{ }C_{f }= 1.00 E3

4B = C_{i}V_{i} = C_{f}V_{f }
2.00E3 M X 3.00 = 6.00 X C_{f} _{ }C_{f }= 1.00 E3

4B = C_{i}V_{i} = C_{f}V_{f }
2.00E3 M X 2.00 = 6.00 X C_{f} _{ }C_{f }= 6.67 E4

5B = C_{i}V_{i} = C_{f}V_{f }
2.00E3 M X 3.00 = 6.00 X C_{f} _{ }C_{f }= 1.00 E3

5B = C_{i}V_{i} = C_{f}V_{f }
2.00E3 M X 1.00 = 6.00 X C_{f} _{ }C_{f }= 3.333 E4

(b) Show the determination of equilibrium FeSCN2+ concentration from absorbance using Beer’s Law linear regression equation. (5 points)
1B
A = 2622.52 x [FeSCN^{2+}] 2.691E4
0.105 = 2622.52 x [FeSCN^{2+}] 2.691E4
0.105 – 0.0002691 = 2622.52[FeSCN^{2+}]
[FeSCN^{2+}] = = 3.99 E5
2B
A = 2622.52 x [FeSCN^{2+}] 2.691E4
0.225 = 2622.52 x [FeSCN^{2+}] 2.691E4
0.225 – 0.0002691 = 2622.52[FeSCN^{2+}]
[FeSCN^{2+}] = = 8.57E5
3B
A = 2622.52 x [FeSCN^{2+}] 2.691E4
0.345 = 2622.52 x [FeSCN^{2+}] 2.691E4
0.345 – 0.0002691 = 2622.52[FeSCN^{2+}]
[FeSCN^{2+}] = = 1.31E4
4B
A = 2622.52 x [FeSCN^{2+}] 2.691E4
0.451 = 2622.52 x [FeSCN^{2+}] 2.691E4
0.451 – 0.0002691 = 2622.52[FeSCN^{2+}]
[FeSCN^{2+}] = = 1.72E4
5B
A = 2622.52 x [FeSCN^{2+}] 2.691E4
0.564= 2622.52 x [FeSCN^{2+}] 2.691E4
0.564 – 0.0002691 = 2622.52[FeSCN^{2+}]
[FeSCN^{2+}] = = 2.15E4
(c) Show the ce table to calculate the equilibrium concentrations of Fe^{3+} and SCN^{–} and K_{f} calculation. (10 points)
From, Fe^{3+} + SCN^{1} <=> FeSCN^{2+}
^{ }
Trial 1B
Fe^{3+} SCN^{1 }FeSCN^{2+}
Initial  0.00033  0.001  0 
Change  0.000330.0000399  0.0010.0000399  0.0000399 
Equilibrium  0.0002901  0.0009601  0.0000399 
K_{f} = [FeSCN^{2+}]/[Fe^{3+}][ SCN^{1}] = (0.0000399)/ (0.002901)X (0.0009601)
K_{f = }14.33
Trial 2B
Fe^{3+} SCN^{1 }FeSCN^{2+}
Initial  0.000667  0.001  0 
Change  0.0006670.0000857  0.0010.0000857  0.0000857 
Equilibrium  0.0005813  0.0009143  0.0000857 
K_{f} = [FeSCN^{2+}]/[Fe^{3+}][ SCN^{1}] = (0.0000857)/ (0.0005813)X (0.0009143)
K_{f = }161.25
Trial 3B
Fe^{3+} SCN^{1 }FeSCN^{2+}
Initial  0.001  0.001  0 
Change  0.0010.000131  0.0010.000131  0.000131 
Equilibrium  0.000869  0.000869  0.000131 
K_{f} = [FeSCN^{2+}]/[Fe^{3+}][ SCN^{1}] = (0.000131)/ (0.000869)X (0.000869)
K_{f = }173.47
Trial 4B
Fe^{3+} SCN^{1 }FeSCN^{2+}
Initial  0.001  0.000667  0 
Change  0.0010.000172  0.0006670.000172  0.000172 
Equilibrium  0.000828  0.000495  0.000172 
K_{f} = [FeSCN^{2+}]/[Fe^{3+}][ SCN^{1}] = (0.000172)/ (0.000495)X (0.000828)
K_{f = }419.66
Trial 5B
Fe^{3+} SCN^{1 }FeSCN^{2+}
Initial  0.001  0.000333  0 
Change  0.0010.000215  0.0003330.000215  0.000215 
Equilibrium  0.000785  0.000118  0.000215 
K_{f} = [FeSCN^{2+}]/[Fe^{3+}][ SCN^{1}] = (0.000215)/ (0.000785)X (0.000118)
K_{f = }2321.06
(d) Write a summary of the argument on the formation of FeSCN^{2+} complex (10 points)
The deductions from the experiments above indicate that absorbance is directly proportional to the concentration of the reagents. Trials 2B and 3B have close results. The Kf value for ttrials1B and 5B differs from others by a big margin. These are the unwanted values since they are out of range, even if you consider the standard deviation. Therefore, trials 2 and 3 give a clear hypothesis to illustrate the relationship between the Beers regression law and how they relate to absorbance and concentrations of different solutions.
(II) Standard Cell Potentials (Voltaic Cells):
Galvanic Cell
A galvanic cell is an important electrochemical cell. It is named after Luigi Galvani, an Italian physicist. It is also called a Voltaic cell, after an Italian physicist, Alessandro Volta. A galvanic cell generally consists of two different metal rods called electrodes. Each electrode is immersed in a solution containing its onions, and these form a halfcell. Each halfcell is connected by a salt bridge or separated by a porous membrane. The solutions in which the electrodes are immersed are called electrolytes.
The chemical reaction that takes place in a galvanic cell is the redox reaction. One electrode acts as the anode in which oxidation takes place, and the other acts as the cathode in which reduction takes place. The best example of a galvanic cell is the Daniell cell.
The cell voltage is proportional to ΔG, the change in Gibbs free energy. The ΔG of the reaction could be harnessed to do useful electrical work. This thermodynamic equation gives the value of ΔG.
ΔG = ΔG° + RT ln Q — (1)
The relationship between cell voltage, E, and ΔG for the cell reaction is given by the following equation.
DG = nFE, or E = DG/nF —(2), where n is number of electrons, F = 96485 J/mol.V
The Nernst equation showing the cell potential can be finally written as
(or)
Where E° = E°_{1/2}(reduction) − E°_{1/2}(oxidation)
E° = E°_{1/2}(cathode) − E°_{1/2}(anode)
In this investigation, you will make the following voltaic cells by using the following simulation:
http://amrita.olabs.edu.in/?sub=73&brch=8&sim=153&cnt=4
Activity series: Use cell notation and perform the virtual lab.
Part I  Part II  
Anode
(oxidation) 
Cathode
(reduction) 
Anode
(oxidation) 
Cathode
(reduction) 
(1) Mg/Mg^{2+}(0.05 M)  Zn^{2+} (0.1 M)/Zn (2) Mg/Mg^{2+}( (0.05 M)  2Ag^{+} (0.1 M)/2Ag (3) Mg/Mg^{2+}( (0.05 M)  Cu^{2+}(0.1 M)/Cu (4) Mg/Mg^{2+}( (0.05 M)  Fe^{2+}(0.1 M)/Fe (5) Mg/Mg^{2+}(0.05 M)  Ba^{2+} (0.1 M)/Ba

(6) Zn/Zn^{2+}(0.05 M)  Ba^{2+} (0.1 M)/Ba (7) Zn/Zn^{2+}(0.05 M)  Mg^{2+} (0.1 M)/Mg (8) Zn/Zn^{2+} (0.05 M)  2Ag^{+} (0.1 M)/2Ag (9) Zn/Zn^{2+} (0.05 M)  Cu^{2+}(0.1 M)/Cu (10) Zn/Zn^{2+} (0.05 M)  Fe^{2+}(0.1 M)/Fe 

Procedure:
You will create a series of voltaic cell models in two different parts: (i) Mg involves oxidation consistently, and other ions involve reduction, and (ii) Zn involves oxidation consistently, and other ions involve reduction.
Set the concentration of solution in the anode part (left) as 0.05 M, whereas the concentration of the solution in the cathode part is 0.1 M.
This setup is consistent for all 10 cell models where you can see the type of solutions (e.g., aqueous BaCl_{2}, MgCl_{2}, AgNO_{3}, ZnSO_{4}, CuSO_{4} and FeSO_{4}) where cations are ionized in the solution.
Part I:
 Select the electrode Mg in the anode part (top), where MgCl_{2} (the corresponding ionic solution) will appear in the beaker by default.
 Select the electrode Zn in the cathode part (bottom), where ZnSO_{4} (the corresponding ionic solution) will appear in the beaker by default. This is model #1.
 Observe the cell potential (E_{cell}) from the digital voltmeter.
 Repeat the same procedure with the anode (Mg) and various cathodes (Ag, Cu, Fe and Ba) to complete the rest of the models (Models 25).
Part II:
 Select the electrode Zn in the anode part (top), where ZnSO_{4} (the corresponding ionic solution) will appear in the beaker by default.
 Select the electrode Ba in the cathode part (bottom), where BaCl_{2} (the corresponding ionic solution) will appear in the beaker by default. This is model #6…
 Observe the cell potential (E_{cell}) from the digital voltmeter.
 Repeat the same procedure with the anode (Zn) and various cathodes (Ag, Cu, Fe and Ba) to complete the rest of the models (Models 710).
Table: Cell potential measured for various voltaic cells (by using simulation) in comparison with the theoretical values. (10 points)
Voltaic cell  Electrodes  Anode  Cathode  E_{cell}, V
(Exp.) 
E_{cell}, V
(Pred.)^{a} 
% Error 
1  Mg & Zn  Mg/Mg^{2+}(0.05 M)  Zn^{2+} (0.1 M)/Zn  1.619  1.619  0 
2  Mg & Ag  Mg/Mg^{2+}( (0.05 M)  2Ag^{+} (0.1M)/2Ag  3.17  3.25  2.52 
3  Mg & Cu  Mg/Mg^{2+}( (0.05 M)  Cu^{2+}(0.1 M)/Cu  2.71  2.79  2.95 
4  Mg & Fe  Mg/Mg^{2+}( (0.05 M)  Fe^{2+}(0.1 M)/Fe  1.93  2.01  4.14 
5  Mg & Ba  Mg/Mg^{2+}(0.05 M)  Ba^{2+} (0.1 M)/Ba  0.53  0.44  2.00 
6  Zn & Ba  Zn/Zn^{2+}(0.05 M)  Ba^{2+} (0.1 M)/Ba  2.14  2.05  16.9 
7  Zn & Mg  Zn/Zn^{2+}(0.05 M)  Mg^{2+} (0.1 M)/Mg  1.61  1.52  5.59 
8  Zn & Ag  Zn/Zn^{2+} (0.05 M)  2Ag^{+} (0.1 M)/2Ag  1.56  1.64  5.12 
9  Zn & Cu  Zn/Zn^{2+} (0.05 M)  Cu^{2+}(0.1 M)/Cu  1.1  1.18  7.27 
10  Zn & Fe  Zn/Zn^{2+} (0.05 M)  Fe^{2+}(0.1 M)/Fe  0.32  0.40  25.0 
^{a} Finish the answering question (a) and enter values here.
(a) Calculate the predicted E_{cell} value for each cell by the Nernst Equation. (20 points)
where n = 2 and Q = [0.05]/[0.1] à both are consistent in all 10 models
(1) Model calculation for Model #1: Collect the reduction potential from the table at the bottom of the PPT or other resources.
E^{o} = E^{o}(cathode) – E^{o}(anode) = (0.76 V) – (2.37 V) = +1.61 V
E = E^{o} – (0.0591/n) (log Q) = 1.61 – (0.0591/2) (log [0.05/0.1]) = 1.619 V
From the Desmos scientific calculator
From the Virtual Lab: 
The calculation for Model #2:
E^{o} = E^{o}(cathode) – E^{o}(anode) = (+0.80V) – (2.37 V) = +3.17 V
E = E^{o} – (0.0591/n) (log Q) = 3.17 – (0.0591/2) (log [0.05/0.1]) = 3.25 V
Calculation for Model #3:
E^{o} = E^{o}(cathode) – E^{o}(anode) = (+0.34V) – (2.37 V) = +2.71 V
E = E^{o} – (0.0591/n) (log Q) = 2.71 – (0.0591/2) (log [0.05/0.1]) = 2.79 V
The calculation for Model #4:
E^{o} = E^{o}(cathode) – E^{o}(anode) = (0.44V) – (2.37 V) = +1.93 V
E = E^{o} – (0.0591/n) (log Q) = 1.93 – (0.0591/2) (log [0.05/0.1]) = 2.01V
The calculation for Model #5:
E^{o} = E^{o}(cathode) – E^{o}(anode) = (2.90V) – (2.37 V) = 0.53 V
E = E^{o} – (0.0591/n) (log Q) = 0.53 – (0.0591/2) (log [0.05/0.1]) = 0.44V
Calculation for Model #6:
E^{o} = E^{o}(cathode) – E^{o}(anode) = (2.90V) – (0.76 V) = 2.14 V
E = E^{o} – (0.0591/n) (log Q) = 2.14 – (0.0591/2) (log [0.05/0.1]) = 2.05 V
Calculation for Model #7:
E^{o} = E^{o}(cathode) – E^{o}(anode) = (2.37V) – (0.76 V) = 1.61 V
E = E^{o} – (0.0591/n) (log Q) = 1.61 – (0.0591/2) (log [0.05/0.1]) = 1.52 V
Calculation for Model #8:
E^{o} = E^{o}(cathode) – E^{o}(anode) = (+0.80V) – (0.76 V) = +1.56 V
E = E^{o} – (0.0591/n) (log Q) = 1.56 – (0.0591/2) (log [0.05/0.1]) = 1.64V
Calculation for Model #9:
E^{o} = E^{o}(cathode) – E^{o}(anode) = (+0.34V) – (0.76 V) = +1.1 V
E = E^{o} – (0.0591/n) (log Q) = 1.1 – (0.0591/2) (log [0.05/0.1]) = 1.18V
Calculation for Model #10:
E^{o} = E^{o}(cathode) – E^{o}(anode) = (0.44V) – (0.76 V) = +0.32 V
E = E^{o} – (0.0591/n) (log Q) = 0.32 – (0.0591/2) (log [0.05/0.1]) = 0.40V
(b) Write an overall redox reaction for each voltaic cell first and then halfreactions, which are labeled “oxidation” and “reduction.” (10 points)
Model 1: Mg/Mg^{2+ } Zn^{2+}/Zn
Redox: Mg_{(s)} + Zn^{2+}_{(aq)} à Mg^{2+}_{(aq)} + Zn_{(s)} _{ } Oxidation: Mg à Mg ^{2+} + 2e^{–} Reduction: Zn^{2+} + 2e^{–} àZn 
Model 6: Zn/Zn^{2+ } Ba^{2+}/Ba
Oxidation: Zn à Zn ^{2+} + 2e^{–}
Reduction: Ba^{2+} + 2e^{–} àBa

Model 2: Mg/Mg^{2+ } 2Ag^{+}/ 2Ag
Oxidation: Mg à Mg ^{2+} + 2e^{–}
Reduction: 2Ag^{2+} + 2e^{–} à2Ag

Model 7: Zn/Zn^{2+ } Mg^{2+}/Mg
Oxidation: Zn à Zn ^{2+} + 2e^{–}
Reduction: Mg^{2+} + 2e^{–} àMg 
Model 3: Mg/Mg^{2+ } Cu^{2+}/Cu
Oxidation: Mg à Mg ^{2+} + 2e^{–}
Reduction: Cu^{2+} + 2e^{–} àCu 
Model 8: Zn/Zn^{2+ } 2Ag^{+}/2Ag
Oxidation: Zn à Zn ^{2+} + 2e^{–}
Reduction: 2Ag^{2+} + 2e^{–} à2Ag

Model 4: Mg/Mg^{2+ } Fe^{2+}/Fe
Oxidation: Mg à Mg ^{2+} + 2e^{–}
Reduction: Fe^{2+} + 2e^{–} àFe

Model 9: Zn/Zn^{2+ } Cu^{2+}/Cu
Oxidation: Zn à Zn ^{2+} + 2e^{–}
Reduction: Cu^{2+} + 2e^{–} àCu

Model 5: Mg/Mg^{2+ } Ba^{2+}/Ba
Oxidation: Mg à Mg ^{2+} + 2e^{–}
Reduction: Ba^{2+} + 2e^{–} àBa

Model 10: Zn/Zn^{2+ } Fe^{2+}/Fe
Oxidation: Zn à Zn ^{2+} + 2e^{–}
Reduction: Fe^{2+} + 2e^{–} àFe

(c) Write a summary of the argument based on this simulation lab. (10 points)
From the simulations, it is evident that the standard electrode potentials provided determine whether a reaction will take place or not. Also, the most reactive metals from the table undergo oxidation, thus strong reducing agents. The least reactive metals undergo reduction, hence strong oxidizing agents. The value of the electrode potential determines the position of a metal in a reactivity series. For example, copper has an electrode potential of +0.34 V, while magnesium has 2.37V. Therefore, copper cannot displace magnesium from a solution of its ions, but magnesium can displace copper. Overall, negative electrode potential from a complete cell indicates that a reaction cannot take place and vice versa.
Appendix Table: Reduction potential for various electrode and reduction reactions.
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Question
This is a lab practical; please complete all pages and show calculations and graphs as asked.