Electrochemistry Analytical Chemistry
Essay by Sally Zhang • November 6, 2017 • Lab Report • 1,566 Words (7 Pages) • 1,104 Views
Electrochemistry
Introduction
The purpose for this lab was to understand reduction oxidation, fundamental electrochemical cells and electrochemical processes. In the lab, we will use electrochemistry to generate sufficient voltage to light a LED bulb, explore the factors that affect voltage by constructing simple galvanic cells, and use electroplating to deposit a shiny coating of copper on to stainless steel.
Data/Results
Voltaic Piles:
(a)Three different variables
Configuration | Original | The order of the voltaic pile was reverse | Adding more filter paper into the voltaic pile | The number of the cells in voltaic pile were halved |
ΔV (V) | 0.391 | -0.661 | 0.598 | 0.327 |
(b)
1. Two units separated by salt water soaked paper: | 2. Two units stacked Al to Al/Cu to Cu: | 3. Two units stacked Cu to Al: |
ΔV =0.728V | ΔV=0.371V | ΔV=0.291V |
Galvanic Cells:
Part 1:
Anode | Cathode | Predicted E(V) | Measured E(V) |
Zn | Cu | 1.1 | 0.833 |
Al | Zn | 0.9 | 0.703 |
Mg | Al | 0.71 | 0.604 |
Al | Cu | 2.00 | 0.777 |
Zn | Fe | 0.32 | 0.323 |
Fe | Cu | 0.78 | 0.732 |
The actual ranking of relative potential: Cu2+ >Fe2+>Zn2+>Al3+>Mg2+
Part 2
Solution 1 | Solution 2 | Predicted E(V) | Measured E(V) |
1M ZnSO4 | 1M CuSO4 | 1.1 | 0.833 |
1M ZnSO4 | 0.1M CuSO4 | 1.07 | 0.743 |
0.1M ZnSO4 | 1M CuSO4 | 1.13 | 0.804 |
0.1M ZnSO4 | 0.1M CuSO4 | 1.1 | 0.72 |
1M CuSO4 | 0.01M CuSO4 | 0.0592 | 0.0704 |
Electroplating:
Initial mass of copper strip (g) | 9.3081 |
Final mass of copper strip (g) | 9.2350 |
Initial mass of stainless steel strip (g) | 0.4248 |
Final mass of stainless steel strip (g) | 0.4938 |
Average current (A) | 0.249 |
Time of current application (s) | 600 |
Total current applied (A) | 0.249 |
Discussion
Voltaic Piles:
In the experiment of voltaic piles, first we test the voltage of original configuration, which gave us 0.391 V. Then we tried to light the LED bulb, but it was not light. The LED bulb usually need about 1.5 volts, but the voltage difference of the voltaic pile we measured in the lab was 0.591V. Therefore, the voltaic pile didn’t have enough voltage to light the bulb. The reason of this effect might be the oxide layer of the aluminum foil, which caused a reacting barrier between aluminum and electrolyte.
After that, we determined three variables: reversing the order of the voltaic pile, adding more filter paper into the voltaic pile, and halving the number of the cells in voltaic pile were halved. First, when the order of voltaic pile was reversed, the voltage difference became negative because the electrons flew reversely. Secondly, when we add more filter paper into the voltaic pile, the voltage difference had increased. The electric potentials of two electrodes were combined to generate the voltage and the voltage would increase to a certain level due to the concentration increased. Third, when we halved the number of the cells in voltaic pile, we got the similar voltage as the original configuration. What we expected was that the voltage would be halved if the number of the cells in voltaic pile were halved. If we considered voltage of half number of the cells in voltaic pile was measured correct. We should get the voltage of original configuration was around 0.65V. Based on the voltage of reverse configuration, the voltage of half number of the cells was reasonable. Then we could figure out that the expected voltage of original configuration was similar as the value of the reverse configuration. Therefore, there existed some errors when we measured the original configuration, which might due the the setting up the voltaic pile. For some filter paper might not cover the coin well, which caused lower voltage. Based on the results of these three variable we got from the lab, we could figure out that reverse configuration of the voltaic pile had the greatest effect on voltage because the voltage of the pile had changed from positive to negative when the order of voltaic pile was reversed.
In the lab, aluminum was oxidized and hydrogen ions were reduced. If copper ions are present, the copper will be involved in the redox reaction (Cu2+ + 2e-→ Cu)for the voltaic pile. However, insteading of reacting with the copper, the aluminum reacted with hydrogen because oxidability of hydrogen was stronger than copper. Therefore, the copper would be consider as a conductor.
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