Electrochemical Cells and Cell Potentials
Hands-On Labs, Inc.
Version 42-0153-00-02
Lab Report Assistant
This document is not meant to be a substitute for a formal laboratory report. The Lab Report Assistant is simply a summary of the experiment’s questions, diagrams if needed, and data tables that should be addressed in a formal lab report. The intent is to facilitate students’ writing of lab reports by providing this information in an editable file which can be sent to an instructor.
Exercise 1: Construction of a Galvanic Cell
Data Table 1. Spontaneous Reaction Observations.
Metal in Solution |
Observations |
Zinc in Copper Sulfate |
Zinc turned into black |
Copper in Zinc Sulfate |
No change occured |
Data Table 2. Voltmeter Readings.
Time (minutes) |
Voltmeter Reading (Volts) |
0 |
-0.98 |
15 |
-0.99 |
30 |
-0.99 |
45 |
-0.99 |
60 |
-0.99 |
75 |
-0.99 |
90 |
-0.98 |
105 |
-0.98 |
120 |
-0.95 |
135 |
-0.94 |
Data Table 3. Standard Cell Potential.
Equation |
E°(Volts) |
|
Oxidation Half-Reaction |
Please tell me about this ? |
? |
Reduction Half-Reaction |
Please tell me about this ? |
? |
Redox Reaction |
Please tell me about this ? |
? |
Data Table 4. Galvanic Cell Setup. Include a piece of paper with your name and date on it.
Photograph of galvanic cell |
Questions A
What were the concentrations of the solutions (zinc solution, copper solution, and salt bridge)? Were the concentrations consistent with those of standard state conditions? Explain your answer.
Question B
Was the amount of electric energy produced in your galvanic cell consistent with the standard cell potential of the reaction (as calculated in Data Table 3)? Hypothesize why it was or was not consistent.
Question C
Was there evidence of electron transfer from the anode to the cathode? Use your data in Data Table 2 to explain your answer.
Part D
For the following redox reaction in a galvanic cell, write the oxidation half-reaction and the reduction-half reaction, and calculate the standard cell potential of the reaction. Use Table 1 in the Background as needed. Explain how you identified which half-reaction is the oxidizer and which is the reducer. Show all of your work.
Cu + Fe +3 ----- > Cu +2 + Fe +2
Please answer the reactions part and All the four question by helping from tble data.
Thank you so much in advance ....
Oxidation half cell
Zn(s) ------> Zn2+(aq) + 2e- E0 = -0.76 V
Reduction half cell
Cu2+(aq) + 2e- -------> Cu(s) E0 = + 0.34 V
Redox reaction Zn(s) + CuSO4(aq) -----------> ZnSO4(aq) + Cu(s) E0 = + 1.10 V
a) The concentrations will not change and will remain the same
[ZnSO4] = 1M
[CuSO4] = 1M
Due to equilibrium the number of ions of Zinc gained from Zn metal will get converted to Zn(s).
so net concentration will remain the same
b) The potential values are same in the experiment, so the concentration are consistent throughout the experiment
c) the negative value shows that the cell is opposite, so the electron transfer is from cathode to anode (which is actually from anode to cathode)
d) The emf values are consistent and galvanic readings are consistent
So amount of energy produced is also consistent throughout the experiment
The half cell which has low reduction potential value will act as anode and the one with more reduction potential value will act as cathode
Anode half cell (oxidation) : Cu ---- > Cu+2 + 2e E0 = +0.34 V
Cathode half cell (reduction) : Fe+3 + e ----- > Fe +2 E0 = +0.77 V
Overall
Cu + Fe+3 ----- > Cu+2 + Fe +2
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