# https://www.homeworkmarket.com/questions/two-lab-reports-19245609

Electrochemical Cells and Cell Potentials

Objective:
The purpose of this experiment is to create and experiment galvanic cell and collect/interpret data by using a multimeter to describe the flow of electrons. The we g=had to determine how it is calculated by using the formulas given.
Procedure:
Exercise 1: Construction of a Galvanic Cell
Exercise 1: Construction of a Galvanic Cell

Data Table 1. Spontaneous Reaction Observations.

Metal in Solution

Observations

Zinc in Copper Sulfate

The solution turns colorless and black precipitate forms

Copper in Zinc Sulfate

No change

Data Table 2. Voltmeter Readings.

Time (minutes)

0

1.1

15

1.057

30

0.984

45

0.975

60

0.968

75

0.957

90

0.953

105

0.950

120

0.949

135

0.947

Data Table 3. Standard Cell Potential.

Equation

E°(Volts)

Oxidation Half-Reaction

+0.76

Reduction Half-Reaction

+0.34

Redox Reaction

+1.10

Data Table 4. Galvanic Cell Setup.

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.

Concentration of copper solution and zinc solution was taken as 1 M each. And the salt bridge is not an issue to be considered.
As the reaction proceeds, although movement of anion allows the overall charges in solution to remain neutral, the net movement of anion produces a concentration gradient across the two solutions. In other words, after a while the net concentration of ions in the zinc oxidation side will be greater than in the coper reduction side. This concentration gradient will oppose movement of anion. In consideration of keeping the overall concentration of ions in balance between the two sides, cations will also be moving to the right.

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.

There is a little drop of resulting electrical energy from the standard cell potential of reaction.
The cause of voltage drop is: The actual electrical resistances in the “circuits”, the cells themselves”. The cell potentials that you calculate are the “ideal” situation and you would get those if there was not some electrical resistance. But like every machine has some friction, every circuit has some resistance, and the effect of the resistance is to lower the potential difference, the voltage of the cell. The internal resistance is not something you can get rid of. You can minimize it by using “clean” electrodes and keeping the distances between electrodes short.
You may also have an error in your voltmeter. You want to use a voltmeter with a high input impedance (resistance), ideally around 10 megohms. The voltmeter itself becomes part of the circuit and gives false readings. Avoid cheap voltmeters with lower input impedances
C. Was there evidence of electron transfer from the anode to the cathode? Use your data in Data Table 2 to explain your answer.

The cell potential is a measure of the tendency of the anode metal to be oxidized (lose electrons) and the tendency of the metal ions in the cathode compartment to be reduced (gain electrons). Thus, cell potentials vary with the composition of the substances being used as electrodes. Cell potentials also vary with the concentrations of ions and pressures of gases, as well as the temperature at which the reaction occurs. As the redox reaction proceeds, and the electrons travel from anode to the cathode, the total cell potential will decrease. Another evidence is that the direction of electron transfer is tested by dipping the copper electrode directly in zinc solution and zinc electrode directly in copper solution to see which electrode becomes plated with the ion of the solution.

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.

Equation

Eo(volts)

-0.34

Fe+3 (aq) + e- Fe+2(s)

+0.77

+0.78V

()*1 Eo = -0.34

(Fe+3 (aq) + e- Fe+2(s))*2 Eo= 0.77*2=1.54

Eo= Ereduction + Eoxidation = 1.54- (-0.34) = 1.88

Reducers lose electrons , so is reducer

And oxidizer gains electrons so Fe+3 (aq) + e- Fe+2(s) is oxidizer

3+2+2+
(s)(aq)(aq)(s)
Redox reaction: Cu + Fe Cu + Fe
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