Equilibrium and le chatelier's principle lab answers

Two Lab Reports

I would like you to write two lab reports about two different experiments. I will upload you the instructions on how does the report have to be done. Also, I will upload you the experiment file ( this file will contain everything data, question and answers, purpose, etc..) I would like you to rewrite this answers I wrote on your own words. I will also upload you a sample for an old lab report. I want two pages for each experiment four in total. Thank you!

Mohammed Alnaimi

Student ID 48408 Lesson Electrochemical Cells and Cell Potentials Institution Ocean County College Session 2019L3 CHEM 182 DL1 Course CHEM 182 DL1 Instructor Nancy Marashi

Final Report

Exercise 1

CHEM 182 DL1 Electrochemical Cells and Cell Potentials

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.

1. What were the concentrations of the solutions (zinc solution, copper solution, and salt bridge)? Were the concentrations consistent with those of standard state condit ions? Explain your answer.

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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

2. Was the electric potential f ound for 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.

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.

3. Was there evidence of electron transfer f rom the anode to the cathode? Use your data in Data Table 2 to explain your answer.

4. 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 f rom the background as needed. Explain how you identif ied which substance was oxidized and which was reduced. Show all of your work.

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( Cu ---> Cu + 2e ) E = -0.34

( Fe + e ---> Fe )*2 E = 0.77*2=1.54

( Cu + Fe ---> Cu + Fe ) E = E + E = 1.54- (-0.34) = 1.88

Reducers lose electrons, so ( Cu ---> Cu + 2e ) is a reducer

And oxidizer gains electrons so ( Fe + e ---> Fe ) is an oxidizer

Table 1. Standard reduction potentials.

0(s ) 2+ (aq) - o

+3 (aq)

- +2 (s )

o

0 (s )

+3 (aq)

2+ (aq)

+2 (s )

o reducti on oxidati on

0(s ) 2+ (aq) -

+3(aq) - +2(s )

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Data Table 1: Spontaneous Reaction Observations

Data Table 2: Multimeter Readings

Metal in Solution: Observations

Zinc in Copper (II) Sulfate Solution

Copper in Zinc Sulfate Solution

The solution turns colorless and black precipitate forms

No change

Time (minutes) Multimeter Reading (Volts)

0

15

30

45

60

75

90

105

120

135

150

1.1

1.057

0.984

0.975

0.968

0.957

0.953

0.950

0.949

0.947

0.945

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Data Table 3: Standard Cell Potential

Photo 1: Galvanic Cell Set-up

Photo 2: Metal Strips After Experiment

Equation

Oxidation Half-Reaction

Reduction Half-Reaction

Redox Reaction

E°(Volts)

Oxidation Half-Reaction

Reduction Half-Reaction

Redox Reaction

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