CuSO4-1.0m ZnSO4-1.0M Kcl-1.0M One beakers with 45mL of zinc second beakerwith c

Question

CuSO4-1.0m

ZnSO4-1.0M

Kcl-1.0M

One beakers with 45mL of zinc second beakerwith copper sulfate
Put the salt bridge(1.5 inches folded of filter paper soak in potassium chloride) into place by submerging 1 end in copper sulfate and the other end in zinc sulfate. Clip a piece of zinch to one end of the jumper cable and clip a piece of copper onto the other. Place the copper into copper sulfate solution ans the zinc into the zinc sulfate solution.

Exercise 1: Construction of a Galvanic Cell

Data Table 1. Spontaneous Reaction Observations.

Metal in Solution

Observations

Zinc in Copper Sulfate

Zinc turns black

Copper in Zinc Sulfate

No change occured

Data Table 2. Voltmeter Readings.

Time (minutes)

Voltmeter Reading (Volts)

0

1.07

15

1.08

30

1.08

45

1.08

60

1.08

75

1.08

90

1.08

105

1.08

120

1.05

135

1.04

Data Table 3. Standard Cell Potential.

Equation

E°(Volts)

Oxidation Half-Reaction

Zn(s) ------> Zn2+(aq) + 2e-

0.76V

Reduction Half-Reaction

Cu2+(aq) + 2e- -------> Cu(s)

0.13V

Redox Reaction

Zn(s) + CuSO4(aq) -----------> ZnSO4(aq) + Cu(s)

0.89V

Questions

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.

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.

Redox Reaction: Cu (s) + Fe3+(aq) ----> Cu2+(aq) + Fe2+ (s)

Metal in Solution

Observations

Zinc in Copper Sulfate

Zinc turns black

Copper in Zinc Sulfate

No change occured

Explanation / Answer

Q.1: As stated in the question the concentrations of zinc solution, copper solution, and salt bridge are 1.0 M each.

Q.2: Yes these were consistent with the standard state condition as the standard state condition is also 1.0 M

Q.3: The given redox reaction is

Redox Reaction: Cu (s) + Fe3+(aq) ----> Cu2+(aq) + Fe2+ (s)

Here Cu is oxidizing to Cu2+(aq). Hence

Oxidation-half reaction: Cu(s) ------- > Cu2+(aq) + 2e- : E0(oxi) = - 0.34 V (answer)

Here Fe3+(aq) is reduced to Fe2+(aq). Hence

Reduction-half reaction: Fe3+(aq) + 1e- ------- > Fe2+(aq) : E0(red) = +0.77 V (answer)

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Standard cell potenial, E0(cell) = E0(oxi) + E0(red)

=>  E0(cell) = - 0.34 V + 0.77 V = + 0.43 V (answer)

Note: I have used the E0 values available with me. To get even more accurate answer you can use the values available in Table 1.

Due to higher standard reduction potential (+0.77 V compared to +0.34V) of  Fe3+(aq), it gets reduced and acts as an oxidizer.

Due to higher standard oxidation potential( - 0.34 V compared to - 0.77 V) of Cu(s), it gets oxidized and acts as an reducer.

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