Question

# CAn someone please let me know if my answers are correct Exercise 1: Construction of a...

Exercise 1: Construction of a Galvanic Cell

Data Table 1. Spontaneous Reaction Observations.

 Metal in Solution Observations Zinc in Copper Sulfate Zinc turned black Copper in Zinc Sulfate There was no change

 Time (minutes) Multimeter Reading (Volts) 0 1.08 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) à Zn2+(aq) + 2e- -.76 Reduction Half-Reaction Cu2+(aq) + 2e- à Cu(s) .13 Redox Reaction Zn(s) + CuSO4(aq) à ZnSO4(aq) + Cu(s) .89

E cell = 0.13 – (-0.76) = .89

Data Table 4. Galvanic Cell Setup.

 Photograph of galvanic cell

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.

The concentrations of solutions would not be changed if we are taking Zn/ZnSO4 // CusO4/Cu cell. Because how many ions are coming from Zn to ZnSO4. That many ions again reproduced due to an equilibrium reaction. Therefore, all concentrations are - CuSO4 = 1 M, ZnSO4 = 1M

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.

Yes, the concentrations are consistent throughout the experiment. It was proved by the potential values that occurred in the experiment. The obtained/ galvanic meter values are consistent and supporting, to calculate the values of total cell potentials. If we add, both oxidation and reduction potential values from Data Table 3 we get the cell potential. 0.89 V (0.76+0.13). The amount of energy produced in cell is consistent throughout the experiment and it is proportional to total cell potential.

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

Yes, the voltage reading increased overtime. If there was no electron transfer from anode to cathode in the cell, then there would not have been any generation of electrical energy. The values of the Galvano meter (Data Table 2) show the electrons are transferring from anode to cathode.

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.

Oxidation: Cu(s) à Cu2+(aq) + 2e-   E = 0.34

Reduction: Fe3+(aq) + 1e-à Fe2+(s)   E = 0.77

E cell = 0.77 – 0.34 = 0.43 V

The copper is the oxidizer because it gained two electrons during the reaction. The Iron is the reducer because it lost one electron

It seems to be everything is correct, but why suddenly Fe3+, Fe2+ comes in play at last question! May be you have not shown that redox reaction picture! ok

The answer is incorrect for last question. Reason is lower reduction potential metal always behaves as anode means it loses electrons and we say that is reducer.

Correct answer: The copper is the reducer because it loses two electrons during the reaction. The Iron is the oxidizer because it gains one electron.

But you have written Ecell correctly as Ecell = Ecathode - Eanode. the oxidation half-reaction and the reduction-half reaction are correct.

Cathode is iron (reduction happened here means gain of electrons) anode is copper (oxidation happens here means loss of electrons).

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