Interactive Activity—The Relationship among E°cell, Keq, and Gibbs Free Energy S

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Interactive Activity—The Relationship among E°cell, Keq, and Gibbs Free Energy

Select the image to explore the activity that shows how Ecell, Keq, and G are related to each other.

Constants

The following values may be useful when solving this tutorial.

Part A

Cu2+(aq)+2eCu(s) and Fe(s)Fe2+(aq)+2e

Cu2+(aq)+Fe(s)Cu(s)+Fe2+(aq)

Express your answer numerically to three significant figures.

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Part B

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Part C

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Interactive Activity—The Relationship among E°cell, Keq, and Gibbs Free Energy

In an electrochemical cell, the potential difference between two electrodes under standard conditions is known as the standard cell potential (Ecell). The standard cell potential can be used to identify the overall tendency of a redox reaction to occur spontaneously. The spontaneity of a reaction is identified using the Gibbs free energy G. Gis related to Ecell. Ecell and G are also related to equilibrium constant Keq of the reaction.

Select the image to explore the activity that shows how Ecell, Keq, and G are related to each other.

In the activity, you should see a triangle, whose three vertices represent Keq, Ecell, and G. You can select two vertices and determine the relation between them. You can then reset the activity and select the next two quantities.

Constants

The following values may be useful when solving this tutorial.

Constant Value ECu 0.337 V EFe -0.440 V R 8.314 Jmol1K1 F 96,485 C/mol T 298 K

Part A

In the activity, click on the Ecell and Keq quantities to observe how they are related. Use this relation to calculate Keq for the following redox reaction that occurs in an electrochemical cell having two electrodes: a cathode and an anode. The two half-reactions that occur in the cell are

Cu2+(aq)+2eCu(s) and Fe(s)Fe2+(aq)+2e

The net reaction is

Cu2+(aq)+Fe(s)Cu(s)+Fe2+(aq)

Use the given standard reduction potentials in your calculation as appropriate.

Express your answer numerically to three significant figures.

Keq =

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Part B

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Part C

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Explanation / Answer

Cu2+(aq)+Fe(s) Cu(s)+Fe2+(aq)

We know that Eocell = Eocathode - Eoanode

                               = EoCu/Cu2+ - EoFe2+/Fe

                               = 0.337 - (-0.440)

                               = 0.777 V

Also we have Eocell = [(RT)/(nF) ]lnKeq

Where

R = gas constant = 8.314 J/mol-K

T = temperature = 298 K

n = number of electrons transferred = 2

F = faraday = 96485 C/mol

Keq = equilibrium constant = ?

Plug the values we get lnKeq = (nFEocell ) / (RT)

                                            = 60.52

                                     Keq = e60.52

                                            = 1.92x1026

Therefore the equilibrium constant is 1.92x1026

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