The following values may be useful when solving this tutorial.

Constant Value

E∘Cu 0.337 V

E∘Fe -0.440 V

R 8.314 J⋅mol−1⋅K−1

F 96,485 C/mol

T 298 K

Part A

In the activity, click on the E∘cell 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)+2e−→Cu(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.

what i did – incorrect

Cu2+(aq)+2e- –> Cu(s) E^o=-0.337v

Fe(s) –> Fe2+(aq)+2e- E^o=-0.440v

——-

Cu2+(aq)+2e−→Cu(s) and Fe(s)→Fe2+(aq)+2e− E^o= -0.337-(-0.440) = 0.103v

E^o cell = (RT/nF)ln Keq

ln Keq = E^o cell (RT/nF)

ln Keq = 0.103v (8.314×298)/(2x 96485)

ln Keq = 8.02

Keq = e^8.02

Keq = 3048.28 three significant figures

3.05×10^3

Where did i do wrong? Thanks a lot 🙂

ln K = nFEcell/RT is what you want to use.

Ecell is +0.337 + 0.440 = ?

The Evalue for Cu is given in the problem as 0.337 and that’s for Cu^2+ + 2e = Cu(s). You should not have changed the sign.

E vlue for Fe^2+ + 2e = Fe(s) is given in the problem as -0.440. Since the reaction is the reverse of that in the question you change the sign, then add the two.

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