The following values may be useful when solving this tutorial.
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
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
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
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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