The following values may be useful when solving this tutorial. In the activity,
ID: 829946 • Letter: T
Question
The following values may be useful when solving this tutorial.
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.
Constant Value E?Cu 0.337 V E?Fe -0.440V R 8.314 J?mol?1?K?1 F 96,485 C/mol T 298 KExplanation / Answer
Cu2+(aq) + Fe(s) => Cu(s) + Fe2+(aq)
Eo(cell) = Eo(Cu2+/Cu) - Eo(Fe2+/Fe)
= 0.337 - (-0.440) = 0.777 V
Molar gas constant R = 8.314 J/mol.K
Temperature T = 298 K
Moles of electrons transferred n = 2
Faraday constant F = 96485 C/mol
Delta Go = -nFEo(cell) = -RT ln Keq
Equilibrium constant Keq = exp(nFEo(cell)/RT)
= exp(2 x 96485 x 0.777/(8.314 x 298))
= 1.92 x 10^26 ( = 1.92 x 1026)
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