The following values may be useful when solving this. Constant Value ECu 0.337V
ID: 497042 • Letter: T
Question
The following values may be useful when solving this. Constant Value
ECu 0.337V
ECo -0.277 V
R 8.314 Jmol1K1
F 96,485 C/mol
T 298 K
Part A In the activity, click on the E cell and K quantities to observe how they are related. Use this relation to calculate K 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 Co(s)Co2+(aq)+2e The net reaction is Cu2+(aq)+Co(s)Cu(s)+Co2+(aq) Use the given standard reduction potentials in your calculation as appropriate. Express your answer numerically to three significant figures.
Explanation / 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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