Constants The following values may be useful when solving this tutorial. Part A
ID: 691949 • Letter: C
Question
Constants
The following values may be useful when solving this tutorial.
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 Ni(s)Ni2+(aq)+2e
The net reaction is
Cu2+(aq)+Ni(s)Cu(s)+Ni2+(aq)
Use the given standard reduction potentials in your calculation as appropriate.
Constant Value ECu 0.337 V ENi -0.257 V R 8.314 Jmol1K1 F 96,485 C/mol T 298 KExplanation / Answer
Get E°cell first
Remember that each species will have a specific reduction potential. Remember that this is, as the name implies, a potential to reduce. We use it to compare it (numerical) with other species.
Note that the basis if 2H+ + 2e- -> H2(g) reduction. Therefore E° = 0 V
All other samples are based on this reference.
Find the Reduction Potential of each reaction (Tables)
ECu 0.337 V .
ENi -0.257 V
The most positive has more potential to reduce, it will be reduced
The most negative will be oxidized, since it will donate it selectrons
For total E°cell potential:
E°cell = Ered – Eox
Eox = -Ered of the one being oxidized
E°cell = 0.337 - (-0.257 ) = 0.594V
E°cell = 0.594V
b)
dG = -nF*E°cell
dG = -rT*lnK
K = exp(nFEcell/(RT))
K = exp(2*96500*0.594/(8.314*298))
K = 1.246*10^20