The following values may be useful when solving this tutorial. Part A In the act
ID: 918879 • Letter: T
Question
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
Find Keq
Ke needs G°
G° = -nFE°Cell
E°cell = Ered - Eox
Ered = 0.337 - (-0.257 ) = 0.594 V
F = 96500 and n = 2
threfore
G° = -nFE°Cell
G° = -2*96500*0.594 = -114642
G = -RT*lnK
K = exp(G/(-RT)) = exp(114642 /(8.314*298)) = 1.246*10^20
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