Answer
0.368 V
Work Step by Step
$E^{\circ}_{cell}=\frac{0.0592\,V}{n}\log K$
Number of electrons transferred $n=2$
Equilibrium constant $K=2.69\times10^{12}$
Then, $E^{\circ}_{cell}=\frac{0.0592\,V}{2}\times\log (2.69\times10^{12})=0.368\,V$
Chemistry 12th Edition
by Chang, Raymond; Goldsby, Kenneth
Published by
McGraw-Hill Education
ISBN 10:
0078021510
ISBN 13:
978-0-07802-151-0
Chapter 18 - Electrochemistry - Questions & Problems - Page 850: 18.24
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