Answer
The change in cell voltage is 0.0592V
Work Step by Step
E = E$^0$-$\frac{0.0592}{2}$$\times$log$\frac{[Fe^{2+}]}{[Ag^+]{^2}}$
E = E$^0$-$\frac{0.0592}{2}$$\times$log$\frac{[1]}{[10]{^2}}$=0.0592 V
Chemistry: The Central Science (13th Edition)
by Brown, Theodore E.; LeMay, H. Eugene; Bursten, Bruce E.; Murphy, Catherine; Woodward, Patrick; Stoltzfus, Matthew E.
Published by
Prentice Hall
ISBN 10:
0321910419
ISBN 13:
978-0-32191-041-7
Chapter 20 - Electrochemistry - Exercises - Page 900: 20.8c
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