Chemistry (7th Edition)

Published by Pearson
ISBN 10: 0321943171
ISBN 13: 978-0-32194-317-0

Chapter 18 - Electrochemistry - Section Problems - Page 804: 128


3.22 g

Work Step by Step

Time, t= 20.0 min= $(20.0\times60)s$= 1200 s Current, I= 2.40 A Charge=$ I\times t$=$ 2.40 A\times 1200 s$= 2880 C According to the reaction: $Ag^{+}+ e^{-}→Ag $ we require 1F or 96487 C to deposit 1 mol or 107.87 g of Ag at the cathode. For 2880 C, the mass of Ag obtained= $\frac{107.87gmol^{-1}\times2880 C}{1\times96487Cmol^{-1}}$ = 3.22 g
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