Introductory Chemistry (5th Edition)

Published by Pearson
ISBN 10: 032191029X
ISBN 13: 978-0-32191-029-5

Chapter 16 - Oxidation and Reduction - Exercises - Cumulative Problems - Page 606: 104


0.022 mol

Work Step by Step

Given half reaction is $Au^{3+}$ (aq) + 3e- ---> Au(s) So, we can conclude that 3 mol of electrons can be used to electroplate 1 mole of Au. 1 mol of Au= 196.9 g So to electroplate 1.4g of Au, number of moles electrons required is: = 1.4 g x( 3 mole of electrons/ 196.9 g ) = 0.022 mol
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