Explanation Three electrolytic cells were connected in series. The electrolytes used in the three cells are C u S O 4 , A u 2 ( S O 4 ) 3 , a n d A g N O 3 , respectively. i) In the first electrolytic cell, we are using copper sulfate ( C u S O 4 ) as the electrolyte. Reduction half-reaction taking place at cathode is C u + 2 ( a q ) + 2 e − → C u ( s ) In this reaction, two moles of electrons are utilized to deposit one mole of Cu. That means 2 m o l e − = 1 m o l C u Amount of copper (Cu) deposited = 1.74 g Amount of current flowed in the cell = 2.33 A We have to calculate the time in seconds required to deposit 1.74 g of Cu. Mass of one mol of Cu is 63.55 g. A 96,485 C of the charge is passed through the cell by 1 mol e − . Then, 63.55 g of Cu can be deposited by the 2 m o l e − (or) 63.55 g of Cu can be deposited by the charge of 2(96,485 C). Charge passed through the cell to deposit 1.74 g of Cu is given by 1.74 g C u × 1 m o l C u 63.55 g C u × 2 m o l e − 1 m o l C u × 96 , 485 C 1 m o l e − = 5.28 × 10 3 C So, 5.28 × 10 3 coulombs of charge is passed through the cell to deposit 1.74 g of Cu. Product of amount of current (A) flowing through the cell and time (sec) in which current flowing gives the total charge flowing through the cell (C) are calculated as follows. Total charge flowing through the cell is 5.28 × 10 3 C . Time taken for the plating is 1.74 g o f C u = 5.28 × 10 3 C × 1 2.33 C s = 5.28 × 10 3 C × 1 s 2.33 C = 2.27 × 10 3 s Therefore, 2.27 × 10 3 s of time will be taken to deposit 1.74 g of Cu at a current of 2.33 A. ii) We have to calculate the mass of gold (Au) deposited at the respective electrode by the flow of 5.28 × 10 3 C charge. Amount of current flowing in the cell = 2.33 A Time = 2.27 × 10 3 s Total charge of the cell is given by Total c h a r g e = C u r r e n t ( A ) × T i m e ( s ) Then, the amount of Au deposited in 2

Modified Mastering Chemistry with Pearson eText -- Standalone Access Card -- for Chemistry: A Molecular Approach (4th Edition)

4th Edition

ISBN: 9780134162430

Author: Nivaldo J. Tro

Publisher: PEARSON

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Electrolysis

Electrolysis is defined as the process of breaking down ionic compounds into their constituent elements by passing a direct electric current through the compound in a fluid state. The cathode reduces cations, while the anode oxidizes anions. An electroly…

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Chapter 19, Problem 136E

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Interpretation: The time up to which the current is applied in the given electrolytic cells is to be determined. Along with this, the mass of gold and silver that was deposited is to be calculated.

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Chapter 19, Problem 135E

Chapter 19, Problem 137E

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Modified Mastering Chemistry with Pearson eText -- Standalone Access Card -- for Chemistry: A Molecular Approach (4th Edition)

Ch. 19 - Prob. 1SAQ Ch. 19 - Q2. Which statement is true for voltaic cells? a)...Ch. 19 - Prob. 3SAQ Ch. 19 - Prob. 4SAQ Ch. 19 - Prob. 5SAQ Ch. 19 - Prob. 6SAQ Ch. 19 - Prob. 7SAQ Ch. 19 - Prob. 8SAQ Ch. 19 - Prob. 9SAQ Ch. 19 - Prob. 10SAQ

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