Interpretation Introduction Interpretation: For each of the given reaction, the value of Δ G rxn º is to be determined by the use of tabulated electrode potentials. Explanation The relation between Δ G rxn o and E cell o is given as: Δ G rxn º = - n F E cell º (a) For given reaction: Pb + 2 ( aq ) + Mg ( s ) → Pb ( s ) + Mg + 2 ( aq ) Oxidation : Mg ( s ) → Mg + 2 ( aq ) + 2 e - E oxi º = - E red º = 2.37 V Reduction : Pb + 2 ( aq ) + 2e - → Pb ( s ) E cell º = - 0.13 V Pb + 2 ( aq ) + Mg ( s ) → PB ( s ) + Mg + 2 ( aq ) ¯ E cell º = E oxi º + E red º = 2.37 V - 0.13 V = + 2.24 V But, Δ G rxn º = - n F E cell º = − 2 m o l e − × 96 , 485 C 1 m o l e − × 2.24 J C × 1 k J 1000 J = − 432.25 k J ≈ − 432 k J (b) For given reaction: Br 2 ( l ) + 2Cl - ( aq ) → 2Br - ( aq ) + Cl 2 ( g ) Oxidation : 2Cl - ( aq ) → Cl 2 ( g ) + 2e - E oxi º = - E red º = - 1.36 V Reduction : Br 2 ( aq ) + 2e - → 2 Br - ( aq ) E red º = 1.09 V Br 2 ( l ) + 2Cl - ( aq ) → 2 Br - ( aq ) + Cl 2 ( g ) ¯ E cell º = E oxi º + E red º = - 1

Chemistry: A Molecular Approach Plus Mastering Chemistry with Pearson eText -- Access Card Package (4th Edition) (New Chemistry Titles from Niva Tro)

4th Edition

ISBN: 9780134103976

Author: Nivaldo J. Tro

Publisher: PEARSON

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Chapter 19, Problem 65E

Interpretation Introduction

Interpretation: For each of the given reaction, the value of ΔGrxnº is to be determined by the use of tabulated electrode potentials.

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Chapter 19, Problem 64E

Chapter 19, Problem 66E

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Chemistry: A Molecular Approach Plus Mastering Chemistry with Pearson eText -- Access Card Package (4th Edition) (New Chemistry Titles from Niva Tro)

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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