Explanation Separate the reaction into oxidation and reduction half-reactions. In this case, you can see that P b ( s ) is oxidized. The given balanced equation is as below. 2 C l O 2 ( g ) + P b ( s ) → 2 C l O 2 − ( a q ) + P b 2 + ( a q ) O x i d a t i o n : P b ( s ) → P b 2 + ( a q ) + 2 e − Reduction : 2 C l O 2 ( g ) + 2 e − → 2 C l O 2 − ( a q ) Look up the standard electrode potentials for each half-reaction in Table 19.1. Add the half-cell reactions together to obtain the overall redox equation: A t a n o d e , o x i d a t i o n : P b ( s ) → P b 2 + ( a q ) + 2 e − E º = − 0

Chemistry: A Molecular Approach & Student Solutions Manual for Chemistry: A Molecular Approach, Books a la Carte Edition Package

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ISBN: 9780321955517

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Chapter 18, Problem 3SAQ

Interpretation Introduction

Interpretation: Ecell° for the following reaction is to be calculated.

2 ClO2(g)+Pb(s) →2 ClO2–(aq)+Pb2+(aq)

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Chapter 18, Problem 2SAQ

Chapter 18, Problem 4SAQ

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Chapter 18 Solutions

Chemistry: A Molecular Approach & Student Solutions Manual for Chemistry: A Molecular Approach, Books a la Carte Edition Package

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E c e l l ° for the following reaction is to be calculated. 2 C l O 2 ( g ) + P b ( s ) → 2 C l O 2 – ( a q ) + P b 2 + ( a q )

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Interpretation: Ecell° for the following reaction is to be calculated. 2 ClO2(g)+Pb(s) →2 ClO2–(aq)+Pb2+(aq)

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Chapter 18 Solutions

Interpretation: Ecell° for the following reaction is to be calculated.

2 ClO2(g)+Pb(s) →2 ClO2–(aq)+Pb2+(aq)