Explanation We have following redox reaction: M n O 4 − ( a q ) + B r − ( a q ) → M n O 2 ( s ) + B r O 3 − ( a q ) Step 1: Separate the overall reaction into two half-reactions: one for oxidation and one for reduction. O x i d a t i o n : B r − ( a q ) → B r O 3 − ( a q ) R e d u c t i o n : M n O 4 − ( a q ) → M n O 2 ( s ) Step 2: Balance the elements, excluding H, O. O x i d a t i o n : B r − ( a q ) → B r O 3 − ( a q ) R e d u c t i o n : M n O 4 − ( a q ) → M n O 2 ( s ) Step 3: Balance the elements H, O. O x i d a t i o n : B r − ( a q ) + 6 H O − ( a q ) → B r O 3 − ( a q ) + 3 H 2 O ( l ) R e d u c t i o n : M n O 4 − ( a q ) + 2 H 2 O ( l ) → M n O 2 ( s ) + 4 H O − ( a q ) In these half-reactions, all elements are balanced, including H, O. Hence, in the next step we need to balance the charge. Step 4: Balance each half-reaction with respect to charge by adding electrons. O x i d a t i o n : B r − ( a q ) + 6 H O − ( a q ) → B r O 3 − ( a q ) + 3 H 2 O ( l ) + 6 e − R e d u c t i o n : M n O 4 − ( a q ) + 2 H 2 O ( l ) + 3 e − → M n O 2 ( s ) + 4 H O − ( a q ) Step 5: Make the number of electrons in both half-reactions equal by multiplying one or both half-reactions by a small whole number. O x i d a t i o n : [ B r − ( a q ) + 6 H O − ( a q ) → B r O 3 − ( a q ) + 3 H 2 O ( l ) + 6 e − ] × 1 R e d u c t i o n : [ M n O 4 − ( a q ) + 2 H 2 O ( l ) + 3 e − → M n O 2 ( s ) + 4 H O − ( a q ) ] × 2 Add these two reactions to get the balanced reaction. 2 M n O 4 − ( a q ) + B r − ( a q ) + H 2 O ( l ) → 2 M n O 2 ( s ) + 2 H O − ( a q ) + B r O 3 − ( a q ) We have following redox reaction: A g ( s ) + C N − ( a q ) + O 2 ( g ) → A g ( C N ) 2 − ( a q ) Step 1: Separate the overall reaction into two half-reactions: one for oxidation and one for reduction. O x i d a t i o n : A g ( s ) + C N − ( a q ) → A g ( C N ) 2 − ( a q ) Reduction : O 2 ( g ) → H O − ( a q ) Step 2: Balance the elements, excluding H, O. O x i d a t i o n : A g ( s ) + 2 C N − ( a q ) → A g ( C N ) 2 − ( a q ) Reduction : O 2 ( g ) + 2 H 2 O ( l ) → 4 H O − ( a q ) Step 3: Balance the elements H, O. O x i d a t i o n : A g ( s ) + 2 C N − ( a q ) → A g ( C N ) 2 − ( a q ) + 1 e − Reduction : O 2 ( g ) + 2 H 2 O ( l ) + 4 e − → 4 H O − ( a q ) In these half-reactions, all elements are balanced, including H, O

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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Question

Chapter 19, Problem 42E

Interpretation Introduction

Interpretation: Each of the given redox reaction that occurs in aqueous basic medium is to be balanced.

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Chapter 19, Problem 41E

Chapter 19, Problem 43E

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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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