Interpretation Introduction Interpretation: Each of the given redox reaction taking place in aqueous acidic medium is t be balanced. Explanation We have the following redox reaction: Z n ( s ) + S n + 2 ( a q ) → Z n + 2 ( a q ) + S n ( s ) The above redox reaction is already balanced both in terms of charge and elements on both sides. So, there is no need to balance. We have the following redox reaction: M g ( s ) + C r + 3 ( a q ) → M g + 2 ( a q ) + C r ( s ) 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 : M g ( s ) → M g + 2 ( a q ) Reduction : C r + 3 ( a q ) → C r ( s ) In these half-reactions, all elements are balanced. Hence, in the next step we need to balance the charge. Step 2: Balance each half-reaction with respect to charge by adding electrons. O x i d a t i o n : M g ( s ) → M g + 2 ( a q ) + 2 e − Reduction : C r + 3 ( a q ) + 3 e − → C r ( s ) Step 3: 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 : 3 × [ M g ( s ) → M g + 2 ( a q ) + 2 e − ] Reduction : 2 × [ C r + 3 ( a q ) + 3 e − → C r ( s ) ] Add these two reactions 3 M g ( s ) → 3 M g + 2 ( a q ) + 6 e − 2 C r + 3 ( a q ) + 6 e − → 2 C r ( s ) 3 M g ( s ) + 2 C r + 3 ( a q ) → 3 M g + 2 ( a q ) _ ¯ + 2 C r ( s ) Therefore, the balanced equation of the reaction is 3 M g ( s ) + 2 C r + 3 ( a q ) → 3 M g + 2 ( a q ) + 2 C r ( s ) We have the following redox reaction: M n O 4 − ( a q ) + A l ( s ) → M n + 2 ( a q ) + A l + 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 : A l ( s ) → A l + 3 ( a q ) Reduction : M n O 4 − ( a q ) → M n + 2 ( a q ) Step 2: Balance the elements excluding H, O. O x i d a t i o n : A l ( s ) → A l + 3 ( a q ) Reduction : M n O 4 − ( a q ) → M n + 2 ( a q ) Step 3: Balance the elements H, O

Chemistry: A Molecular Approach, Books a la Carte Plus Mastering Chemistry with Pearson eText -- Access Card Package (4th Edition)

4th Edition

ISBN: 9780134162454

Author: Nivaldo J. Tro

Publisher: PEARSON

See similar textbooks

Videos

Question

Chapter 19, Problem 38E

Interpretation Introduction

Interpretation: Each of the given redox reaction taking place in aqueous acidic medium is t be balanced.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 19, Problem 37E

Chapter 19, Problem 39E

Students have asked these similar questions

4. Provide a clear arrow-pushing mechanism for the following reactions. Do not skip proton transfers, do not combine steps, and make sure your arrows are clear enough to be interpreted without ambiguity. a) NHBoc ⚫OBn HO. H3C CO2CH3 -OBn H3C H3C. H3C. NHBOC CI CO2CH3

Draw structures of the following compounds and identify their role: mCPBA (MCPBA) DMS Py 9-BBN LAH Sia₂BH TsCI PCC t-BuOK LDA MeLi n-BuLi DMSO DMF Sodium Borohydride Lithium DiisopropylAmide 2

Using Luther's rule, calculate the reference potential of the Hg2+/Hg redox electrode. DATA: Electrode potentials E° = 0,854 V y E 0,788 V Hg2+/Hg 2+ Hg2/Hg

Chapter 19 Solutions

Chemistry: A Molecular Approach, Books a la Carte Plus Mastering Chemistry with Pearson eText -- Access Card Package (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

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.