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

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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Chapter 19, Problem 38E

Interpretation Introduction

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

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Chapter 19, Problem 37E

Chapter 19, Problem 39E

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(28 pts.) 7. Propose a synthesis for each of the following transformations. You must include the reagents and product(s) for each step to receive full credit. The number of steps is provided. (OC 4) 4 steps 4 steps OH b.

LTS Solid: AT=Te-Ti Trial 1 Trial 2 Trial 3 Average ΔΗ Mass water, g 24.096 23.976 23.975 Moles of solid, mol 0.01763 001767 0101781 Temp. change, °C 2.9°C 11700 2.0°C Heat of reaction, J -292.37J -170.473 -193.26J AH, kJ/mole 16.58K 9.647 kJ 10.85 kr 16.58K59.64701 KJ mol 12.35k Minimum AS, J/mol K 41.582 mol-k Remember: q = mCsAT (m = mass of water, Cs=4.184J/g°C) & qsin =-qrxn & Show your calculations for: AH in J and then in kJ/mole for Trial 1: qa (24.0969)(4.1845/g) (-2.9°C)=-292.37J qsin = qrxn = 292.35 292.37J AH in J = 292.375 0.2923kJ 0.01763m01 =1.65×107 AH in kJ/mol = = 16.58K 0.01763mol mol qrx Minimum AS in J/mol K (Hint: use the average initial temperature of the three trials, con Kelvin.) AS=AHIT (1.65×10(9.64×103) + (1.0 Jimai

Chapter 19 Solutions

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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