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Answer: Balance each redox reaction occurring in basic

Introductory Chemistry | 5th Edition | ISBN: 9780321910295 | Authors: Nivaldo J Tro ISBN: 9780321910295 34

Solution for problem 70P Chapter 16

Introductory Chemistry | 5th Edition

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Introductory Chemistry | 5th Edition | ISBN: 9780321910295 | Authors: Nivaldo J Tro

Introductory Chemistry | 5th Edition

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Problem 70P

Balance each redox reaction occurring in basic solution.

(a) NO2− (aq) + Al(s) → NH3(g) + AlO2− (aq)

(b) Al(s) + MnO4− (aq) → MnO2(s) + Al(OH)4− (aq)

Step-by-Step Solution:
Step 1 of 3

Solution: Here, we are going to balance the given redox reactions.

Step1:

NO2-(aq) + Al(s) → NH3(g) + AlO2-(aq)

The two half reactions are:

Oxidation: Al(s) → AlO2-(aq)         [ Al oxidized from 0 to +3]

Reduction: NO2- (aq) → NH3 (g)         [ N reduced from +3 to -3 ]

Step2:

To balance the O atoms in the reduction half reaction, we add two water molecule on the right:

NO2- (aq) → NH3 (g) + 2H2O

To balance the H atoms, we add seven H+ ions on the left:

NO2- (aq) + 7H+ → NH3 (g) + 2H2O

As the reaction takes place in a basic solution, therefore, for seven H+ ions, we add seven OH– ions to both sides of the equation:

         NO2- (aq) + 7H+ + 7OH- → NH3 (g) + 2H2O + 7OH-

Replacing the H+ and OH– ions with water, the resultant equation is:

 NO2- (aq) + 7H2O → NH3 (g) + 2H2O + 7OH-

  NO2- (aq) + 5H2O → NH3 (g) + 7OH-

Similarly, to balance the two oxygen atoms in the oxidation half reaction, we add two water molecules on the left and so on:  

                                Al(s) + 2H2O → AlO2-(aq)

                                Al(s) + 2H2O → AlO2-(aq) + 4H+

                        Al(s) + 2H2O + 4OH- → AlO2-(aq) + 4H+  + 4OH-

                        Al(s) + 2H2O + 4OH- → AlO2-(aq) + 4H2O

                        Al(s)...

Step 2 of 3

Chapter 16, Problem 70P is Solved
Step 3 of 3

Textbook: Introductory Chemistry
Edition: 5
Author: Nivaldo J Tro
ISBN: 9780321910295

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Answer: Balance each redox reaction occurring in basic

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