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Balance each redox reaction occurring in basic solution.(a) ClO? (aq) + Cr(OH)4? (aq) ?

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

Solution for problem 69P 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 69P

Problem 69P

Balance each redox reaction occurring in basic solution.

(a) ClO− (aq) + Cr(OH)4− (aq) → CrO42− (aq) + Cl− (aq)

(b) MnO4− (aq) + Br− (aq) → MnO2(s) + BrO3− (aq)

Step-by-Step Solution:
Step 1 of 3

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

Step1:

  1. ClO-(aq) + Cr(OH)4-(aq) → CrO42-(aq) + Cl-(aq)

The two half reactions are:

Oxidation: Cr(OH)4- (aq) → CrO42- (aq)         [ Cr oxidized from +3 to +6]

Reduction: ClO- (aq) → Cl- (aq)         [ Cl reduced from +1 to -1 ]

Step2:

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

ClO- (aq) → Cl- (aq) + H2O

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

ClO- (aq) + 2H+ → Cl- (aq) + H2O

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

         ClO- (aq) + 2H+ + 2OH- → Cl- (aq) + H2O + 2OH-

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

  ClO- (aq) + 2H2O → Cl- (aq) + H2O + 2OH-

 ClO- (aq) + H2O → Cl- (aq) + 2OH-

Similarly, to balance the four hydrogen atoms in the oxidation half reaction, we add four H+ ions on the right:  

                                Cr(OH)4- (aq) + 4OH- → CrO42- (aq) + 4H+ + 4OH-

                                Cr(OH)4- (aq) + 4OH- → CrO42- (aq) + 4H2O

Step3:

In this step we balance the charges of the two half-reactions by using e- in the manner depicted as:

 ClO- (aq) + H2O + 2e- → Cl- (aq) + 2OH-

Cr(OH)4- (aq) + 4OH- → CrO42- (aq) + 4H2O + 3e-

Step4:

Now to equalise the number of electrons, we multiply the oxidation half-reaction by 2 and the reduction half-reaction by 3.        

                         3ClO- (aq) + 3H2O + 6e- → 3Cl- (aq) + 6OH-

2Cr(OH)4- (aq) + 8OH- → 2CrO42- (aq) + 8H2O + 6e-

Step5:

Add two half-reactions to obtain the net reactions after cancelling electrons on both sides.

 3ClO- (aq) + 2Cr(OH)4- (aq) + 3H2O + 8OH-  → 2CrO42- (aq) + 3Cl- (aq) + 8H2O + 6OH-

 3ClO- (aq) + 2Cr(OH)4- (aq) + 2OH-  → 2CrO42- (aq) + 3Cl- (aq) + 5H2O

Step6:

A final verification shows that the equation is balanced in respect of the number of atoms and charges on both sides. Thus, the balanced redox reaction is:

                 3ClO- (aq) + 2Cr(OH)4- (aq) + 2OH-  → 2CrO42- (aq) + 3Cl- (aq) + 5H2O 

Step7:

     b)        MnO4-(aq) + Br-(aq) MnO2(s) + BrO3-(aq)

Following the same steps above, the balanced redox reaction is:

                2MnO4-(aq) + Br-(aq) + H2O → 2MnO2(s) + BrO3-(aq) + 2OH-

                                        --------------------------------

                        

Step 2 of 3

Chapter 16, Problem 69P is Solved
Step 3 of 3

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

This full solution covers the following key subjects: MNO, balance, bro, clo, cro. This expansive textbook survival guide covers 19 chapters, and 2046 solutions. Introductory Chemistry was written by and is associated to the ISBN: 9780321910295. Since the solution to 69P from 16 chapter was answered, more than 423 students have viewed the full step-by-step answer. The full step-by-step solution to problem: 69P from chapter: 16 was answered by , our top Chemistry solution expert on 05/06/17, 06:45PM. This textbook survival guide was created for the textbook: Introductory Chemistry, edition: 5. The answer to “Balance each redox reaction occurring in basic solution.(a) ClO? (aq) + Cr(OH)4? (aq) ? CrO42? (aq) + Cl? (aq)(b) MnO4? (aq) + Br? (aq) ? MnO2(s) + BrO3? (aq)” is broken down into a number of easy to follow steps, and 29 words.

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Balance each redox reaction occurring in basic solution.(a) ClO? (aq) + Cr(OH)4? (aq) ?