Introductory Chemistry: An Active Learning Approach
Introductory Chemistry: An Active Learning Approach
6th Edition
ISBN: 9781305079250
Author: Mark S. Cracolice, Ed Peters
Publisher: Cengage Learning
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Chapter 19, Problem 51E
Interpretation Introduction

Interpretation:

The oxidizer and reducer with oxidized and reduced products are to be stated. The oxidation and reduction half-reaction equations are to be stated. The balanced redox equation is to be stated.

Concept introduction:

The oxidizer is the species whose oxidation state falls during the course of reaction and reducer is the species whose oxidation number increases. Oxidized product is the oxidation product of the reducer and reduced product is the reduction product of the oxidizer.

Expert Solution & Answer
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Answer to Problem 51E

The oxidizer is NO3 and reducer is As2O3 and their corresponding reduced and oxidized product are NO and AsO43.

The oxidation half-reaction equation is shown below.

As2O3+5H2O2AsO43+10H++4e

The reduction half-reaction equation is shown below.

NO3+4H++3eNO+2H2O

The balanced redox equation is shown below.

4NO3+3As2O3+7H2O4NO+14H++6AsO43

Explanation of Solution

The given redox reaction equation to be balanced is shown below.

As2O3+NO3AsO43+NO

The oxidation state of the central metal atom is calculated by knowing the standard oxidation states of few elements.

The oxidation state of arsenic in As2O3 is calculated below.

Step-1: Write down the oxidation number of every element and for unknown take “n”.

As2   O3n     2

Step-2: Multiply the oxidation state with their number of atoms of an element.

As2   O32n     3(2)

Step-3: Add the oxidation numbers and set them equal to the charge of the species.

As2   O32n+3(2)=0

Calculate the value of n by simplifying the equation as shown below.

2n+3(2)=02n+(6)=02n=0+62n=+6

Divide the equation by two on both sides and simplify as shown below.

2n2=+62n=+3

The oxidation state of arsenic in As2O3 is +3.

The oxidation state of arsenic in AsO43 is calculated below.

Step-1: Write down the oxidation number of every element and for unknown take “n”.

As   O4n     2

Step-2: Multiply the oxidation state with their number of atoms of an element.

As       O4n     4(2)

Step-3: Add the oxidation numbers and set them equal to the charge of the species.

As       O4n+4(2)=3

Calculate the value of n by simplifying the equation as shown below.

n+4(2)=3n+(8)=3n=3+8n=+5

The oxidation state of arsenic in AsO43 is +5.

The oxidation state of the nitrogen in NO3 is calculated below.

Step-1: Write down the oxidation number of every element and for unknown take “n”.

N     O3n   2

Step-2: Multiply the oxidation state with their number of atoms of an element.

N     O3n   3(2)

Step-3: Add the oxidation numbers and set them equal to the charge of the species.

N     O3n+3(2)=1

Calculate the value of n by simplifying the equation as shown below.

n+3(2)=1n+(6)=1n=1+6n=+5

The oxidation state of nitrogen is +5 in NO3.

The oxidation number of nitrogen in NO is calculated below.

Step-1: Write down the oxidation number of every element and for unknown take “n”.

N     On   2

Step-2: Multiply the oxidation state with their number of atoms of an element.

N     On   2

Step-3: Add the oxidation numbers and set them equal to the charge of the species.

N     On+(2)=0

Calculate the value of n by simplifying the equation as shown below.

n+(2)=0n2=0n=+2

The oxidation state of nitrogen in NO is +2.

The nitrogen in NO3(N=+5) is getting reduced because the oxidation state of nitrogen in the product NO(N=+2) is less than in the reactant. Similarly, the arsenic is getting oxidized from the reactant As2O3(As=+3) to the product AsO43(As=+5).

Therefore, the oxidizer is NO3 and reducer is As2O3 and their corresponding reduced and oxidized product are NO and AsO43.

The oxidation half-reaction equation for the above equation is shown below.

As2O3AsO43

The balancing of the half-reactions is done by the following the steps shown below.

Step-1: Identify and balance the element getting oxidized or reduced.

The arsenic is getting oxidized and the number of atoms of that is not balanced on both sides. Multiply AsO42 by two on the right-hand side of the equation.

As2O32AsO43

Step-2: Balance elements other than oxygen and hydrogen if any.

As2O32AsO43

Step-3: Balance oxygen atoms by adding water on the appropriate side.

Oxygen atoms are balanced by adding five water molecule to the left-hand side of the equation.

As2O3+5H2O2AsO43

Step-4: Balance the hydrogen atoms by adding H+ to the relevant side when necessary.

The number of hydrogen atoms is balanced by adding the ten H+ on the right-hand side of the equation.

As2O3+5H2O2AsO43+10H+

Step-5: Balance the charge by adding electrons to the appropriate side.

Six electrons are added to the right-hand side in order to balance the charge.

As2O3+5H2O2AsO43+10H++4e

Step-6: Recheck the equation to be sure that it is perfectly balanced.

The equation is completely balanced and is shown below.

As2O3+5H2O2AsO43+10H++4e…(1)

The reduction half-reaction for the above reaction is shown below.

NO3NO

The balancing of the half-reactions is done by the following the steps shown below.

Step-1: Identify and balance the element getting oxidized or reduced.

The nitrogen is getting reduced and its number of atoms are balanced on both sides.

NO3NO

Step-2: Balance elements other than oxygen and hydrogen if any.

NO3NO

Step-3: Balance oxygen atoms by adding water on the appropriate side.

The number of oxygen atoms is balanced by adding two water molecules on the right-hand side of the equation.

NO3NO+2H2O

Step-4: Balance the hydrogen atoms by adding H+ to the relevant side when necessary.

The number of hydrogen atoms is balanced by adding four H+ ions on the left-hand side of the equation.

NO3+4H+NO+2H2O

Step-5: Balance the charge by adding electrons to the appropriate side.

The charge is balanced by adding three electrons on the left-hand side of the equation.

NO3+4H++3eNO+2H2O

Step-6: Recheck the equation to be sure that it is perfectly balanced.

The equation is completely balanced and is shown below.

NO3+4H++3eNO+2H2O…(2)

The balanced redox equation is obtained by adding equation (1) and (2) in such a way that electrons are canceled out.

Multiply equation (1) by three and equation (2) by four in order to cancel out the number of electrons as shown below.

4NO3+16H++12e4NO+8H2O3As2O3+15H2O6AsO43+30H++12e

Add equation (3) and (1) to get a balanced redox equation as shown below.

4NO3+16H++12e4NO+8H2O+3As2O3+15H2O6AsO43+30H++12e

The common things on both sides of the equation canceled out to give the balanced redox equation.

The balanced redox equation after adding these equations is shown below.

4NO3+3As2O3+7H2O4NO+14H++6AsO43

Conclusion

The oxidizer and reducer with oxidized and reduced products, oxidation and reduction half-reaction equations, and balanced redox equation are rightfully stated above.

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Chapter 19 Solutions

Introductory Chemistry: An Active Learning Approach

Ch. 19 - Prob. 11ECh. 19 - Identify each of the following half-reaction as...Ch. 19 - Prob. 13ECh. 19 - Prob. 14ECh. 19 - Prob. 15ECh. 19 - Prob. 16ECh. 19 - Prob. 17ECh. 19 - Prob. 18ECh. 19 - Prob. 19ECh. 19 - Prob. 20ECh. 19 - Prob. 21ECh. 19 - Prob. 22ECh. 19 - Prob. 23ECh. 19 - Prob. 24ECh. 19 - Prob. 25ECh. 19 - Prob. 26ECh. 19 - Prob. 27ECh. 19 - Prob. 28ECh. 19 - Prob. 29ECh. 19 - Prob. 30ECh. 19 - Prob. 31ECh. 19 - Prob. 32ECh. 19 - Prob. 33ECh. 19 - Prob. 34ECh. 19 - Prob. 35ECh. 19 - Prob. 36ECh. 19 - Prob. 37ECh. 19 - Prob. 38ECh. 19 - Prob. 39ECh. 19 - Prob. 40ECh. 19 - Prob. 41ECh. 19 - Prob. 42ECh. 19 - Prob. 43ECh. 19 - In this section, each equation identifies an...Ch. 19 - Prob. 45ECh. 19 - Prob. 46ECh. 19 - Prob. 47ECh. 19 - Prob. 48ECh. 19 - Prob. 49ECh. 19 - Prob. 50ECh. 19 - Prob. 51ECh. 19 - Prob. 52ECh. 19 - Prob. 53ECh. 19 - Prob. 54ECh. 19 - Prob. 55ECh. 19 - Prob. 56ECh. 19 - Prob. 57ECh. 19 - Prob. 58ECh. 19 - As an example of an electrolytic cell, the text...Ch. 19 - Prob. 60ECh. 19 - Prob. 61ECh. 19 - Prob. 62ECh. 19 - Prob. 19.1TCCh. 19 - Prob. 19.2TCCh. 19 - Prob. 19.3TCCh. 19 - Prob. 1CLECh. 19 - Prob. 2CLECh. 19 - Prob. 3CLECh. 19 - Prob. 4CLECh. 19 - Prob. 5CLECh. 19 - Prob. 1PECh. 19 - Prob. 2PECh. 19 - Prob. 3PECh. 19 - Prob. 4PECh. 19 - Prob. 5PECh. 19 - Prob. 6PECh. 19 - Consider the reaction of copper and nitric acid:...Ch. 19 - Prob. 8PECh. 19 - Prob. 9PECh. 19 - Prob. 10PECh. 19 - Prob. 11PECh. 19 - Aqueous chromate ion, CrO42(aq), and hydrogen...Ch. 19 - Prob. 13PE
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