Whether the given reaction is a redox process is to be explained. In the given reaction acid, base, conjugate acid and the conjugate base are to be identified. Concept introduction: An acid is defined as a compound which donates a proton and a proton acceptor is known as base according to Bronsted-Lowry theory. The reaction that shows the change of oxidation state between reactant and product is known as redox reaction . To determine: If the given reaction is a redox process.

Question

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Answer 1

Question

Chapter 16, Problem 16.174AP

(a)

Interpretation Introduction

Interpretation: Whether the given reaction is a redox process is to be explained. In the given reaction acid, base, conjugate acid and the conjugate base are to be identified.

Concept introduction: An acid is defined as a compound which donates a proton and a proton acceptor is known as base according to Bronsted-Lowry theory. The reaction that shows the change of oxidation state between reactant and product is known as redox reaction.

To determine: If the given reaction is a redox process.

(a)

Expert Solution

Answer to Problem 16.174AP

Solution

No, the given reaction is not a redox process.

Explanation of Solution

Explanation

The given reaction is,

$HNO3(aq)+2H2SO4(aq)→NO2+(aq)+H3O+(aq)+2HSO4−(aq)$

The oxidation number of nitrogen and sulfur atoms in reactant and product side is calculated to check if it is a redox reaction. When it shows the change in oxidation state between reactant and product then it is a redox reaction. Therefore, the oxidation number of nitrogen in $HNO3$ is calculated as,

The oxidation state of hydrogen in a compound is usually $+1$ as it contains only one valence electron and of oxygen is usually $−2$ as it requires two more electrons in order to achieve stable electronic configuration. The sum of all oxidation numbers in a neutral compound is zero. The oxidation number of N is assumed to be $x$ and is calculated by using the formula,

$Charge on HNO3=[(Number of H atoms×oxidation number of H)+(Number of N atoms×oxidation number of N)+(Number of O atoms×oxidation number of O)]$

Since, the sum of all oxidation numbers in a neutral compound is zero that is charge on $HNO3$ is zero.

Substitute the number of atoms and their oxidation number in the above formula to calculate the oxidation number of nitrogen.

$Charge on HNO3=[(1×(+1))+1×(x)+3(−2)]0=1+x−60=x−5x=5$

Hence, the oxidation number of nitrogen is $+5$ .

Similarly, in $NO2+$ . The oxidation number of N is assumed to be $x$ and is calculated by using the formula,

$Charge on NO2+=[(Number of N atoms×oxidation number of N)+(Number of O atoms×oxidation number of O)]$

Substitute the number of atoms and their oxidation number in the above formula to calculate the oxidation number of nitrogen.

$Charge on NO2+=[1×(x)+2(−2)]+1=x−4x=4+1x=5$

Hence, the oxidation number of nitrogen is $+5$ .

The oxidation number of sulfur in $H2SO4$ is calculated as,

The oxidation state of oxygen is usually $−2$ as it requires two more electrons in order to achieve stable electronic configuration and of hydrogen is $+1$ as it contains one valence electron. The oxidation number of $S$ is assumed to be $x$ and is calculated by using the formula,

$Charge on H2SO4=[(Number of H atoms×oxidation number of H)+(Number of S atoms×oxidation number of S)+(Number of O atoms×oxidation number of O)]$

Since, the sum of all oxidation numbers in a neutral compound is zero that is charge on $H2SO4$ is zero.

Substitute the number of atoms and their oxidation number in the above formula to calculate the oxidation number of carbon.

$Charge on H2SO4=[(2×1)+(1×x)+(4×(−2))]0=2+x+(−8)0=x−6x=6$

Hence, the oxidation number of sulfur is $+6$ .

Similarly, in $HSO4−$ . The oxidation number of $S$ is assumed to be $x$ and is calculated by using the formula,

$Charge on HSO4−=[(Number of H atoms×oxidation number of H)+(Number of S atoms×oxidation number of S)+(Number of O atoms×oxidation number of O)]$

Substitute the number of atoms and their oxidation number in the above formula to calculate the oxidation number of carbon.

$Charge on HSO4−=[(1×1)+(1×x)+(4×(−2))]−1=1+x+(−8)−1=x−7x=6$

Hence, the oxidation number of sulfur is $+6$ .

There is no change in the oxidation number of the atoms involved in the reaction. Hence, the given reaction is not a redox reaction.

(b)

Interpretation Introduction

To determine: The acid, base, conjugate acid and conjugate base in the given reaction.

(b)

Expert Solution

Answer to Problem 16.174AP

Solution

The acid is $H2SO4$ , $HNO3$ is a base, $HSO4−$ is a conjugate base and $NO2+$ is a conjugate acid.

Explanation of Solution

Explanation

According to Bronsted-Lowry theory, when an acid donates a proton the species formed is known as conjugate base and when the base accepts a proton the species formed is known as conjugate acid. The given reaction is,

$HNO3(aq)+2H2SO4(aq)→NO2+(aq)+H3O+(aq)+2HSO4−(aq)$

In the reaction, $H2SO4$ act as an acid as it donates a proton forming $HSO4−$ as a conjugate base. Nitric acid $(HNO3)$ is a base and forms $NO2+$ conjugate acid.

Conclusion

a. The given reaction is not a redox process.
b. In the given reaction, $H2SO4$ is an acid, $HNO3$ is a base, $HSO4−$ is a conjugate base and $NO2+$ is a conjugate acid.

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Answer 2

Question

Chapter 16, Problem 16.174AP

(a)

Interpretation Introduction

Interpretation: Whether the given reaction is a redox process is to be explained. In the given reaction acid, base, conjugate acid and the conjugate base are to be identified.

Concept introduction: An acid is defined as a compound which donates a proton and a proton acceptor is known as base according to Bronsted-Lowry theory. The reaction that shows the change of oxidation state between reactant and product is known as redox reaction.

To determine: If the given reaction is a redox process.

(a)

Expert Solution

Answer to Problem 16.174AP

Solution

No, the given reaction is not a redox process.

Explanation of Solution

Explanation

The given reaction is,

$HNO3(aq)+2H2SO4(aq)→NO2+(aq)+H3O+(aq)+2HSO4−(aq)$

The oxidation number of nitrogen and sulfur atoms in reactant and product side is calculated to check if it is a redox reaction. When it shows the change in oxidation state between reactant and product then it is a redox reaction. Therefore, the oxidation number of nitrogen in $HNO3$ is calculated as,

The oxidation state of hydrogen in a compound is usually $+1$ as it contains only one valence electron and of oxygen is usually $−2$ as it requires two more electrons in order to achieve stable electronic configuration. The sum of all oxidation numbers in a neutral compound is zero. The oxidation number of N is assumed to be $x$ and is calculated by using the formula,

$Charge on HNO3=[(Number of H atoms×oxidation number of H)+(Number of N atoms×oxidation number of N)+(Number of O atoms×oxidation number of O)]$

Since, the sum of all oxidation numbers in a neutral compound is zero that is charge on $HNO3$ is zero.

Substitute the number of atoms and their oxidation number in the above formula to calculate the oxidation number of nitrogen.

$Charge on HNO3=[(1×(+1))+1×(x)+3(−2)]0=1+x−60=x−5x=5$

Hence, the oxidation number of nitrogen is $+5$ .

Similarly, in $NO2+$ . The oxidation number of N is assumed to be $x$ and is calculated by using the formula,

$Charge on NO2+=[(Number of N atoms×oxidation number of N)+(Number of O atoms×oxidation number of O)]$

Substitute the number of atoms and their oxidation number in the above formula to calculate the oxidation number of nitrogen.

$Charge on NO2+=[1×(x)+2(−2)]+1=x−4x=4+1x=5$

Hence, the oxidation number of nitrogen is $+5$ .

The oxidation number of sulfur in $H2SO4$ is calculated as,

The oxidation state of oxygen is usually $−2$ as it requires two more electrons in order to achieve stable electronic configuration and of hydrogen is $+1$ as it contains one valence electron. The oxidation number of $S$ is assumed to be $x$ and is calculated by using the formula,

$Charge on H2SO4=[(Number of H atoms×oxidation number of H)+(Number of S atoms×oxidation number of S)+(Number of O atoms×oxidation number of O)]$

Since, the sum of all oxidation numbers in a neutral compound is zero that is charge on $H2SO4$ is zero.

Substitute the number of atoms and their oxidation number in the above formula to calculate the oxidation number of carbon.

$Charge on H2SO4=[(2×1)+(1×x)+(4×(−2))]0=2+x+(−8)0=x−6x=6$

Hence, the oxidation number of sulfur is $+6$ .

Similarly, in $HSO4−$ . The oxidation number of $S$ is assumed to be $x$ and is calculated by using the formula,

$Charge on HSO4−=[(Number of H atoms×oxidation number of H)+(Number of S atoms×oxidation number of S)+(Number of O atoms×oxidation number of O)]$

Substitute the number of atoms and their oxidation number in the above formula to calculate the oxidation number of carbon.

$Charge on HSO4−=[(1×1)+(1×x)+(4×(−2))]−1=1+x+(−8)−1=x−7x=6$

Hence, the oxidation number of sulfur is $+6$ .

There is no change in the oxidation number of the atoms involved in the reaction. Hence, the given reaction is not a redox reaction.

(b)

Interpretation Introduction

To determine: The acid, base, conjugate acid and conjugate base in the given reaction.

(b)

Expert Solution

Answer to Problem 16.174AP

Solution

The acid is $H2SO4$ , $HNO3$ is a base, $HSO4−$ is a conjugate base and $NO2+$ is a conjugate acid.

Explanation of Solution

Explanation

According to Bronsted-Lowry theory, when an acid donates a proton the species formed is known as conjugate base and when the base accepts a proton the species formed is known as conjugate acid. The given reaction is,

$HNO3(aq)+2H2SO4(aq)→NO2+(aq)+H3O+(aq)+2HSO4−(aq)$

In the reaction, $H2SO4$ act as an acid as it donates a proton forming $HSO4−$ as a conjugate base. Nitric acid $(HNO3)$ is a base and forms $NO2+$ conjugate acid.

Conclusion

a. The given reaction is not a redox process.
b. In the given reaction, $H2SO4$ is an acid, $HNO3$ is a base, $HSO4−$ is a conjugate base and $NO2+$ is a conjugate acid.

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Answer 3

Question

Chapter 16, Problem 16.174AP

(a)

Interpretation Introduction

Interpretation: Whether the given reaction is a redox process is to be explained. In the given reaction acid, base, conjugate acid and the conjugate base are to be identified.

Concept introduction: An acid is defined as a compound which donates a proton and a proton acceptor is known as base according to Bronsted-Lowry theory. The reaction that shows the change of oxidation state between reactant and product is known as redox reaction.

To determine: If the given reaction is a redox process.

(a)

Expert Solution

Answer to Problem 16.174AP

Solution

No, the given reaction is not a redox process.

Explanation of Solution

Explanation

The given reaction is,

$HNO3(aq)+2H2SO4(aq)→NO2+(aq)+H3O+(aq)+2HSO4−(aq)$

The oxidation number of nitrogen and sulfur atoms in reactant and product side is calculated to check if it is a redox reaction. When it shows the change in oxidation state between reactant and product then it is a redox reaction. Therefore, the oxidation number of nitrogen in $HNO3$ is calculated as,

The oxidation state of hydrogen in a compound is usually $+1$ as it contains only one valence electron and of oxygen is usually $−2$ as it requires two more electrons in order to achieve stable electronic configuration. The sum of all oxidation numbers in a neutral compound is zero. The oxidation number of N is assumed to be $x$ and is calculated by using the formula,

$Charge on HNO3=[(Number of H atoms×oxidation number of H)+(Number of N atoms×oxidation number of N)+(Number of O atoms×oxidation number of O)]$

Since, the sum of all oxidation numbers in a neutral compound is zero that is charge on $HNO3$ is zero.

Substitute the number of atoms and their oxidation number in the above formula to calculate the oxidation number of nitrogen.

$Charge on HNO3=[(1×(+1))+1×(x)+3(−2)]0=1+x−60=x−5x=5$

Hence, the oxidation number of nitrogen is $+5$ .

Similarly, in $NO2+$ . The oxidation number of N is assumed to be $x$ and is calculated by using the formula,

$Charge on NO2+=[(Number of N atoms×oxidation number of N)+(Number of O atoms×oxidation number of O)]$

Substitute the number of atoms and their oxidation number in the above formula to calculate the oxidation number of nitrogen.

$Charge on NO2+=[1×(x)+2(−2)]+1=x−4x=4+1x=5$

Hence, the oxidation number of nitrogen is $+5$ .

The oxidation number of sulfur in $H2SO4$ is calculated as,

The oxidation state of oxygen is usually $−2$ as it requires two more electrons in order to achieve stable electronic configuration and of hydrogen is $+1$ as it contains one valence electron. The oxidation number of $S$ is assumed to be $x$ and is calculated by using the formula,

$Charge on H2SO4=[(Number of H atoms×oxidation number of H)+(Number of S atoms×oxidation number of S)+(Number of O atoms×oxidation number of O)]$

Since, the sum of all oxidation numbers in a neutral compound is zero that is charge on $H2SO4$ is zero.

Substitute the number of atoms and their oxidation number in the above formula to calculate the oxidation number of carbon.

$Charge on H2SO4=[(2×1)+(1×x)+(4×(−2))]0=2+x+(−8)0=x−6x=6$

Hence, the oxidation number of sulfur is $+6$ .

Similarly, in $HSO4−$ . The oxidation number of $S$ is assumed to be $x$ and is calculated by using the formula,

$Charge on HSO4−=[(Number of H atoms×oxidation number of H)+(Number of S atoms×oxidation number of S)+(Number of O atoms×oxidation number of O)]$

Substitute the number of atoms and their oxidation number in the above formula to calculate the oxidation number of carbon.

$Charge on HSO4−=[(1×1)+(1×x)+(4×(−2))]−1=1+x+(−8)−1=x−7x=6$

Hence, the oxidation number of sulfur is $+6$ .

There is no change in the oxidation number of the atoms involved in the reaction. Hence, the given reaction is not a redox reaction.

(b)

Interpretation Introduction

To determine: The acid, base, conjugate acid and conjugate base in the given reaction.

(b)

Expert Solution

Answer to Problem 16.174AP

Solution

The acid is $H2SO4$ , $HNO3$ is a base, $HSO4−$ is a conjugate base and $NO2+$ is a conjugate acid.

Explanation of Solution

Explanation

According to Bronsted-Lowry theory, when an acid donates a proton the species formed is known as conjugate base and when the base accepts a proton the species formed is known as conjugate acid. The given reaction is,

$HNO3(aq)+2H2SO4(aq)→NO2+(aq)+H3O+(aq)+2HSO4−(aq)$

In the reaction, $H2SO4$ act as an acid as it donates a proton forming $HSO4−$ as a conjugate base. Nitric acid $(HNO3)$ is a base and forms $NO2+$ conjugate acid.

Conclusion

a. The given reaction is not a redox process.
b. In the given reaction, $H2SO4$ is an acid, $HNO3$ is a base, $HSO4−$ is a conjugate base and $NO2+$ is a conjugate acid.

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Answer 4

Question

Chapter 16, Problem 16.174AP

(a)

Interpretation Introduction

Interpretation: Whether the given reaction is a redox process is to be explained. In the given reaction acid, base, conjugate acid and the conjugate base are to be identified.

Concept introduction: An acid is defined as a compound which donates a proton and a proton acceptor is known as base according to Bronsted-Lowry theory. The reaction that shows the change of oxidation state between reactant and product is known as redox reaction.

To determine: If the given reaction is a redox process.

(a)

Expert Solution

Answer to Problem 16.174AP

Solution

No, the given reaction is not a redox process.

Explanation of Solution

Explanation

The given reaction is,

$HNO3(aq)+2H2SO4(aq)→NO2+(aq)+H3O+(aq)+2HSO4−(aq)$

The oxidation number of nitrogen and sulfur atoms in reactant and product side is calculated to check if it is a redox reaction. When it shows the change in oxidation state between reactant and product then it is a redox reaction. Therefore, the oxidation number of nitrogen in $HNO3$ is calculated as,

The oxidation state of hydrogen in a compound is usually $+1$ as it contains only one valence electron and of oxygen is usually $−2$ as it requires two more electrons in order to achieve stable electronic configuration. The sum of all oxidation numbers in a neutral compound is zero. The oxidation number of N is assumed to be $x$ and is calculated by using the formula,

$Charge on HNO3=[(Number of H atoms×oxidation number of H)+(Number of N atoms×oxidation number of N)+(Number of O atoms×oxidation number of O)]$

Since, the sum of all oxidation numbers in a neutral compound is zero that is charge on $HNO3$ is zero.

Substitute the number of atoms and their oxidation number in the above formula to calculate the oxidation number of nitrogen.

$Charge on HNO3=[(1×(+1))+1×(x)+3(−2)]0=1+x−60=x−5x=5$

Hence, the oxidation number of nitrogen is $+5$ .

Similarly, in $NO2+$ . The oxidation number of N is assumed to be $x$ and is calculated by using the formula,

$Charge on NO2+=[(Number of N atoms×oxidation number of N)+(Number of O atoms×oxidation number of O)]$

Substitute the number of atoms and their oxidation number in the above formula to calculate the oxidation number of nitrogen.

$Charge on NO2+=[1×(x)+2(−2)]+1=x−4x=4+1x=5$

Hence, the oxidation number of nitrogen is $+5$ .

The oxidation number of sulfur in $H2SO4$ is calculated as,

The oxidation state of oxygen is usually $−2$ as it requires two more electrons in order to achieve stable electronic configuration and of hydrogen is $+1$ as it contains one valence electron. The oxidation number of $S$ is assumed to be $x$ and is calculated by using the formula,

$Charge on H2SO4=[(Number of H atoms×oxidation number of H)+(Number of S atoms×oxidation number of S)+(Number of O atoms×oxidation number of O)]$

Since, the sum of all oxidation numbers in a neutral compound is zero that is charge on $H2SO4$ is zero.

Substitute the number of atoms and their oxidation number in the above formula to calculate the oxidation number of carbon.

$Charge on H2SO4=[(2×1)+(1×x)+(4×(−2))]0=2+x+(−8)0=x−6x=6$

Hence, the oxidation number of sulfur is $+6$ .

Similarly, in $HSO4−$ . The oxidation number of $S$ is assumed to be $x$ and is calculated by using the formula,

$Charge on HSO4−=[(Number of H atoms×oxidation number of H)+(Number of S atoms×oxidation number of S)+(Number of O atoms×oxidation number of O)]$

Substitute the number of atoms and their oxidation number in the above formula to calculate the oxidation number of carbon.

$Charge on HSO4−=[(1×1)+(1×x)+(4×(−2))]−1=1+x+(−8)−1=x−7x=6$

Hence, the oxidation number of sulfur is $+6$ .

There is no change in the oxidation number of the atoms involved in the reaction. Hence, the given reaction is not a redox reaction.

(b)

Interpretation Introduction

To determine: The acid, base, conjugate acid and conjugate base in the given reaction.

(b)

Expert Solution

Answer to Problem 16.174AP

Solution

The acid is $H2SO4$ , $HNO3$ is a base, $HSO4−$ is a conjugate base and $NO2+$ is a conjugate acid.

Explanation of Solution

Explanation

According to Bronsted-Lowry theory, when an acid donates a proton the species formed is known as conjugate base and when the base accepts a proton the species formed is known as conjugate acid. The given reaction is,

$HNO3(aq)+2H2SO4(aq)→NO2+(aq)+H3O+(aq)+2HSO4−(aq)$

In the reaction, $H2SO4$ act as an acid as it donates a proton forming $HSO4−$ as a conjugate base. Nitric acid $(HNO3)$ is a base and forms $NO2+$ conjugate acid.

Conclusion

a. The given reaction is not a redox process.
b. In the given reaction, $H2SO4$ is an acid, $HNO3$ is a base, $HSO4−$ is a conjugate base and $NO2+$ is a conjugate acid.

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Answer 5

Chapter 16, Problem 16.174AP

(a)

Interpretation Introduction

Interpretation: Whether the given reaction is a redox process is to be explained. In the given reaction acid, base, conjugate acid and the conjugate base are to be identified.

Concept introduction: An acid is defined as a compound which donates a proton and a proton acceptor is known as base according to Bronsted-Lowry theory. The reaction that shows the change of oxidation state between reactant and product is known as redox reaction.

To determine: If the given reaction is a redox process.

(a)

Expert Solution

Answer to Problem 16.174AP

Solution

No, the given reaction is not a redox process.

Explanation of Solution

Explanation

The given reaction is,

$HNO3(aq)+2H2SO4(aq)→NO2+(aq)+H3O+(aq)+2HSO4−(aq)$

The oxidation number of nitrogen and sulfur atoms in reactant and product side is calculated to check if it is a redox reaction. When it shows the change in oxidation state between reactant and product then it is a redox reaction. Therefore, the oxidation number of nitrogen in $HNO3$ is calculated as,

The oxidation state of hydrogen in a compound is usually $+1$ as it contains only one valence electron and of oxygen is usually $−2$ as it requires two more electrons in order to achieve stable electronic configuration. The sum of all oxidation numbers in a neutral compound is zero. The oxidation number of N is assumed to be $x$ and is calculated by using the formula,

$Charge on HNO3=[(Number of H atoms×oxidation number of H)+(Number of N atoms×oxidation number of N)+(Number of O atoms×oxidation number of O)]$

Since, the sum of all oxidation numbers in a neutral compound is zero that is charge on $HNO3$ is zero.

Substitute the number of atoms and their oxidation number in the above formula to calculate the oxidation number of nitrogen.

$Charge on HNO3=[(1×(+1))+1×(x)+3(−2)]0=1+x−60=x−5x=5$

Hence, the oxidation number of nitrogen is $+5$ .

Similarly, in $NO2+$ . The oxidation number of N is assumed to be $x$ and is calculated by using the formula,

$Charge on NO2+=[(Number of N atoms×oxidation number of N)+(Number of O atoms×oxidation number of O)]$

Substitute the number of atoms and their oxidation number in the above formula to calculate the oxidation number of nitrogen.

$Charge on NO2+=[1×(x)+2(−2)]+1=x−4x=4+1x=5$

Hence, the oxidation number of nitrogen is $+5$ .

The oxidation number of sulfur in $H2SO4$ is calculated as,

The oxidation state of oxygen is usually $−2$ as it requires two more electrons in order to achieve stable electronic configuration and of hydrogen is $+1$ as it contains one valence electron. The oxidation number of $S$ is assumed to be $x$ and is calculated by using the formula,

$Charge on H2SO4=[(Number of H atoms×oxidation number of H)+(Number of S atoms×oxidation number of S)+(Number of O atoms×oxidation number of O)]$

Since, the sum of all oxidation numbers in a neutral compound is zero that is charge on $H2SO4$ is zero.

Substitute the number of atoms and their oxidation number in the above formula to calculate the oxidation number of carbon.

$Charge on H2SO4=[(2×1)+(1×x)+(4×(−2))]0=2+x+(−8)0=x−6x=6$

Hence, the oxidation number of sulfur is $+6$ .

Similarly, in $HSO4−$ . The oxidation number of $S$ is assumed to be $x$ and is calculated by using the formula,

$Charge on HSO4−=[(Number of H atoms×oxidation number of H)+(Number of S atoms×oxidation number of S)+(Number of O atoms×oxidation number of O)]$

Substitute the number of atoms and their oxidation number in the above formula to calculate the oxidation number of carbon.

$Charge on HSO4−=[(1×1)+(1×x)+(4×(−2))]−1=1+x+(−8)−1=x−7x=6$

Hence, the oxidation number of sulfur is $+6$ .

There is no change in the oxidation number of the atoms involved in the reaction. Hence, the given reaction is not a redox reaction.

(b)

Interpretation Introduction

To determine: The acid, base, conjugate acid and conjugate base in the given reaction.

(b)

Expert Solution

Answer to Problem 16.174AP

Solution

The acid is $H2SO4$ , $HNO3$ is a base, $HSO4−$ is a conjugate base and $NO2+$ is a conjugate acid.

Explanation of Solution

Explanation

According to Bronsted-Lowry theory, when an acid donates a proton the species formed is known as conjugate base and when the base accepts a proton the species formed is known as conjugate acid. The given reaction is,

$HNO3(aq)+2H2SO4(aq)→NO2+(aq)+H3O+(aq)+2HSO4−(aq)$

In the reaction, $H2SO4$ act as an acid as it donates a proton forming $HSO4−$ as a conjugate base. Nitric acid $(HNO3)$ is a base and forms $NO2+$ conjugate acid.

Conclusion

a. The given reaction is not a redox process.
b. In the given reaction, $H2SO4$ is an acid, $HNO3$ is a base, $HSO4−$ is a conjugate base and $NO2+$ is a conjugate acid.

Answer 6

Chapter 16, Problem 16.174AP

(a)

Interpretation Introduction

Interpretation: Whether the given reaction is a redox process is to be explained. In the given reaction acid, base, conjugate acid and the conjugate base are to be identified.

Concept introduction: An acid is defined as a compound which donates a proton and a proton acceptor is known as base according to Bronsted-Lowry theory. The reaction that shows the change of oxidation state between reactant and product is known as redox reaction.

To determine: If the given reaction is a redox process.

(a)

Expert Solution

Answer to Problem 16.174AP

Solution

No, the given reaction is not a redox process.

Explanation of Solution

Explanation

The given reaction is,

$HNO3(aq)+2H2SO4(aq)→NO2+(aq)+H3O+(aq)+2HSO4−(aq)$

The oxidation number of nitrogen and sulfur atoms in reactant and product side is calculated to check if it is a redox reaction. When it shows the change in oxidation state between reactant and product then it is a redox reaction. Therefore, the oxidation number of nitrogen in $HNO3$ is calculated as,

The oxidation state of hydrogen in a compound is usually $+1$ as it contains only one valence electron and of oxygen is usually $−2$ as it requires two more electrons in order to achieve stable electronic configuration. The sum of all oxidation numbers in a neutral compound is zero. The oxidation number of N is assumed to be $x$ and is calculated by using the formula,

$Charge on HNO3=[(Number of H atoms×oxidation number of H)+(Number of N atoms×oxidation number of N)+(Number of O atoms×oxidation number of O)]$

Since, the sum of all oxidation numbers in a neutral compound is zero that is charge on $HNO3$ is zero.

Substitute the number of atoms and their oxidation number in the above formula to calculate the oxidation number of nitrogen.

$Charge on HNO3=[(1×(+1))+1×(x)+3(−2)]0=1+x−60=x−5x=5$

Hence, the oxidation number of nitrogen is $+5$ .

Similarly, in $NO2+$ . The oxidation number of N is assumed to be $x$ and is calculated by using the formula,

$Charge on NO2+=[(Number of N atoms×oxidation number of N)+(Number of O atoms×oxidation number of O)]$

Substitute the number of atoms and their oxidation number in the above formula to calculate the oxidation number of nitrogen.

$Charge on NO2+=[1×(x)+2(−2)]+1=x−4x=4+1x=5$

Hence, the oxidation number of nitrogen is $+5$ .

The oxidation number of sulfur in $H2SO4$ is calculated as,

The oxidation state of oxygen is usually $−2$ as it requires two more electrons in order to achieve stable electronic configuration and of hydrogen is $+1$ as it contains one valence electron. The oxidation number of $S$ is assumed to be $x$ and is calculated by using the formula,

$Charge on H2SO4=[(Number of H atoms×oxidation number of H)+(Number of S atoms×oxidation number of S)+(Number of O atoms×oxidation number of O)]$

Since, the sum of all oxidation numbers in a neutral compound is zero that is charge on $H2SO4$ is zero.

Substitute the number of atoms and their oxidation number in the above formula to calculate the oxidation number of carbon.

$Charge on H2SO4=[(2×1)+(1×x)+(4×(−2))]0=2+x+(−8)0=x−6x=6$

Hence, the oxidation number of sulfur is $+6$ .

Similarly, in $HSO4−$ . The oxidation number of $S$ is assumed to be $x$ and is calculated by using the formula,

$Charge on HSO4−=[(Number of H atoms×oxidation number of H)+(Number of S atoms×oxidation number of S)+(Number of O atoms×oxidation number of O)]$

Substitute the number of atoms and their oxidation number in the above formula to calculate the oxidation number of carbon.

$Charge on HSO4−=[(1×1)+(1×x)+(4×(−2))]−1=1+x+(−8)−1=x−7x=6$

Hence, the oxidation number of sulfur is $+6$ .

There is no change in the oxidation number of the atoms involved in the reaction. Hence, the given reaction is not a redox reaction.

Answer 7

Chapter 16, Problem 16.174AP

(a)

Interpretation Introduction

Interpretation: Whether the given reaction is a redox process is to be explained. In the given reaction acid, base, conjugate acid and the conjugate base are to be identified.

Concept introduction: An acid is defined as a compound which donates a proton and a proton acceptor is known as base according to Bronsted-Lowry theory. The reaction that shows the change of oxidation state between reactant and product is known as redox reaction.

To determine: If the given reaction is a redox process.

Answer 8

(a)

Expert Solution

Answer to Problem 16.174AP

Solution

No, the given reaction is not a redox process.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Explanation

The given reaction is,

$HNO3(aq)+2H2SO4(aq)→NO2+(aq)+H3O+(aq)+2HSO4−(aq)$

The oxidation number of nitrogen and sulfur atoms in reactant and product side is calculated to check if it is a redox reaction. When it shows the change in oxidation state between reactant and product then it is a redox reaction. Therefore, the oxidation number of nitrogen in $HNO3$ is calculated as,

The oxidation state of hydrogen in a compound is usually $+1$ as it contains only one valence electron and of oxygen is usually $−2$ as it requires two more electrons in order to achieve stable electronic configuration. The sum of all oxidation numbers in a neutral compound is zero. The oxidation number of N is assumed to be $x$ and is calculated by using the formula,

$Charge on HNO3=[(Number of H atoms×oxidation number of H)+(Number of N atoms×oxidation number of N)+(Number of O atoms×oxidation number of O)]$

Since, the sum of all oxidation numbers in a neutral compound is zero that is charge on $HNO3$ is zero.

Substitute the number of atoms and their oxidation number in the above formula to calculate the oxidation number of nitrogen.

$Charge on HNO3=[(1×(+1))+1×(x)+3(−2)]0=1+x−60=x−5x=5$

Hence, the oxidation number of nitrogen is $+5$ .

Similarly, in $NO2+$ . The oxidation number of N is assumed to be $x$ and is calculated by using the formula,

$Charge on NO2+=[(Number of N atoms×oxidation number of N)+(Number of O atoms×oxidation number of O)]$

Substitute the number of atoms and their oxidation number in the above formula to calculate the oxidation number of nitrogen.

$Charge on NO2+=[1×(x)+2(−2)]+1=x−4x=4+1x=5$

Hence, the oxidation number of nitrogen is $+5$ .

The oxidation number of sulfur in $H2SO4$ is calculated as,

The oxidation state of oxygen is usually $−2$ as it requires two more electrons in order to achieve stable electronic configuration and of hydrogen is $+1$ as it contains one valence electron. The oxidation number of $S$ is assumed to be $x$ and is calculated by using the formula,

$Charge on H2SO4=[(Number of H atoms×oxidation number of H)+(Number of S atoms×oxidation number of S)+(Number of O atoms×oxidation number of O)]$

Since, the sum of all oxidation numbers in a neutral compound is zero that is charge on $H2SO4$ is zero.

Substitute the number of atoms and their oxidation number in the above formula to calculate the oxidation number of carbon.

$Charge on H2SO4=[(2×1)+(1×x)+(4×(−2))]0=2+x+(−8)0=x−6x=6$

Hence, the oxidation number of sulfur is $+6$ .

Similarly, in $HSO4−$ . The oxidation number of $S$ is assumed to be $x$ and is calculated by using the formula,

$Charge on HSO4−=[(Number of H atoms×oxidation number of H)+(Number of S atoms×oxidation number of S)+(Number of O atoms×oxidation number of O)]$

Substitute the number of atoms and their oxidation number in the above formula to calculate the oxidation number of carbon.

$Charge on HSO4−=[(1×1)+(1×x)+(4×(−2))]−1=1+x+(−8)−1=x−7x=6$

Hence, the oxidation number of sulfur is $+6$ .

There is no change in the oxidation number of the atoms involved in the reaction. Hence, the given reaction is not a redox reaction.

Answer 13

(b)

Interpretation Introduction

To determine: The acid, base, conjugate acid and conjugate base in the given reaction.

(b)

Expert Solution

Answer to Problem 16.174AP

Solution

The acid is $H2SO4$ , $HNO3$ is a base, $HSO4−$ is a conjugate base and $NO2+$ is a conjugate acid.

Explanation of Solution

Explanation

According to Bronsted-Lowry theory, when an acid donates a proton the species formed is known as conjugate base and when the base accepts a proton the species formed is known as conjugate acid. The given reaction is,

$HNO3(aq)+2H2SO4(aq)→NO2+(aq)+H3O+(aq)+2HSO4−(aq)$

In the reaction, $H2SO4$ act as an acid as it donates a proton forming $HSO4−$ as a conjugate base. Nitric acid $(HNO3)$ is a base and forms $NO2+$ conjugate acid.

Conclusion

a. The given reaction is not a redox process.
b. In the given reaction, $H2SO4$ is an acid, $HNO3$ is a base, $HSO4−$ is a conjugate base and $NO2+$ is a conjugate acid.

Answer 14

(b)

Interpretation Introduction

To determine: The acid, base, conjugate acid and conjugate base in the given reaction.

Answer 15

(b)

Expert Solution

Answer to Problem 16.174AP

Solution

The acid is $H2SO4$ , $HNO3$ is a base, $HSO4−$ is a conjugate base and $NO2+$ is a conjugate acid.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Explanation

According to Bronsted-Lowry theory, when an acid donates a proton the species formed is known as conjugate base and when the base accepts a proton the species formed is known as conjugate acid. The given reaction is,

$HNO3(aq)+2H2SO4(aq)→NO2+(aq)+H3O+(aq)+2HSO4−(aq)$

In the reaction, $H2SO4$ act as an acid as it donates a proton forming $HSO4−$ as a conjugate base. Nitric acid $(HNO3)$ is a base and forms $NO2+$ conjugate acid.

Conclusion

a. The given reaction is not a redox process.
b. In the given reaction, $H2SO4$ is an acid, $HNO3$ is a base, $HSO4−$ is a conjugate base and $NO2+$ is a conjugate acid.