Chapter 4, Problem 4.141AP

(a)

Interpretation Introduction

Interpretation: The questions corresponding to the reaction of sulfuric acid and calcium carbonate are to be answered.

Concept introduction: The reaction in which reduction and oxidation process occurs simultaneously is termed as redox reaction.

In oxidation there is loss of electrons occur which increases the oxidation state of an atom whereas in reduction process there is gaining of electrons occur which decreases the oxidation state of an atom.

To determine: If the given reactions are redox reactions or not and the number of electrons transferred.

Answer to Problem 4.141AP

Solution

The reaction between ${\text{H}}_{\text{2}}\text{S}$ and ${\text{O}}_{\text{2}}$ is a redox reaction. The reaction is,

${\text{H}}_{\text{2}}\text{S}\left(aq\right)+2{\text{O}}_2\left(g\right)\to {\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)$

There are total eight electrons are transferred.

Explanation of Solution

Explanation

The given reactions are,

${\text{H}}_{\text{2}}\text{S}\left(aq\right)+2{\text{O}}_2\left(g\right)\to {\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)$ (1)

${\text{H}}_{\text{2}}{\text{SO}}_4\left(aq\right)+{\text{CaCO}}_3\left(s\right)\to {\text{CaSO}}_{\text{4}}\left(s\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)+{\text{CO}}_{\text{2}}\left(g\right)$ (2)

The identification of the above reactions that they are whether redox reaction or not is done by calculating the oxidation states of sulfur atom in both reactant and product side.

The oxidation state of sulfur atom in reaction (1) is calculated as follows.

To calculate the oxidation number sulfur in the given compound ${\text{H}}_{\text{2}}\text{S}$ the oxidation state hydrogen is usually $+1$. The oxidation number of $\text{S}$ in ${\text{H}}_{\text{2}}\text{S}$ is assumed to be $x$ and is calculated by using the formula,

${\text{Charge on H}}_{\text{2}}\text{S}=\left[\begin{array}{c}\left(\text{Number of}\text{S}\text{atoms}\times \text{oxidation number of S}\right)\\ +\left(\text{Number of H atoms}\times \text{oxidation number of H}\right)\end{array}\right]$

Since, the charge on ${\text{H}}_{\text{2}}\text{S}$ is $0$ .

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

$\begin{array}{c}{\text{Charge on H}}_{\text{2}}\text{S}=\left[\left(2\times \left(1\right)\right)+\left(x\right)\right]\\ 0=\left[\left(2\right)+x\right]\\ x=-2\end{array}$

Thus, the oxidation number of $\text{S}$ in ${\text{H}}_{\text{2}}\text{S}$ is $-2$

The given element is ${\text{H}}_2{\text{SO}}_{\text{4}}$. The oxidation state of oxygen in a compound is usually $-2$. The common oxidation state for hydrogen atom is $+1$ and the oxidation number of sulfur atom in ${\text{H}}_2{\text{SO}}_{\text{4}}$ is assumed to be $x$ which will be calculated by formula,

${\text{Charge on H}}_2{\text{SO}}_{\text{4}}=\left[\begin{array}{c}\left(\text{Number of S atoms}\times \text{oxidation number of S}\right)+\\ \left(\text{Number of H atoms}\times \text{oxidation number of H}\right)+\\ \left(\text{Number of O atoms}\times \text{oxidation number of O}\right)\end{array}\right]$

The charge on ${\text{H}}_2{\text{SO}}_{\text{4}}$ is $0$.

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

$\begin{array}{c}\text{Charge}\text{on}{\text{H}}_2{\text{SO}}_{\text{4}}=\left[\left(x\right)+2\left(1\right)+4\left(-2\right)\right]\\ 0=x-6\\ x=+6\end{array}$

Therefore, the oxidation number of sulfur atom in ${\text{H}}_2{\text{SO}}_{\text{4}}$ is $+6$.

Hence, the oxidation number of sulfur atom in ${\text{H}}_{\text{2}}\text{S}$ is $-2$ and in ${\text{H}}_2{\text{SO}}_{\text{4}}$ is $+6$.

Hence, the oxidation number of sulfur atom in ${\text{H}}_{\text{2}}\text{S}$ is $-2$ and the oxidation number of sulfur atom in ${\text{H}}_2{\text{SO}}_{\text{4}}$ is $+6$. There is increase in the oxidation number of sulfur atom that represents there is oxidation process is taking place.

Therefore, the given reaction of sulfur compound is a redox reaction.

The oxidation state of sulfur atom in reaction (2) is calculated as follows.

The given element is ${\text{H}}_2{\text{SO}}_{\text{4}}$. The oxidation state of oxygen in a compound is usually $-2$. The common oxidation state for hydrogen atom is $+1$ and the oxidation number of sulfur atom in ${\text{H}}_2{\text{SO}}_{\text{4}}$ is assumed to be $x$ which will be calculated by formula,

${\text{Charge on H}}_2{\text{SO}}_{\text{4}}=\left[\begin{array}{c}\left(\text{Number of S atoms}\times \text{oxidation number of S}\right)+\\ \left(\text{Number of H atoms}\times \text{oxidation number of H}\right)+\\ \left(\text{Number of O atoms}\times \text{oxidation number of O}\right)\end{array}\right]$

The charge on ${\text{H}}_2{\text{SO}}_{\text{4}}$ is $0$.

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

$\begin{array}{c}\text{Charge}\text{on}{\text{H}}_2{\text{SO}}_{\text{4}}=\left[\left(x\right)+2\left(1\right)+4\left(-2\right)\right]\\ 0=x-6\\ x=+6\end{array}$

Therefore, the oxidation number of sulfur atom in ${\text{H}}_2{\text{SO}}_{\text{4}}$ is $+6$.

The given element is ${\text{CaSO}}_{\text{4}}$. The oxidation state of oxygen in a compound is usually $-2$. The oxidation state of calcium atom is $+2$ and the oxidation number of sulfur atom in ${\text{CaSO}}_{\text{4}}$ is assumed to be $x$ which will be calculated by formula,

${\text{Charge on CaSO}}_{\text{4}}=\left[\begin{array}{c}\left(\text{Number of Ca atoms}\times \text{oxidation number of Ca}\right)+\\ \left(\text{Number of O atoms}\times \text{oxidation number of O}\right)\end{array}\right]$

The charge on ${\text{CaSO}}_{\text{4}}$ is $0$.

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

$\begin{array}{c}\text{Charge}\text{on}{\text{CaSO}}_{\text{4}}=\left[2+\left(x\right)+4\left(-2\right)\right]\\ 0=2+x-8\\ x=+6\end{array}$

Therefore, the oxidation number of sulfur atom in ${\text{CaSO}}_{\text{4}}$ is $+6$.

Hence, the oxidation number of sulfur atom in ${\text{H}}_2{\text{SO}}_{\text{4}}$ is $+6$. And in ${\text{CaSO}}_{\text{4}}$ is $+6$. There is no change in the oxidation number of sulfur atom that represents there is no oxidation-reduction process is taking place.

Therefore, the given reaction of sulfur compound is not a redox reaction.

Thus, the reaction between ${\text{H}}_{\text{2}}\text{S}$ and ${\text{O}}_{\text{2}}$ is a redox reaction. The reaction is,

${\text{H}}_{\text{2}}\text{S}\left(aq\right)+2{\text{O}}_2\left(g\right)\to {\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)$

In the above reaction there are total $8$ electrons transferred from $-2$ to $+6$ oxidation state of sulfur atom present in ${\text{H}}_{\text{2}}\text{S}$ and ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$.

(b)

Interpretation Introduction

To determine: A net ionic equation for the reaction between ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ with ${\text{CaCO}}_{\text{3}}$.

Answer to Problem 4.141AP

Solution

The net ionic equation for the above reaction is,

${\text{2H}}^{+}\left(aq\right){\text{+SO}}_4{}^{2-}\left(aq\right)+{\text{CaCO}}_3\left(s\right)\to {\text{CaSO}}_{\text{4}}\left(s\right)+{\text{H}}_2\text{O}\left(l\right)+{\text{CO}}_2\left(g\right)$

Explanation of Solution

Explanation

The reaction between ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ with ${\text{CaCO}}_{\text{3}}$ is given as,

${\text{H}}_{\text{2}}{\text{SO}}_4\left(aq\right)+{\text{CaCO}}_3\left(s\right)\to {\text{CaSO}}_{\text{4}}\left(s\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)+{\text{CO}}_{\text{2}}\left(g\right)$

This given reaction is chemically balanced. Therefore, there is no need to balance the above equation as it consists of equal number of reactants and products.

The net ionic equation for the above reaction is ,

${\text{2H}}^{+}\left(aq\right){\text{+SO}}_4{}^{2-}\left(aq\right)+{\text{CaCO}}_3\left(s\right)\to {\text{CaSO}}_{\text{4}}\left(s\right)+{\text{H}}_2\text{O}\left(l\right)+{\text{CO}}_2\left(g\right)$

(c)

Interpretation Introduction

To determine: The difference between the net ionic equation and the given equation.

Answer to Problem 4.141AP

Solution

The difference between the net ionic equation and the given equation is stated below.

Explanation of Solution

Explanation

The given equation is,

${\text{H}}_{\text{2}}{\text{SO}}_4\left(aq\right)+{\text{CaCO}}_3\left(s\right)\to {\text{CaSO}}_{\text{4}}\left(s\right)+{\text{H}}_{\text{2}}{\text{CO}}_3\left(aq\right)$

The overall net ionic equation for the above reaction is ,

${\text{2H}}^{+}\left(aq\right){\text{+SO}}_4{}^{2-}\left(aq\right)+{\text{Ca}}^{2+}\left(s\right){\text{+CO}}_3{}^{-}\left(aq\right)\to {\text{CaSO}}_{\text{4}}\left(s\right)+{\text{2H}}^{+}\left(aq\right)+{\text{CO}}_{\text{3}}{}^{-}\left(aq\right)$

After removing the removing spectator ions the net ionic equation is obtained is

${\text{SO}}_4{}^{2-}\left(aq\right)+{\text{Ca}}^{2+}\left(s\right)\to {\text{CaSO}}_{\text{4}}\left(s\right)$

Therefore, there is a difference of spectator ions in both the equations that means in the given reaction spectator ions are present and in the net ionic equation the spectator ions are absent.

Conclusion

1. a) The reaction between ${\text{H}}_{\text{2}}\text{S}$ and ${\text{O}}_{\text{2}}$ is a redox reaction. The reaction is,

${\text{H}}_{\text{2}}\text{S}\left(aq\right)+2{\text{O}}_2\left(g\right)\to {\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)$

There are total eight electrons are transferred.

1. b) The net ionic equation for the above reaction is,

${\text{2H}}^{+}\left(aq\right){\text{+SO}}_4{}^{2-}\left(aq\right)+{\text{CaCO}}_3\left(s\right)\to {\text{CaSO}}_{\text{4}}\left(s\right)+{\text{H}}_2\text{O}\left(l\right)+{\text{CO}}_2\left(g\right)$

1. c) There is a difference of spectator ions in both the equations that means in the given reaction spectator ions are present and in the net ionic equation the spectator ions are absent