Chapter 4, Problem 4.111QP

(a)

Interpretation Introduction

Interpretation: The oxidation numbers of the given reactants and products are to be calculated. The given equation is to be balanced. The total number of electrons which are transferred for each chromium atom is to be calculated.

Concept introduction: The reactants and the products in a. given ionic equation must be balanced for the given number of ions or atoms of each element. A compound is said to be oxidized when it loses electrons which increases its oxidation number. On the other hand reduction is the process of gaining electrons which decreases the oxidation number of the compound. Usually the common oxidation numbers of elements are used to determine whether the given compounds are oxidized or reduced.

To determine: The oxidation numbers of the given reactants and products.

Answer to Problem 4.111QP

Solution

The oxidation number of chromium atom in ${\text{HCrO}}_{\text{4}}{}^{-}$ is $\underset{\_}{+6}$ and in ${\text{Cr}}_{\text{2}}{\text{O}}_{\text{3}}$ is $\underset{\_}{+3}$.

The common oxidation number of oxygen atom is $\underset{\_}{-2}$.

The common oxidation number of hydrogen atom is $\underset{\_}{+1}$.

The oxidation number of sulfur atom in ${\text{H}}_{\text{2}}\text{S}$ is $\underset{\_}{-2}$ and in ${\text{SO}}_{\text{4}}{}^{\text{2-}}$ is $\underset{\_}{+6}$.

Explanation of Solution

Explanation

Given

The given unbalanced ionic reaction between hydrogen chromate ion $\left({\text{HCrO}}_{\text{4}}{}^{\text{-}}\right)$ and hydrogen sulfide gas $\left({\text{H}}_{\text{2}}\text{S}\right)$ is,

${\text{HCrO}}_{\text{4}}{}^{-}\left(aq\right)+{\text{H}}_{\text{2}}\text{S}\left(g\right)\to {\text{Cr}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)+{\text{SO}}_{\text{4}}{}^{2-}$

The common oxidation numbers of the oxygen atom and hydrogen atom are $\underset{\_}{-2}$ and $\underset{\_}{+1}$ respectively which will be used to calculate the oxidation numbers of chromium and sulfur atom in the given ionic equation.

The oxidation number of chromium atom in ${\text{HCrO}}_{\text{4}}{}^{\text{-}}$ is assumed to be $x$ and the common oxidation number of oxygen is $\underset{\_}{-2}$

The oxidation number of chromium atom is calculated by the formula,

$\left(\begin{array}{l}\left(\begin{array}{l}\text{Oxidation}\text{number}\\ \text{of}\text{hydrogen}\end{array}\right)+\left(\begin{array}{l}\text{Oxidation}\text{number}\\ \text{of}\text{chromium}\end{array}\right)\\ +\left(4\left(\begin{array}{l}\text{Oxidation}\text{number}\\ \text{of}\text{oxygen}\end{array}\right)\right)\end{array}\right)=\left(\begin{array}{l}\text{Overall}\text{charge}\\ \text{on}\text{the}\text{complex}\end{array}\right)$

Substitute the required values in the above equation.

$\begin{array}{c}1+x+4\left(-2\right)=-1\\ 1+x-8=-1\\ x=+8-2\\ x=+6\end{array}$

Therefore, the oxidation number of chromium atom in ${\text{HCrO}}_{\text{4}}{}^{\text{-}}$ is $\underset{\_}{+6}$.

The given element is ${\text{Cr}}_{\text{2}}{\text{O}}_{\text{3}}$. The oxidation state of oxygen in a compound is usually $-2$ and the oxidation state of chromium is assumed to be $x$ and is calculated by using the formula,

${\text{Charge on Cr}}_{\text{2}}{\text{O}}_{\text{3}}=\left[\begin{array}{c}\left(\text{Number of Cr atoms}\times \text{oxidation number of Cr}\right)+\\ \left(\text{Number of O atoms}\times \text{oxidation number of O}\right)\end{array}\right]$

The charge on ${\text{Cr}}_{\text{2}}{\text{O}}_{\text{3}}$ is $0$.

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

$\begin{array}{c}\text{Charge}\text{on}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{3}}=\left[\left(2x\right)+3\left(-2\right)\right]\\ 0=2x-6\\ 2x=+6\\ x=+3\end{array}$

Therefore, the oxidation number of chromium atom in ${\text{Cr}}_{\text{2}}{\text{O}}_{\text{3}}$ is $\underset{\_}{+3}$.

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,

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

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.

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

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

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

${\text{Charge on SO}}_4{}^{2-}=\left[\begin{array}{c}\left(\text{Number of S atoms}\times \text{oxidation number of S}\right)+\\ \left(\text{Number of O atoms}\times \text{oxidation number of O}\right)\end{array}\right]$

The charge on ${\text{SO}}_{\text{4}}{}^{\text{2-}}$ is $-2$.

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

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

Therefore, the oxidation number of sulfur atom in ${\text{SO}}_{\text{4}}{}^{\text{2-}}$ is $\underset{\_}{+6}$.

Hence, the oxidation number of chromium atom in ${\text{HCrO}}_{\text{4}}{}^{\text{-}}$ is $\underset{\_}{+6}$ and in ${\text{Cr}}_{\text{2}}{\text{O}}_{\text{3}}$ is $\underset{\_}{+3}$. The oxidation number of sulfur atom in ${\text{H}}_{\text{2}}\text{S}$ is $\underset{\_}{-2}$ and in ${\text{SO}}_{\text{4}}{}^{\text{2-}}$ is $\underset{\_}{+6}$.

(b)

Interpretation Introduction

To determine: The balanced ionic equation from the given unbalanced ionic equation.

Answer to Problem 4.111QP

Solution

The balanced ionic equation between hydrogen chromate ion $\left({\text{HCrO}}_{\text{4}}{}^{\text{-}}\right)$ and hydrogen sulfide gas $\left({\text{H}}_{\text{2}}\text{S}\right)$ is,

${\text{8HCrO}}_{\text{4}}{}^{-}\left(aq\right)+3{\text{H}}_{\text{2}}\text{S}\left(g\right)+2{\text{H}}^{+}\left(aq\right)\to 4{\text{Cr}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)+3{\text{SO}}_{\text{4}}{}^{2-}\left(aq\right)+8{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Explanation of Solution

Explanation

Given

The given unbalanced ionic reaction between hydrogen chromate ion $\left({\text{HCrO}}_{\text{4}}{}^{\text{-}}\right)$ and hydrogen sulfide gas $\left({\text{H}}_{\text{2}}\text{S}\right)$ is,

${\text{HCrO}}_{\text{4}}{}^{-}\left(aq\right)+{\text{H}}_{\text{2}}\text{S}\left(g\right)\to {\text{Cr}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)+{\text{SO}}_{\text{4}}{}^{2-}$

During the process of oxidation each sulfur atom present in the above equation will lose eight electrons and each chromium atom will gain three electrons during the reduction process. Thus the ratio of electrons will be $8:3$. Therefore, to balance the equation coefficient of $8$ will be placed before ${\text{HCrO}}_{\text{4}}{}^{\text{-}}$ and a coefficient $3$ will be placed before ${\text{H}}_{\text{2}}\text{S}$. The equation obtained is,

${\text{8HCrO}}_{\text{4}}{}^{-}\left(aq\right)+3{\text{H}}_{\text{2}}\text{S}\left(g\right)\to {\text{Cr}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)+{\text{SO}}_{\text{4}}{}^{2-}\left(aq\right)$

The coefficient $4$ and $3$ will be placed in front of ${\text{Cr}}_{\text{2}}{\text{O}}_{\text{3}}$ and ${\text{SO}}_{\text{4}}{}^{\text{2-}}$ respectively to balance the chromium and sulfur atoms. To balance the hydrogen atoms in the given equation water molecules will be added to the right hand side of the equation. The equation obtained is,

${\text{8HCrO}}_{\text{4}}{}^{-}\left(aq\right)+3{\text{H}}_{\text{2}}\text{S}\left(g\right)\to 4{\text{Cr}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)+3{\text{SO}}_{\text{4}}{}^{2-}\left(aq\right)+8{\text{H}}_{\text{2}}\text{O}\left(l\right)$

To balance the oxygen atom put the coefficient $8$ in front of oxygen atoms. Finally, balance the net charge by adding hydrogen ions to the left hand side of the equation which will give the whole balanced equation as,

${\text{8HCrO}}_{\text{4}}{}^{-}\left(aq\right)+3{\text{H}}_{\text{2}}\text{S}\left(g\right)+2{\text{H}}^{+}\left(aq\right)\to 4{\text{Cr}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)+3{\text{SO}}_{\text{4}}{}^{2-}\left(aq\right)+8{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Thus, the net charge and the whole equation are balanced.

(c)

Interpretation Introduction

To determine: The total number of electrons which are transferred for each chromium atom.

Answer to Problem 4.111QP

Solution

The total number of electrons which are transferred for each chromium atom are $\underset{\_}{3}$.

Explanation of Solution

Explanation

The difference between the oxidation number on the chromium atom is equal to the number of electrons transferred for each chromium atom which is calculated as,

$+6-\left(+3\right)=3$

Therefore the total number of electrons which are transferred for each chromium atom is $\underset{\_}{3}$

Conclusion

The conclusion for each part of the question is as follows,

1. a) The oxidation number of chromium atom in ${\text{HCrO}}_{\text{4}}{}^{\text{-}}$ is $\underset{\_}{+6}$ and in ${\text{Cr}}_{\text{2}}{\text{O}}_{\text{3}}$ is $\underset{\_}{+3}$. The oxidation number of oxygen atom is $\underset{\_}{-2}$. The oxidation number of hydrogen atom is $\underset{\_}{+1}$. The oxidation number of sulfur atom in ${\text{H}}_{\text{2}}\text{S}$ is $\underset{\_}{-2}$ and in ${\text{SO}}_{\text{4}}{}^{\text{2-}}$ is $\underset{\_}{+6}$
2. b) The balanced ionic equation between hydrogen chromate ion $\left({\text{HCrO}}_{\text{4}}{}^{\text{-}}\right)$ and hydrogen sulfide gas $\left({\text{H}}_{\text{2}}\text{S}\right)$ is,

   ${\text{8HCrO}}_{\text{4}}{}^{-}\left(aq\right)+3{\text{H}}_{\text{2}}\text{S}\left(g\right)+2{\text{H}}^{+}\left(aq\right)\to 4{\text{Cr}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)+3{\text{SO}}_{\text{4}}{}^{2-}\left(aq\right)+8{\text{H}}_{\text{2}}\text{O}\left(l\right)$

3. c) The total number of electrons which are transferred for each chromium atom is $\underset{\_}{3}$