Chapter 4, Problem 4.125AP

(a)

Interpretation Introduction

Interpretation: The ionic equation for the given redox reaction is to be written. The names of the elements which are oxidized and reduced are to be stated. The oxidizing and reducing agents are to be identified in the given reaction. The amount of sodium dithionate in grams is to be calculated.

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.

A reducing agent will lose electrons and will oxidize itself whereas an oxidizing agent will gain electrons and will reduce itself.

To determine: The ionic equation between solution of dithionate ions $\left({\text{S}}_{\text{2}}{\text{O}}_{\text{4}}{}^{\text{2-}}\right)$ and chromate ions $\left({\text{CrO}}_{\text{4}}{}^{\text{2-}}\right)$.

Answer to Problem 4.125AP

Solution

The ionic equation between solution of dithionate ions $\left({\text{S}}_{\text{2}}{\text{O}}_{\text{4}}{}^{\text{2-}}\right)$ and chromate ions $\left({\text{CrO}}_{\text{4}}{}^{\text{2-}}\right)$ is,

$3{\text{S}}_{\text{2}}{\text{O}}_{\text{4}}{}^{\text{2-}}\left(aq\right)+2{\text{CrO}}_{\text{4}}{}^{\text{2-}}\left(aq\right)+{\text{2OH}}^{\text{-}}\left(aq\right)+{\text{2H}}_{\text{2}}\text{O}\left(l\right)\to 6{\text{SO}}_{\text{3}}{}^{\text{2-}}\left(aq\right)+\text{2Cr}{\left(\text{OH}\right)}_{\text{3}}\left(s\right)$

Explanation of Solution

Explanation

This stated redox reaction is,

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

In the above redox reaction, the oxidation state of hydrogen is $+1$.

To balance the number of hydrogen atoms on both the sides one water molecule is added to the reactant side of the equation.

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

To balance the sulfur and oxygen atoms on both the sides the $\frac{3}{2}$ is placed in front of dithionate ions $\left({\text{S}}_{\text{2}}{\text{O}}_{\text{4}}{}^{\text{2-}}\right)$ ions and coefficient 3 is placed in front of sulfite ions $\left({\text{SO}}_{\text{3}}{}^{\text{2-}}\right)$ ions. The chemical equation thus obtained is,

$\frac{3}{2}{\text{S}}_{\text{2}}{\text{O}}_{\text{4}}{}^{\text{2-}}\left(aq\right)+1{\text{CrO}}_{\text{4}}{}^{\text{2-}}\left(aq\right)+{\text{OH}}^{\text{-}}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\to 3{\text{SO}}_{\text{3}}{}^{\text{2-}}\left(aq\right)+1\text{Cr}{\left(\text{OH}\right)}_{\text{3}}\left(s\right)$

To remove the fraction coefficient both the sides are multiplied by $2$. The equation obtained is,

$2\times \left(\frac{3}{2}{\text{S}}_{\text{2}}{\text{O}}_{\text{4}}{}^{\text{2-}}\left(aq\right)+1{\text{CrO}}_{\text{4}}{}^{\text{2-}}\left(aq\right)+{\text{OH}}^{\text{-}}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\to 3{\text{SO}}_{\text{3}}{}^{\text{2-}}\left(aq\right)+1\text{Cr}{\left(\text{OH}\right)}_{\text{3}}\left(s\right)\right)$

The balanced net ionic equation thus obtained is,

${\text{3S}}_{\text{2}}{\text{O}}_{\text{4}}{}^{\text{2-}}\left(aq\right)+2{\text{CrO}}_{\text{4}}{}^{\text{2-}}\left(aq\right)+2{\text{OH}}^{\text{-}}\left(aq\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)\to 6{\text{SO}}_{\text{3}}{}^{\text{2-}}\left(aq\right)+2\text{Cr}{\left(\text{OH}\right)}_{\text{3}}\left(s\right)$

(b)

Interpretation Introduction

To determine: The element which is oxidized and the element which is reduced.

Answer to Problem 4.125AP

Solution

In the given reaction sulfur is oxidized and chromium is reduced.

Explanation of Solution

Explanation

The given element is ${\text{S}}_2{\text{O}}_{\text{4}}{}^{2-}$ in the redox reaction.

The oxidation state of oxygen is usually $-2$ as it requires two more electrons in order to achieve stable electronic configuration.

The oxidation state of $\text{S}$ in ${\text{S}}_2{\text{O}}_{\text{4}}{}^{2-}$ is assumed to be $x$.

The oxidation state of S is calculated using the formula,

${\text{Charge on S}}_2{\text{O}}_{\text{4}}{}^{2-}=\left[\begin{array}{c}\left(\text{Number of}\text{S}\text{atoms}\times \text{oxidation number of S}\right)\\ +\left(\text{Number of O atoms}\times \text{oxidation number of O}\right)\end{array}\right]$

The charge on ${\text{S}}_2{\text{O}}_{\text{4}}{}^{2-}$ is $-2$.

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

$\begin{array}{c}{\text{Charge on S}}_2{\text{O}}_{\text{4}}{}^{2-}=\left[\left(2\times \left(x\right)\right)+\left(4\times \left(-2\right)\right)\right]\\ -2=\left[\left(2x\right)+\left(-8\right)\right]\\ 2x=-2+8\\ x=+3\end{array}$

Thus, the oxidation number of $\text{S}$ in ${\text{S}}_2{\text{O}}_{\text{4}}{}^{2-}$ is $+3$.

The given element is ${\text{SO}}_3{}^{2-}$ in the redox reaction.

The oxidation state of oxygen is usually $-2$ as it requires two more electrons in order to achieve stable electronic configuration.

The oxidation state of $\text{S}$ in ${\text{SO}}_3{}^{2-}$ is assumed to be $x$.

The oxidation state of S is calculated using the formula,

${\text{Charge on SO}}_3{}^{2-}=\left[\begin{array}{c}\left(\text{Number of}\text{S}\text{atoms}\times \text{oxidation number of S}\right)\\ +\left(\text{Number of O atoms}\times \text{oxidation number of O}\right)\end{array}\right]$

The charge on ${\text{SO}}_3{}^{2-}$ is $-2$.

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

$\begin{array}{c}{\text{Charge on SO}}_3{}^{2-}=\left[\left(1\times \left(x\right)\right)+\left(3\times \left(-2\right)\right)\right]\\ -2=\left[\left(x\right)+\left(-6\right)\right]\\ x=-2+6\\ x=+4\end{array}$

Thus, the oxidation number of $\text{S}$ in ${\text{SO}}_3{}^{2-}$ is $+4$.

Therefore from the above calculations, it is stated that sulfur is oxidized because the oxidation state of sulfur increases from $+3$ to $+4$.

The given element is ${\text{CrO}}_4{}^{2-}$.

The oxidation state of oxygen in a compound is usually $-2$.

The oxidation state of chromium is assumed to be $x$.

It is calculated by using the formula,

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

The charge on ${\text{CrO}}_4{}^{2-}$ is $-2$.

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

$\begin{array}{c}{\text{Charge on CrO}}_4{}^{2-}=\left[\left(1\times x\right)+\left(4\times \left(-2\right)\right)\right]\\ -2=x-8\\ x=+6\end{array}$

Thus, the oxidation number of $\text{Cr}$ is $+6$.

The given element is $\text{Cr}{\left(\text{OH}\right)}_3$.

The oxidation state of hydroxide in a compound is usually $-1$.

The oxidation state of chromium is assumed to be $x$ and is calculated by using the formula,

$\text{Charge on Cr}{\left(\text{OH}\right)}_3=\left[\begin{array}{c}\left(\text{Number of Cr atoms}\times \text{oxidation number of Cr}\right)+\\ \left(\text{Number of OH atoms}\times \text{oxidation number of OH}\right)\end{array}\right]$

The charge on $\text{Cr}{\left(\text{OH}\right)}_3$ is zero because it is a neutral compound.

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

$\begin{array}{c}\text{Charge on Cr}{\left(\text{OH}\right)}_3=\left[\left(1\times x\right)+\left(3\times \left(-1\right)\right)\right]\\ 0=x-3\\ x=+3\end{array}$

Thus, the oxidation number of $\text{Cr}$ in $\text{Cr}{\left(\text{OH}\right)}_3$ is $+3$.

Therefore from the above calculations, it is concluded that chromium gets reduced because the oxidation state of chromium decreases from $+6$ to $+3$.

(c)

Interpretation Introduction

To determine: The oxidizing and reducing agent in the above reaction.

Answer to Problem 4.125AP

Solution

The oxidizing agent in the above reaction is ${\text{CrO}}_4{}^{2-}$ and the reducing agent is ${\text{S}}_{\text{2}}{\text{O}}_{\text{4}}{}^{\text{2-}}$.

Explanation of Solution

Explanation

The sulfur atom present in ${\text{S}}_{\text{2}}{\text{O}}_{\text{4}}{}^{\text{2-}}$ is oxidized in the given reaction of dithionate ions $\left({\text{S}}_{\text{2}}{\text{O}}_{\text{4}}{}^{\text{2-}}\right)$ and chromate ions $\left({\text{CrO}}_{\text{4}}{}^{\text{2-}}\right)$.

Thus, ${\text{S}}_{\text{2}}{\text{O}}_{\text{4}}{}^{\text{2-}}$ is a reducing agent because it loses an electron to get oxidized and its oxidation state increases, whereas ${\text{CrO}}_4{}^{2-}$ is the oxidizing agent because it gains electrons and gets reduced.

(d)

Interpretation Introduction

To determine: The mass of ${\text{Na}}_{\text{2}}{\text{S}}_2{\text{O}}_{\text{4}}$ needed for the removal of $\text{Cr}\left(\text{VI}\right)$.

Answer to Problem 4.125AP

Solution

The required mass of ${\text{Na}}_{\text{2}}{\text{S}}_2{\text{O}}_{\text{4}}$ needed for the removal of $\text{Cr}\left(\text{VI}\right)$ is $\underset{\_}{17.24g}$.

Explanation of Solution

Explanation

Given

The volume $\left( V\right)$ of $\text{Cr}\left(\text{VI}\right)$ is $100.0\text{L}$.

Molarity of chromate ions is $0.00148\text{M}$.

The moles of a substance is calculated by the formula,

$\text{Moles}=M\times V$

Where,

• $M$ is the molarity of the substance.
• $V$ is the volume of that substance.

Substitute the given values of molarity and volume of $\text{Cr}\left(\text{VI}\right)$ in the above equation.

$\begin{array}{c}\text{Moles}\text{of}\text{Cr}\left(\text{VI}\right)=0.00148\text{mol/L}\times 100.0\text{L}\\ =\text{0}\text{.0148}\text{moles}\end{array}$

The above net ionic balanced reaction is,

${\text{3S}}_{\text{2}}{\text{O}}_{\text{4}}{}^{\text{2-}}\left(aq\right)+2{\text{CrO}}_{\text{4}}{}^{\text{2-}}\left(aq\right)+2{\text{OH}}^{\text{-}}\left(aq\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)\to 6{\text{SO}}_{\text{3}}{}^{\text{2-}}\left(aq\right)+2\text{Cr}{\left(\text{OH}\right)}_{\text{3}}\left(s\right)$

According to the stated balanced ionic equation, $\text{3}\text{mol}$ of ${\text{Na}}_{\text{2}}{\text{S}}_2{\text{O}}_{\text{4}}$ $\left({\text{S}}_{\text{2}}{\text{O}}_{\text{4}}{}^{\text{2-}}\right)$ react with $2\text{mol}$ of ${\text{CrO}}_{\text{4}}{}^{\text{2-}}$ ions.

Hence,

$\begin{array}{c}3\text{Moles}\text{of}{\text{Na}}_{\text{2}}{\text{S}}_2{\text{O}}_{\text{4}}\text{react}\text{with}\text{moles}\text{of}{\text{CrO}}_{\text{4}}{}^{\text{2-}}=2\\ \text{1}\text{Mole}\text{of}{\text{Na}}_{\text{2}}{\text{S}}_2{\text{O}}_{\text{4}}\text{react}\text{with}\text{moles}\text{of}{\text{CrO}}_{\text{4}}{}^{\text{2-}}=\frac{2}{3}\text{Moles}\text{of}{\text{CrO}}_{\text{4}}{}^{\text{2-}}\end{array}$

Substitute the value of the moles of ${\text{CrO}}_{\text{4}}{}^{\text{2-}}$ in the above expression.

$\begin{array}{c}\text{Moles}\text{of}{\text{Na}}_{\text{2}}{\text{S}}_2{\text{O}}_{\text{4}}=\frac{2}{3}\times \left(\text{0}\text{.0148}\text{mol}\text{of}{\text{CrO}}_{\text{4}}{}^{\text{2-}}\right)\\ =0.\text{099}\text{mole}{\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}\end{array}$

The molar mass is calculated by the formula,

$\text{Mass}\left(\text{g}\right)=\text{Molar}\text{mass}\times \text{Number}\text{of}\text{mole}$ (1)

The molar mass of $\text{Na}$ is $\text{22}\text{.99}\text{g}$.

The molar mass of sulfur is $\text{32}\text{.065}\text{g}$.

The molar mass of oxygen is $16.00\text{g}$.

The molar mass of ${\text{Na}}_{\text{2}}{\text{S}}_2{\text{O}}_{\text{4}}$ $\begin{array}{c}=2\text{Na}+2\text{S}+4\text{O}\\ =\left(\left(2\times \text{22}\text{.99}\right)+\left(2\times 32.\text{065}\right)+\left(4\times \text{16}\text{.00}\right)\right)\text{g/mol}\\ =\text{174}\text{.11}\text{g/mol}\end{array}$

Substitute the value of the number of moles and that of the molar mass in equation (1).

$\begin{array}{c}\text{Mass}\left(\text{g}\right)=\text{Molar}\text{mass}\times \text{Number}\text{of}\text{mole}\\ =174.11\text{g/mol}{\text{Na}}_{\text{2}}{\text{S}}_2{\text{O}}_{\text{4}}\times 0.099\text{mol}\\ =\underset{\_}{17.24g}\end{array}$

The mass of ${\text{Na}}_{\text{2}}{\text{S}}_2{\text{O}}_{\text{4}}$ needed for the removal of $\text{Cr}\left(\text{VI}\right)$ is $\underset{\_}{17.24g}$.

Conclusion

The given questions have been rightfully answered