Chapter 17, Problem 11QRT

(a)

Interpretation Introduction

Interpretation:

The half cell reaction for reduction of ${\text{MnO}}_4^{-}$ to ${\text{Mn}}^{\text{2+}}$ ion in acidic solution has to be stated.

Concept Introduction:

Redox reactions are the reactions in which oxidation and reduction occur simultaneously.  Loss of electrons or gain of oxygen is termed as oxidation.  On the other hand, addition of electrons or hydrogen is known as reduction.  Increase in oxidation number signifies oxidation and decrease in oxidation number signifies reduction.  The galvanic cell, voltaic cell electrolytic cell are the types of electrochemical cells.  In each cell, the oxidation half-reaction takes place at the anode while half-reaction reaction takes place at the cathode.

Answer to Problem 11QRT

The half cell reaction for reduction of ${\text{MnO}}_4^{-}$ to ${\text{Mn}}^{\text{2+}}$ ion in acidic solution is shown below.

    ${\text{MnO}}_4^{-}\left(\text{aq}\right)+8{\text{H}}^{\text{+}}\left(\text{aq}\right)+5e^{-}\to {\text{Mn}}^{\text{2+}}\left(\text{aq}\right)+{\text{4H}}_{\text{2}}\text{O}\left(l\right)$

Explanation of Solution

The reaction for the conversion of ${\text{MnO}}_4^{-}$ to ${\text{Mn}}^{\text{2+}}$ ion is shown below.

    ${\text{MnO}}_4^{-}\left(\text{aq}\right)\to {\text{Mn}}^{\text{2+}}\left(\text{aq}\right)$

The number of oxygen atoms is balanced by adding four molecules of water on product side as shown below.

    ${\text{MnO}}_4^{-}\left(\text{aq}\right)\to {\text{Mn}}^{\text{2+}}\left(\text{aq}\right)+{\text{4H}}_{\text{2}}\text{O}\left(l\right)$

In acidic medium, the number of hydrogen atoms is balanced by adding ${\text{H}}^{\text{+}}$ ions on reactant side as shown below.

    ${\text{MnO}}_4^{-}\left(\text{aq}\right)+8{\text{H}}^{\text{+}}\left(\text{aq}\right)\to {\text{Mn}}^{\text{2+}}\left(\text{aq}\right)+{\text{4H}}_{\text{2}}\text{O}\left(l\right)$

The charge on reactant side is $+7$ and on product side is $+2$.  Therefore, in order to balance the charge, five electrons are added on reactant side as follows.

    ${\text{MnO}}_4^{-}\left(\text{aq}\right)+8{\text{H}}^{\text{+}}\left(\text{aq}\right)+5e^{-}\to {\text{Mn}}^{\text{2+}}\left(\text{aq}\right)+{\text{4H}}_{\text{2}}\text{O}\left(l\right)$

The balanced half cell reaction represents reduction is shown below.

    ${\text{MnO}}_4^{-}\left(\text{aq}\right)+8{\text{H}}^{\text{+}}\left(\text{aq}\right)+5e^{-}\to {\text{Mn}}^{\text{2+}}\left(\text{aq}\right)+{\text{4H}}_{\text{2}}\text{O}\left(l\right)$

(b)

Interpretation Introduction

Interpretation:

The half cell reaction for reduction of ${\text{Cr}}_2{\text{O}}_7^{2-}$ to ${\text{Cr}}^{\text{3+}}$ ion in acidic solution has to be stated.

Concept Introduction:

Refer to part (a).

Answer to Problem 11QRT

The half cell reaction for reduction of ${\text{Cr}}_2{\text{O}}_7^{2-}$ to ${\text{Cr}}^{\text{3+}}$ ion in acidic solution is shown below.

    ${\text{Cr}}_2{\text{O}}_7^{2-}\left(\text{aq}\right)+14{\text{H}}^{\text{+}}\left(\text{aq}\right)+9e^{-}\to {\text{Cr}}^{\text{3+}}\left(\text{aq}\right)+7{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Explanation of Solution

The reaction for the conversion of ${\text{Cr}}_2{\text{O}}_7^{2-}$ to ${\text{Cr}}^{\text{3+}}$ ion is shown below.

    ${\text{Cr}}_2{\text{O}}_7^{2-}\left(\text{aq}\right)\to {\text{Cr}}^{\text{3+}}\left(\text{aq}\right)$

The number of oxygen atoms is balanced by adding seven molecules of water on product side as shown below.

    ${\text{Cr}}_2{\text{O}}_7^{2-}\left(\text{aq}\right)\to {\text{Cr}}^{\text{3+}}\left(\text{aq}\right)+7{\text{H}}_{\text{2}}\text{O}\left(l\right)$

In acidic medium, the number of hydrogen atoms is balanced by adding ${\text{H}}^{\text{+}}$ ions on reactant side as shown below.

    ${\text{Cr}}_2{\text{O}}_7^{2-}\left(\text{aq}\right)+14{\text{H}}^{\text{+}}\left(\text{aq}\right)\to {\text{Cr}}^{\text{3+}}\left(\text{aq}\right)+7{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The charge on reactant side is $+12$ and on product side is $+3$.  Therefore, in order to balance the charge, nine electrons are added on reactant side as follows.

    ${\text{Cr}}_2{\text{O}}_7^{2-}\left(\text{aq}\right)+14{\text{H}}^{\text{+}}\left(\text{aq}\right)+9e^{-}\to {\text{Cr}}^{\text{3+}}\left(\text{aq}\right)+7{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The balanced half cell reaction represents reduction is shown below.

    ${\text{Cr}}_2{\text{O}}_7^{2-}\left(\text{aq}\right)+14{\text{H}}^{\text{+}}\left(\text{aq}\right)+9e^{-}\to {\text{Cr}}^{\text{3+}}\left(\text{aq}\right)+7{\text{H}}_{\text{2}}\text{O}\left(l\right)$

(c)

Interpretation Introduction

Interpretation:

The half cell reaction for oxidation of chlorine gas to ${\text{ClO}}^{-}$ ions has to be stated.

Concept Introduction:

Refer to part (a).

Answer to Problem 11QRT

The half cell reaction for oxidation of chlorine gas to ${\text{ClO}}^{-}$ ions is shown below.

    $\frac{1}{2}{\text{Cl}}_{\text{2}}\left(\text{g}\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\to {\text{ClO}}^{-}\left(\text{aq}\right)+2{\text{H}}^{\text{+}}\left(\text{aq}\right)+e^{-}$

Explanation of Solution

The reaction for the conversion of chlorine gas to ${\text{ClO}}^{-}$ ions is shown below.

    $\frac{1}{2}{\text{Cl}}_{\text{2}}\left(\text{g}\right)\to {\text{ClO}}^{-}\left(\text{aq}\right)$

The number of oxygen atoms is balanced by adding one molecule of water on reactant side as shown below.

    $\frac{1}{2}{\text{Cl}}_{\text{2}}\left(\text{g}\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\to {\text{ClO}}^{-}\left(\text{aq}\right)$

The number of hydrogen atoms is balanced by adding ${\text{H}}^{\text{+}}$ ions on product side as shown below.

    $\frac{1}{2}{\text{Cl}}_{\text{2}}\left(\text{g}\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\to {\text{ClO}}^{-}\left(\text{aq}\right)+2{\text{H}}^{\text{+}}\left(\text{aq}\right)$

The charge on reactant side is $0$ and on product side is $+1$.  Therefore, in order to balance the charge, one electron is added on product side as follows.

    $\frac{1}{2}{\text{Cl}}_{\text{2}}\left(\text{g}\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\to {\text{ClO}}^{-}\left(\text{aq}\right)+2{\text{H}}^{\text{+}}\left(\text{aq}\right)+e^{-}$

The balanced half cell reaction represents oxidation is shown below.

    $\frac{1}{2}{\text{Cl}}_{\text{2}}\left(\text{g}\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\to {\text{ClO}}^{-}\left(\text{aq}\right)+2{\text{H}}^{\text{+}}\left(\text{aq}\right)+e^{-}$

(d)

Interpretation Introduction

Interpretation:

The half cell reaction for reduction of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ to ${\text{H}}_{\text{2}}\text{O}$ in acidic solution has to be stated.

Concept Introduction:

Refer to part (a).

Answer to Problem 11QRT

The half cell reaction for reduction of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ to ${\text{H}}_{\text{2}}\text{O}$ in acidic solution is shown below.

    ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(\text{aq}\right)+2{\text{H}}^{\text{+}}\left(\text{aq}\right)+2e^{-}\to 2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Explanation of Solution

The reaction for the conversion of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ to ${\text{H}}_{\text{2}}\text{O}$ is shown below.

    ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(\text{aq}\right)\to {\text{H}}_{\text{2}}\text{O}\left(l\right)$

The number of oxygen atoms is balanced by adding one molecule of water on product side as shown below.

    ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(\text{aq}\right)\to {\text{H}}_{\text{2}}\text{O}\left(l\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

In acidic medium, the number of hydrogen atoms is balanced by adding ${\text{H}}^{\text{+}}$ ions on reactant side as shown below.

    ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(\text{aq}\right)+2{\text{H}}^{\text{+}}\left(\text{aq}\right)\to {\text{H}}_{\text{2}}\text{O}\left(l\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The charge on reactant side is $+2$ and on product side is $0$.  Therefore, in order to balance the charge, two electrons are added on reactant side as follows.

    ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(\text{aq}\right)+2{\text{H}}^{\text{+}}\left(\text{aq}\right)+2e^{-}\to {\text{H}}_{\text{2}}\text{O}\left(l\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The balanced half cell reaction represents reduction is shown below.

    ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(\text{aq}\right)+2{\text{H}}^{\text{+}}\left(\text{aq}\right)+2e^{-}\to 2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

(e)

Interpretation Introduction

Interpretation:

The half cell reaction for oxidation of ${\text{HNO}}_{\text{2}}$ to ${\text{NO}}_3^{-}$ ion in acidic condition has to be stated.

Concept Introduction:

Refer to part (a).

Answer to Problem 11QRT

The half cell reaction for oxidation of ${\text{HNO}}_{\text{2}}$ to ${\text{NO}}_3^{-}$ ion in acidic condition is shown below.

    ${\text{HNO}}_{\text{2}}+{\text{H}}_{\text{2}}\text{O}\to {\text{NO}}_3^{-}+3{\text{H}}^{\text{+}}+2e^{-}$

Explanation of Solution

The reaction for the conversion of ${\text{HNO}}_{\text{2}}$ to ${\text{NO}}_3^{-}$ ion is shown below.

    ${\text{HNO}}_{\text{2}}\to {\text{NO}}_3^{-}$

The number of oxygen atoms is balanced by adding one molecule of water on reactant side as shown below.

    ${\text{HNO}}_{\text{2}}+{\text{H}}_{\text{2}}\text{O}\to {\text{NO}}_3^{-}$

In acidic medium, the number of hydrogen atoms is balanced by adding ${\text{H}}^{\text{+}}$ ions on product side as shown below.

    ${\text{HNO}}_{\text{2}}+{\text{H}}_{\text{2}}\text{O}\to {\text{NO}}_3^{-}+3{\text{H}}^{\text{+}}$

The charge on reactant side is zero.  Therefore, in order to balance the charge, two electrons are added on product side as follows.

    ${\text{HNO}}_{\text{2}}+{\text{H}}_{\text{2}}\text{O}\to {\text{NO}}_3^{-}+3{\text{H}}^{\text{+}}+2e^{-}$

The half cell reaction for oxidation of ${\text{HNO}}_{\text{2}}$ to ${\text{NO}}_3^{-}$ ion in acidic condition is shown below.

    ${\text{HNO}}_{\text{2}}+{\text{H}}_{\text{2}}\text{O}\to {\text{NO}}_3^{-}+3{\text{H}}^{\text{+}}+2e^{-}$