Chapter 17, Problem 28AE

(a)

Interpretation Introduction

Interpretation:

The oxidizing agent has to be identified in the indicated unbalanced equation.

  ${\text{KMnO}}_4+\text{HCl}\to {\text{Cl}}_{\text{2}}+{\text{MnCl}}_2+\text{KCl}+{\text{H}}_{\text{2}}\text{O}$

Concept Introduction:

• In order to balance any redox reaction method used in acidic medium is summarized below.
• In the first step oxidation number of each element is identified.
• The overall reaction is divided into oxidation and reduction held cell based upon changes observed in oxidation states.
• Each side of half-cell equation is balanced except $\text{H}$ and $\text{O}$.  The electrons lost are made equal to electron gained so as to add two half-cells and formulate overall reaction equation.
• In next step $\text{O}$ atoms are balanced by addition of water molecule.
• For the acidic medium, excess electrons are neutralized by addition of ${\text{H}}^{+}$ ions to each side.

Oxidation refers to phenomenon of loss of electrons; loss of electropositive $\text{H}$ and gain of electronegative $\text{O}$ and substance oxidized is termed as reducing agent.

Reduction is process that involves gain of electrons, gain of $\text{H}$ or loss of electronegative $\text{O}$ atom and substance reduced is termed as oxidizing agent.

Explanation of Solution

The oxidation number of each element is assigned as follows:

  $\stackrel{+1}{\overbrace{\text{K}}}\stackrel{+7}{\overbrace{\text{Mn}}}{\stackrel{-2}{\overbrace{\text{O}}}}_{\text{4}}+\stackrel{+1}{\overbrace{\text{H}}}\stackrel{-1}{\overbrace{\text{Cl}}}\to \stackrel{+1}{\overbrace{\text{K}}}\stackrel{-1}{\overbrace{\text{Cl}}}+\stackrel{+2}{\overbrace{\text{Mn}}}\stackrel{-1}{\overbrace{{\text{Cl}}_{\text{2}}}}+\stackrel{+1}{\overbrace{{\text{H}}_2}}\stackrel{-2}{\overbrace{\text{O}}}+{\stackrel{0}{\overbrace{\text{Cl}}}}_{\text{2}}$        (1)

Since increase in oxidation state of $\text{Cl}$ on transformation from $\text{HCl}$ to ${\text{Cl}}_2$ indicate oxidation thus oxidation half reaction is formulated as follows:

  $\text{2H}\stackrel{-1}{\overbrace{\text{Cl}}}\to {\stackrel{0}{\overbrace{\text{Cl}}}}_{\text{2}}+2{\text{e}}^{-}$        (2)

Since decrease in oxidation state of $\text{Mn}$ on transformation from ${\text{KMnO}}_4$ to ${\text{MnCl}}_2$ indicates reduction thus corresponding reduction half reaction is formulated as follows:

  $\mathrm{K}\stackrel{+7}{\overbrace{Mn}}{O}_4+3\text{HCl}+{\text{5e}}^{-}\to \stackrel{+5}{\overbrace{\text{Mn}}}{\text{Cl}}_{\text{2}}+\text{KCl}+{\text{4H}}_{\text{2}}\text{O}$        (3)

Multiply equation (2) by 5 and balance excess electronic charge by addition of ${\text{H}}^{+}$ ion to right side to obtain equation as follows:

  $\text{10H}\stackrel{-1}{\overbrace{\text{Cl}}}\to 5{\stackrel{0}{\overbrace{\text{Cl}}}}_{\text{2}}+10{\text{e}}^{-}+10{\text{H}}^{+}$        (4)

Multiply equation (2) by 5 and balance excess electronic charge by addition of ${\text{H}}^{+}$ ion to left side to obtain equation as follows:

  $2\mathrm{K}\stackrel{+7}{\overbrace{Mn}}{O}_4+6\text{HCl}+10{\text{e}}^{-}+10{\text{H}}^{+}\to 2\stackrel{+5}{\overbrace{\text{Mn}}}{\text{Cl}}_{\text{2}}+2\text{KCl}+8{\text{H}}_{\text{2}}\text{O}$        (5)

Add equation (4) and equation (6) to get overall reaction as follows:

  ${\text{2KMnO}}_4+16\text{HCl}\to 5{\text{Cl}}_{\text{2}}+2{\text{MnCl}}_2+2\text{KCl}+8{\text{H}}_{\text{2}}\text{O}$

Since substance reduced acts as oxidizing agent hence ${\text{KMnO}}_4$ that undergoes reduction behaves as oxidizing agent.

(b)

Interpretation Introduction

Interpretation:

The reducing agent has to be identified in indicated unbalanced equation.

  ${\text{KMnO}}_4+\text{HCl}\to {\text{Cl}}_{\text{2}}+{\text{MnCl}}_2+\text{KCl}+{\text{H}}_{\text{2}}\text{O}$

Concept Introduction:

Refer to part (a).

Explanation of Solution

The oxidation number of each element is assigned as follows:

  $\stackrel{+1}{\overbrace{\text{K}}}\stackrel{+7}{\overbrace{\text{Mn}}}{\stackrel{-2}{\overbrace{\text{O}}}}_{\text{4}}+\stackrel{+1}{\overbrace{\text{H}}}\stackrel{-1}{\overbrace{\text{Cl}}}\to \stackrel{+1}{\overbrace{\text{K}}}\stackrel{-1}{\overbrace{\text{Cl}}}+\stackrel{+2}{\overbrace{\text{Mn}}}\stackrel{-1}{\overbrace{{\text{Cl}}_{\text{2}}}}+\stackrel{+1}{\overbrace{{\text{H}}_2}}\stackrel{-2}{\overbrace{\text{O}}}+{\stackrel{0}{\overbrace{\text{Cl}}}}_{\text{2}}$        (1)

Since increase in oxidation state of $\text{Cl}$ on transformation from $\text{HCl}$ to ${\text{Cl}}_2$ indicate oxidation thus oxidation half reaction is formulated as follows:

  $\text{2H}\stackrel{-1}{\overbrace{\text{Cl}}}\to {\stackrel{0}{\overbrace{\text{Cl}}}}_{\text{2}}+2{\text{e}}^{-}$        (2)

Since decrease in oxidation state of $\text{Mn}$ on transformation from ${\text{KMnO}}_4$ to ${\text{MnCl}}_2$ indicates reduction thus corresponding reduction half reaction is formulated as follows:

  $\mathrm{K}\stackrel{+7}{\overbrace{Mn}}{O}_4+3\text{HCl}+{\text{5e}}^{-}\to \stackrel{+5}{\overbrace{\text{Mn}}}{\text{Cl}}_{\text{2}}+\text{KCl}+{\text{4H}}_{\text{2}}\text{O}$        (3)

Multiply equation (2) by 5 and balanceexcess electronic charge by addition of ${\text{H}}^{+}$ ion to right side to obtain equation as follows:

  $\text{10H}\stackrel{-1}{\overbrace{\text{Cl}}}\to 5{\stackrel{0}{\overbrace{\text{Cl}}}}_{\text{2}}+10{\text{e}}^{-}+10{\text{H}}^{+}$        (4)

Multiply equation (2) by 5 and balance excess electronic charge by addition of ${\text{H}}^{+}$ ion to left side to obtain equation as follows:

  $2\mathrm{K}\stackrel{+7}{\overbrace{Mn}}{O}_4+6\text{HCl}+10{\text{e}}^{-}+10{\text{H}}^{+}\to 2\stackrel{+5}{\overbrace{\text{Mn}}}{\text{Cl}}_{\text{2}}+2\text{KCl}+8{\text{H}}_{\text{2}}\text{O}$        (5)

Add equation (4) and equation (6) to cancel electrons nod proton from each side and overall reaction as follows:

  ${\text{2KMnO}}_4+16\text{HCl}\to 5{\text{Cl}}_{\text{2}}+2{\text{MnCl}}_2+2\text{KCl}+8{\text{H}}_{\text{2}}\text{O}$

Since substance oxidized acts as reducing agent hence $\text{HCl}$ that undergoes oxidation behaves as reducing agent.

(c)

Interpretation Introduction

Interpretation:

Electrons transferred per mole of ${\text{KMnO}}_4$ have to be calculatedfor indicated unbalanced equation.

  ${\text{KMnO}}_4+\text{HCl}\to {\text{Cl}}_{\text{2}}+{\text{MnCl}}_2+\text{KCl}+{\text{H}}_{\text{2}}\text{O}$

Concept Introduction:

Refer to part (a).

Explanation of Solution

The oxidation number of each element is assigned as follows:

  $\stackrel{+1}{\overbrace{\text{K}}}\stackrel{+7}{\overbrace{\text{Mn}}}{\stackrel{-2}{\overbrace{\text{O}}}}_{\text{4}}+\stackrel{+1}{\overbrace{\text{H}}}\stackrel{-1}{\overbrace{\text{Cl}}}\to \stackrel{+1}{\overbrace{\text{K}}}\stackrel{-1}{\overbrace{\text{Cl}}}+\stackrel{+2}{\overbrace{\text{Mn}}}\stackrel{-1}{\overbrace{{\text{Cl}}_{\text{2}}}}+\stackrel{+1}{\overbrace{{\text{H}}_2}}\stackrel{-2}{\overbrace{\text{O}}}+{\stackrel{0}{\overbrace{\text{Cl}}}}_{\text{2}}$        (1)

Since increase in oxidation state of $\text{Cl}$ on transformation from $\text{HCl}$ to ${\text{Cl}}_2$ indicate oxidation thus oxidation half reaction is formulated as follows:

  $\text{2H}\stackrel{-1}{\overbrace{\text{Cl}}}\to {\stackrel{0}{\overbrace{\text{Cl}}}}_{\text{2}}+2{\text{e}}^{-}$        (2)

Since decrease in oxidation state of $\text{Mn}$ on transformation from ${\text{KMnO}}_4$ to ${\text{MnCl}}_2$ indicates reduction thus corresponding reduction half reaction is formulated as follows:

  $\mathrm{K}\stackrel{+7}{\overbrace{Mn}}{O}_4+3\text{HCl}+{\text{5e}}^{-}\to \stackrel{+5}{\overbrace{\text{Mn}}}{\text{Cl}}_{\text{2}}+\text{KCl}+{\text{4H}}_{\text{2}}\text{O}$        (3)

Multiply equation (2) by 5 and balance excess electronic charge by addition of ${\text{H}}^{+}$ ion to right side to obtain equation as follows:

  $\text{10H}\stackrel{-1}{\overbrace{\text{Cl}}}\to 5{\stackrel{0}{\overbrace{\text{Cl}}}}_{\text{2}}+10{\text{e}}^{-}+10{\text{H}}^{+}$        (4)

Multiply equation (2) by 5 and balance excess electronic charge by addition of ${\text{H}}^{+}$ ion to left side to obtain equation as follows:

  $2\mathrm{K}\stackrel{+7}{\overbrace{Mn}}{O}_4+6\text{HCl}+10{\text{e}}^{-}+10{\text{H}}^{+}\to 2\stackrel{+5}{\overbrace{\text{Mn}}}{\text{Cl}}_{\text{2}}+2\text{KCl}+8{\text{H}}_{\text{2}}\text{O}$        (5)

Therefore it is evident from equation (5) that overall 10 electrons are transferred for ${\text{2KMnO}}_4$, so electrons transferred per mole of ${\text{KMnO}}_4$ is 5.