Chapter 14, Problem 58QP

Complete and balance the following oxidation-reduction reactions, assuming they occur in acidic solution.

$\begin{array}{l}\left(\text{a}\right){\text{ C}}_{\text{2}}{\text{O}}_4^{2-}\left(aq\right)+{\text{MnO}}_4^{-}\left(aq\right)\to {\text{CO}}_{\text{2}}\left(g\right)+{\text{Mn}}^{2+}\left(aq\right)\hfill \\ \left(\text{b}\right){\text{ ClO}}_3^{-}\left(aq\right)+{\text{Cl}}^{-}\left(aq\right)\to {\text{Cl}}_{\text{2}}\left(aq\right)\hfill \\ \left(\text{c}\right){\text{ S}}_{\text{2}}{\text{O}}_3^{2-}\left(aq\right)+{\text{I}}_{\text{2}}\left(aq\right)\to {\text{SO}}_4^{2-}\left(aq\right)+{\text{I}}^{-}\left(aq\right)\hfill \end{array}$

Interpretation Introduction

Interpretation:

The given redox chemical equation, which takes place in acidic medium, is to be completed and balanced.

Explanation of Solution

The skeleton chemical equation is shown as follows,

${\text{MnO}}_4^{-}\left(aq\right)+{\text{C}}_2{\text{O}}_4^{2-}\left(aq\right)\stackrel{}{\to }{\text{CO}}_2\left(g\right){\text{+Mn}}^{2+}\left(aq\right)$ …(1)

The oxidation states are assigned in the above skeleton chemical equation which are shown as follow,

$\stackrel{+7}{\text{Mn}}{\text{O}}_4^{-}\left(aq\right)+{\stackrel{+3}{\text{C}}}_2{\text{O}}_4^{2-}\left(aq\right)\stackrel{}{\to }\stackrel{+4}{\text{C}}{\text{O}}_2\left(g\right){\text{+Mn}}^{2+}\left(aq\right)$ …(2)

The five electrons are added on the left side of the reduction half-reaction.

$\stackrel{+7}{\text{Mn}}{\text{O}}_4^{-}\left(aq\right)+5e^{-}\stackrel{}{\to }{\text{Mn}}^{2+}\left(aq\right)$ …(3)

Two electrons are added on the right side and change the coefficient of carbon dioxide to $2$ of the oxidation-half of the given reaction.

${\stackrel{+3}{\text{C}}}_2{\text{O}}_4^{2-}\left(aq\right)\stackrel{}{\to }2\stackrel{+4}{\text{C}}{\text{O}}_2\left(g\right)+2e^{-}$ …(4)

To balance the electrons on both sides of the overall reaction, equation (3) is multiplied with $2$ and equation (4) is multiplied with $5$ and are added as shown below:

$\begin{array}{l}2\stackrel{+7}{\text{Mn}}{\text{O}}_4^{-}\left(aq\right)+10e^{-}\stackrel{}{\to }{\text{2Mn}}^{2+}\left(aq\right)\\ \underset{¯}{5{\stackrel{+3}{\text{C}}}_2{\text{O}}_4^{2-}\left(aq\right)\stackrel{}{\to }10\stackrel{+4}{\text{C}}{\text{O}}_2\left(g\right)+10e^{-}}\\ \underset{¯}{2\stackrel{+7}{\text{Mn}}{\text{O}}_4^{-}\left(aq\right)+5{\stackrel{+3}{\text{C}}}_2{\text{O}}_4^{2-}\left(aq\right)\stackrel{}{\to }{\text{2Mn}}^{2+}\left(aq\right)+10\stackrel{+4}{\text{C}}{\text{O}}_2\left(g\right)}\mathrm{......}\left(5\right)\end{array}$

To balance ionic charge on both of the chemical reactions, $16$ hydrogen ions are added on left side of equation (5).

$2\stackrel{+7}{\text{Mn}}{\text{O}}_4^{-}\left(aq\right)+5{\stackrel{+3}{\text{C}}}_2{\text{O}}_4^{2-}\left(aq\right)+16{\text{H}}^{+}\stackrel{}{\to }{\text{2Mn}}^{2+}\left(aq\right)+10\stackrel{+4}{\text{C}}{\text{O}}_2\left(g\right)$ ...(6)

To balance the hydrogen atom $8$ of the water molecules are added to the right side of chemical equation (6).

$2\stackrel{+7}{\text{Mn}}{\text{O}}_4^{-}\left(aq\right)+5{\stackrel{+3}{\text{C}}}_2{\text{O}}_4^{2-}\left(aq\right)+16{\text{H}}^{+}\stackrel{}{\to }{\text{2Mn}}^{2+}\left(aq\right)+10\stackrel{+4}{\text{C}}{\text{O}}_2\left(g\right)+8{\text{H}}_2\text{O}$ …(7)

The chemical equation (7) represents the balanced chemical equation for the given reaction.

Interpretation Introduction

Interpretation:

The given redox chemical equation, which takes place in acidic medium, is to be completed and balanced.

Explanation of Solution

The skeleton chemical equation is shown as follows,

${\text{ClO}}_3^{-}\left(aq\right)+{\text{Cl}}^{-}\left(aq\right)\stackrel{}{\to }{\text{Cl}}_2\left(aq\right)$ ...(8)

The oxidation states are assigned in the above skeleton chemical equation which are shown as follow,

$\stackrel{+5}{\text{Cl}}{\text{O}}_3^{-}\left(aq\right)+{\text{Cl}}^{-}\left(aq\right)\stackrel{}{\to }{\stackrel{0}{\text{Cl}}}_2\left(aq\right)$ ...(9)

The 10 electrons are added on the left side and change the coefficient of ${\text{ClO}}_3^{-}$ to $2$ in the reduction half-reaction.

$2\stackrel{+5}{\text{Cl}}{\text{O}}_3^{-}\left(aq\right)+10e^{-}\stackrel{}{\to }{\stackrel{0}{\text{Cl}}}_2\left(aq\right)$ …(10)

Two electrons are added on right side and change the coefficient of ${\text{Cl}}^{-}$ to $2$ in the oxidation-half of the given reaction.

${\text{2Cl}}^{-}\left(aq\right)\stackrel{}{\to }{\stackrel{0}{\text{Cl}}}_2\left(aq\right)+2e^{-}$ …(11)

To balance the electrons on both sides of the overall reaction, equation (11) is multiplied with $5$ .

${\text{10Cl}}^{-}\left(aq\right)\stackrel{}{\to }5{\stackrel{0}{\text{Cl}}}_2\left(aq\right)+10e^{-}$ …(12)

Further, equation (10) and equation (12) are added.

$2\stackrel{+5}{\text{Cl}}{\text{O}}_3^{-}\left(aq\right)+{\text{10Cl}}^{-}\left(aq\right)\stackrel{}{\to }6{\stackrel{0}{\text{Cl}}}_2\left(aq\right)$ …(13)

To balance ionic charge on both of the chemical reactions, $12$ hydrogen ions are added on the left side of equation (13).

$2\stackrel{+5}{\text{Cl}}{\text{O}}_3^{-}\left(aq\right)+{\text{10Cl}}^{-}\left(aq\right)+12{\text{H}}^{+}\stackrel{}{\to }6{\stackrel{0}{\text{Cl}}}_2\left(aq\right)$ …(14)

To balance the hydrogen atom $6$ of the water molecules are added to the right side of chemical equation (14).

$2\stackrel{+5}{\text{Cl}}{\text{O}}_3^{-}\left(aq\right)+{\text{10Cl}}^{-}\left(aq\right)+12{\text{H}}^{+}\stackrel{}{\to }6{\stackrel{0}{\text{Cl}}}_2\left(aq\right)+6{\text{H}}_2\text{O}$ …(15)

The chemical equation (15) represents the balanced chemical equation for the given reaction.

Interpretation Introduction

Interpretation:

The given redox chemical equation, takes place in acidic medium, is to be completed and balanced.

Explanation of Solution

The skeleton chemical equation is shown as follows,

${\text{S}}_2{\text{O}}_3^{2-}\left(aq\right)+{\text{I}}_2\left(aq\right)\stackrel{}{\to }{\text{SO}}_4^{2-}\left(aq\right)+{\text{I}}^{-}\left(aq\right)$ …(16)

The oxidation states are assigned in the above skeleton chemical equation which are shown as follow,

${\stackrel{+2}{\text{S}}}_2{\text{O}}_3^{2-}\left(aq\right)+{\stackrel{0}{\text{I}}}_2\left(aq\right)\stackrel{}{\to }\stackrel{+6}{\text{S}}{\text{O}}_4^{2-}\left(aq\right)+{\text{I}}^{-}\left(aq\right)$ …(17)

The two electrons are added on the left side and change the coefficient of ${\text{I}}^{-}$ to $2$ in the reduction half-reaction.

${\stackrel{0}{\text{I}}}_2\left(aq\right)+2e^{-}\stackrel{}{\to }{\text{2I}}^{-}\left(aq\right)$ …(18)

The eight electrons are added on the right side and change the coefficient of ${\text{SO}}_4^{2-}$ to $2$ in the reduction half-reaction.

${\stackrel{+2}{\text{S}}}_2{\text{O}}_3^{2-}\left(aq\right)\stackrel{}{\to }2\stackrel{+6}{\text{S}}{\text{O}}_4^{2-}\left(aq\right)+8e^{-}$ …(19)

To balance the electrons on both sides of the overall reaction, equation (18) is multiplied with $4$ .

$4{\stackrel{0}{\text{I}}}_2\left(aq\right)+8e^{-}\stackrel{}{\to }{\text{8I}}^{-}\left(aq\right)$ …(20)

Further, equation (19) and equation (20) are added.

${\stackrel{+2}{\text{S}}}_2{\text{O}}_3^{2-}\left(aq\right)+4{\stackrel{0}{\text{I}}}_2\left(aq\right)\stackrel{}{\to }2\stackrel{+6}{\text{S}}{\text{O}}_4^{2-}\left(aq\right)+{\text{8I}}^{-}\left(aq\right)$ …(21)

To balance ionic charge on both sides of the chemical reaction, $10$ hydrogen ions are added on the right side of equation (21).

${\stackrel{+2}{\text{S}}}_2{\text{O}}_3^{2-}\left(aq\right)+4{\stackrel{0}{\text{I}}}_2\left(aq\right)\stackrel{}{\to }2\stackrel{+6}{\text{S}}{\text{O}}_4^{2-}\left(aq\right)+{\text{8I}}^{-}\left(aq\right)+10{\text{H}}^{+}$ …(22)

To balance the hydrogen atom $5$ of the water molecules are added to the left side of chemical equation (22).

${\stackrel{+2}{\text{S}}}_2{\text{O}}_3^{2-}\left(aq\right)+4{\stackrel{0}{\text{I}}}_2\left(aq\right)+5{\text{H}}_2\text{O}\stackrel{}{\to }2\stackrel{+6}{\text{S}}{\text{O}}_4^{2-}\left(aq\right)+{\text{8I}}^{-}\left(aq\right)+10{\text{H}}^{+}$ …(23)

The chemical equation (23) represents the balanced chemical equation for the given reaction.