Chapter 14, Problem 55QP

(a)

Interpretation Introduction

Interpretation:

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

Explanation of Solution

The skeleton equation chemical equation is shown below:

${\text{H}}_2\text{S}\left(aq\right){\text{+Cr}}_2{\text{O}}_7^{2-}\left(aq\right)\stackrel{}{\to }\text{S}\left(s\right){\text{+Cr}}^{3+}\left(aq\right)$ …(1)

The oxidation states are assigned in the above skeleton chemical equation as shown below:

${\text{H}}_2\stackrel{-2}{\text{S}}\left(aq\right)\text{+}{\stackrel{+6}{\text{Cr}}}_2{\text{O}}_7^{2-}\left(aq\right)\stackrel{}{\to }\stackrel{0}{\text{S}}\left(s\right){\text{+Cr}}^{3+}\left(aq\right)$ …(2)

The six electrons are added on the left and change the coefficient of ${\text{Cr}}^{\text{3+}}$ to $2$ in the half-reaction.

${\stackrel{+6}{\text{Cr}}}_2{\text{O}}_7^{2-}\left(aq\right)+6e^{-}\stackrel{}{\to }{\text{2Cr}}^{3+}\left(aq\right)$ …(3)

Two electrons are added on right side of the reduction-half of the given reaction.

${\text{H}}_2\stackrel{-2}{\text{S}}\left(aq\right)\stackrel{}{\to }\stackrel{0}{\text{S}}\left(s\right)+2e^{-}$ …(4)

To balance the electron on the both sides of the overall reaction, equation (4) is multiplied with $3$ .

${\text{3H}}_2\stackrel{-2}{\text{S}}\left(aq\right)\stackrel{}{\to }3\stackrel{0}{\text{S}}\left(s\right)+6e^{-}$ …(5)

Further, equation ( $3$ ) and equation (5) are added.

${\stackrel{+6}{\text{Cr}}}_2{\text{O}}_7^{2-}\left(aq\right)+{\text{3H}}_2\stackrel{-2}{\text{S}}\left(aq\right)\stackrel{}{\to }3\stackrel{0}{\text{S}}\left(s\right)+{\text{2Cr}}^{3+}\left(aq\right)$ …(6)

To balance ionic charge on both of the chemical reactions, $8$ hydrogen ions are added on the left side of equation (6).

${\stackrel{+6}{\text{Cr}}}_2{\text{O}}_7^{2-}\left(aq\right)+{\text{3H}}_2\stackrel{-2}{\text{S}}\left(aq\right)+8{\text{H}}^{+}\stackrel{}{\to }3\stackrel{0}{\text{S}}\left(s\right)+{\text{2Cr}}^{3+}\left(aq\right)$ …(7)

To balance the hydrogen atom $4$ of water molecules are added to the right side of chemical equation (7).

${\stackrel{+6}{\text{Cr}}}_2{\text{O}}_7^{2-}\left(aq\right)+{\text{3H}}_2\stackrel{-2}{\text{S}}\left(aq\right)+8{\text{H}}^{+}\stackrel{}{\to }3\stackrel{0}{\text{S}}\left(s\right)+{\text{2Cr}}^{3+}\left(aq\right)+4{\text{H}}_2\text{O}\left(l\right)$ …(8)

The chemical equation (8) represents the balanced chemical equation for the given reaction.

(b)

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 equation chemical equation is shown as follows,

${\text{V}}^{2+}\left(aq\right){\text{+MnO}}_4^{-}\left(aq\right)\stackrel{}{\to }{\text{VO}}_2^{+}\left(aq\right){\text{+Mn}}^{2+}\left(aq\right)$ …(9)

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

${\text{V}}^{2+}\left(aq\right)\text{+}\stackrel{+7}{\text{Mn}}{\text{O}}_4^{-}\left(aq\right)\stackrel{}{\to }\stackrel{+5}{\text{V}}{\text{O}}_2^{+}\left(aq\right){\text{+Mn}}^{2+}\left(aq\right)$ …(10)

The three electrons are added on the right side in the oxidation half-reaction.

${\text{V}}^{2+}\left(aq\right)\stackrel{}{\to }\stackrel{+5}{\text{V}}{\text{O}}_2^{+}\left(aq\right){\text{+3e}}^{-}$ …(11)

Five electrons are added on the left side of the reduction-half reaction of the one given.

$\stackrel{+7}{\text{Mn}}{\text{O}}_4^{-}\left(aq\right)+5e^{-}\stackrel{}{\to }{\text{Mn}}^{2+}\left(aq\right)$ …(12)

To balance the electrons on both sides of the overall reaction, equation (12) is multiplied with $3$ and equation (11) is multiplied with $5$ and added as shown below:

$\begin{array}{l}{\text{ 5V}}^{2+}\left(aq\right)\to 5\stackrel{+5}{\text{V}}{\text{O}}_2^{+}\left(aq\right){\text{+15e}}^{-}\\ \underset{¯}{3\stackrel{+7}{\text{Mn}}{\text{O}}_4^{-}\left(aq\right)+15e^{-}\to {\text{3Mn}}^{2+}\left(aq\right)}\\ \underset{¯}{{\text{5V}}^{2+}\left(aq\right)+3\stackrel{+7}{\text{Mn}}{\text{O}}_4^{-}\left(aq\right)+4{\text{H}}^{+}\to 5\stackrel{+5}{\text{V}}{\text{O}}_2^{+}\left(aq\right){\text{+3Mn}}^{2+}\left(aq\right)}\cdot \cdot \cdot \cdot \cdot \cdot \left(13\right)\end{array}$

To balance ionic charge on both sides of the chemical reaction, $4$ hydrogen ions are added on the left side of equation (13).

${\text{5V}}^{2+}\left(aq\right)+3\stackrel{+7}{\text{Mn}}{\text{O}}_4^{-}\left(aq\right)+4{\text{H}}^{+}\stackrel{}{\to }5\stackrel{+5}{\text{V}}{\text{O}}_2^{+}\left(aq\right){\text{+3Mn}}^{2+}\left(aq\right)$ …(14)

To balance the hydrogen atom $2$ of the water molecules are added to the right side of the chemical equation (14).

${\text{5V}}^{2+}\left(aq\right)+3\stackrel{+7}{\text{Mn}}{\text{O}}_4^{-}\left(aq\right)+4{\text{H}}^{+}\stackrel{}{\to }5\stackrel{+5}{\text{V}}{\text{O}}_2^{+}\left(aq\right){\text{+3Mn}}^{2+}\left(aq\right)+2{\text{H}}_2\text{O}$ …(15)

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

(c)

Interpretation Introduction

Interpretation:

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

Explanation of Solution

The skeleton equation chemical equation is shown as follows,

${\text{Fe}}^{2+}\left(aq\right)+{\text{ClO}}_3^{-}\left(aq\right)\stackrel{}{\to }{\text{Fe}}^{3+}\left(aq\right)+{\text{Cl}}^{-}\left(aq\right)$ …(16)

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

${\text{Fe}}^{2+}\left(aq\right)+\stackrel{+5}{\text{Cl}}{\text{O}}_3^{-}\left(aq\right)\stackrel{}{\to }{\text{Fe}}^{3+}\left(aq\right)+{\text{Cl}}^{-}\left(aq\right)$ …(17)

Six electrons are added on the left side of the oxidation half-reaction.

$\stackrel{+5}{\text{Cl}}{\text{O}}_3^{-}\left(aq\right)+6e^{-}\stackrel{}{\to }{\text{Cl}}^{-}\left(aq\right)$ ….(18)

An electron is added on the right side of the reduction-half of the given reaction.

${\text{Fe}}^{2+}\left(aq\right)\stackrel{}{\to }{\text{Fe}}^{3+}\left(aq\right)+e^{-}$ …(19)

To balance the electrons on sides both of the overall reaction, equation (19) is multiplied with $6$ .

${\text{6Fe}}^{2+}\left(aq\right)\stackrel{}{\to }{\text{6Fe}}^{3+}\left(aq\right)+6e^{-}$ …(20)

Further, equation (18) and equation (20) are added.

${\text{6Fe}}^{2+}\left(aq\right)+\stackrel{+5}{\text{Cl}}{\text{O}}_3^{-}\left(aq\right)\stackrel{}{\to }{\text{6Fe}}^{3+}\left(aq\right)+{\text{Cl}}^{-}\left(aq\right)$ …(21)

To balance ionic charge on both sides of chemical reaction, $6$ hydrogen ions are added on the left side of equation (21).

${\text{6Fe}}^{2+}\left(aq\right)+\stackrel{+5}{\text{Cl}}{\text{O}}_3^{-}\left(aq\right)+6{\text{H}}^{+}\stackrel{}{\to }{\text{6Fe}}^{3+}\left(aq\right)+{\text{Cl}}^{-}\left(aq\right)$ …(22)

To balance the hydrogen atom $3$ of the water molecules are added to the right side of chemical equation (22).

${\text{6Fe}}^{2+}\left(aq\right)+\stackrel{+5}{\text{Cl}}{\text{O}}_3^{-}\left(aq\right)+6{\text{H}}^{+}\stackrel{}{\to }{\text{6Fe}}^{3+}\left(aq\right)+{\text{Cl}}^{-}\left(aq\right)+3{\text{H}}_2\text{O}\left(l\right)$ …(23)

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