Chapter 14, Problem 56QP

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

$\begin{array}{l}\left(\text{a}\right){\text{ S}}^{2-}\left(aq\right)+{\text{CrO}}_4^{2-}\left(aq\right)\to \text{S}\left(s\right)+\text{Cr}{\left(\text{OH}\right)}_{\text{3}}\left(s\right)\hfill \\ \left(\text{b}\right){\text{ MnO}}_4^{-}\left(aq\right)+{\text{CN}}^{-}\left(aq\right)\to {\text{MnO}}_4^{2-}\left(aq\right)+{\text{CNO}}^{-}\left(aq\right)\hfill \\ \left(\text{C}\right)\text{ A1}\left(s\right)+{\text{OH}}^{-}\left(aq\right)\to \text{Al}{\left(\text{OH}\right)}_4^{-}\left(aq\right)+{\text{H}}_{\text{2}}\left(g\right)\hfill \end{array}$

Interpretation Introduction

Interpretation:

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

Explanation of Solution

The skeleton chemical equation is shown as follows,

${\text{S}}^{2-}\left(aq\right){\text{+CrO}}_4^{2-}\left(aq\right)\stackrel{}{\to }\text{S}\left(s\right)\text{+Cr}{\left(\text{OH}\right)}_3\left(s\right)$ …(1)

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

${\text{S}}^{2-}\left(aq\right)\text{+}\stackrel{+6}{\text{Cr}}{\text{O}}_4^{2-}\left(aq\right)\stackrel{}{\to }\stackrel{0}{\text{S}}\left(s\right)\text{+}\stackrel{+3}{\text{Cr}}{\left(\text{OH}\right)}_3\left(s\right)$ …(2)

Three electrons are added on the left of the half-reaction.

$\stackrel{+6}{\text{Cr}}{\text{O}}_4^{2-}\left(aq\right)+3e^{-}\stackrel{}{\to }\stackrel{+3}{\text{Cr}}{\left(\text{OH}\right)}_3\left(s\right)$ …(3)

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

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

To balance the electrons, equation (3) is multiplied with $2$ .

$2\stackrel{+6}{\text{Cr}}{\text{O}}_4^{2-}\left(aq\right)+6e^{-}\stackrel{}{\to }2\stackrel{+3}{\text{Cr}}{\left(\text{OH}\right)}_3\left(s\right)$ …(5)

To balance the electrons, equation (4) is multiplied with $3$ .

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

Further, equation (5) and equation (6) are added.

$\begin{array}{c}\underset{¯}{\begin{array}{l}2\stackrel{+6}{\text{Cr}}{\text{O}}_4^{2-}\left(aq\right)+6e^{-}\stackrel{}{\to }2\stackrel{+3}{\text{Cr}}{\left(\text{OH}\right)}_3\left(s\right)\\ 3\stackrel{-2}{\text{S}}\left(aq\right)\stackrel{}{\to }3\stackrel{0}{\text{S}}\left(s\right)+6e^{-}\end{array}}\\ \underset{¯}{2\stackrel{+6}{\text{Cr}}{\text{O}}_4^{2-}\left(aq\right)+3\stackrel{-2}{\text{S}}\left(aq\right)\stackrel{}{\to }2\stackrel{+3}{\text{Cr}}{\left(\text{OH}\right)}_3\left(s\right)+3\stackrel{0}{\text{S}}\left(s\right)}\end{array}$ …(7)

To balance the ionic charge on both sides of the chemical reaction, $10$ hydroxyl ions are added on the right side of equation (7).

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

To balance the hydrogen atom $4$ of the water molecules are added to the left side of chemical equation (8).

${\stackrel{+6}{\text{Cr}}}_2{\text{O}}_7^{2-}\left(aq\right)+{\text{3H}}_2\stackrel{-2}{\text{S}}\left(aq\right)+4{\text{H}}_2\text{O}\left(l\right)\stackrel{}{\to }3\stackrel{0}{\text{S}}\left(s\right)+{\text{2Cr}}^{3+}\left(aq\right)+8{\text{OH}}^{-}\left(aq\right)$ …(9)

The chemical equation (9) represents the balanced chemical equation for the given reaction.

Interpretation Introduction

Interpretation:

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

Explanation of Solution

The skeleton chemical equation is shown as follows,

${\text{CN}}^{-}\left(aq\right){\text{+MnO}}_4^{-}\left(aq\right)\stackrel{}{\to }{\text{CNO}}^{-}\left(aq\right){\text{+MnO}}_4^{2-}\left(aq\right)$ …(10)

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

$\stackrel{+2}{\text{C}}{\text{N}}^{-}\left(aq\right)\text{+}\stackrel{+7}{\text{Mn}}{\text{O}}_4^{-}\left(aq\right)\stackrel{}{\to }\stackrel{+4}{\text{C}}{\text{NO}}^{-}\left(aq\right)\text{+}\stackrel{+6}{\text{Mn}}{\text{O}}_4^{2-}\left(aq\right)$ …(11)

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

$\stackrel{+2}{\text{C}}{\text{N}}^{-}\left(aq\right)\stackrel{}{\to }\stackrel{+4}{\text{C}}{\text{NO}}^{-}\left(aq\right)\text{+}2e^{-}$ …(12)

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

$\stackrel{+7}{\text{Mn}}{\text{O}}_4^{-}\left(aq\right)+e^{-}\stackrel{}{\to }\stackrel{+6}{\text{Mn}}{\text{O}}_4^{-2}\left(aq\right)$ …(13)

To balance the electrons on both sides of the overall reaction, equation (13) is multiplied with $2$ .

$2\stackrel{+7}{\text{Mn}}{\text{O}}_4^{-}\left(aq\right)+2e^{-}\stackrel{}{\to }2\stackrel{+6}{\text{Mn}}{\text{O}}_4^{-2}\left(aq\right)$ …(14)

Further, equation (12) and equation (14) are added.

$2\stackrel{+7}{\text{Mn}}{\text{O}}_4^{-}\left(aq\right)+\stackrel{+2}{\text{C}}{\text{N}}^{-}\left(aq\right)\stackrel{}{\to }2\stackrel{+6}{\text{Mn}}{\text{O}}_4^{-2}\left(aq\right)+\stackrel{+4}{\text{C}}{\text{NO}}^{-}\left(aq\right)$ …(15)

To balance ionic charge on both sides of the chemical reaction, $2$ hydroxyl ions are added on the left side of equation (15).

$2\stackrel{+7}{\text{Mn}}{\text{O}}_4^{-}\left(aq\right)+\stackrel{+2}{\text{C}}{\text{N}}^{-}\left(aq\right)+2{\text{OH}}^{-}\stackrel{}{\to }2\stackrel{+6}{\text{Mn}}{\text{O}}_4^{-2}\left(aq\right)+\stackrel{+4}{\text{C}}{\text{NO}}^{-}\left(aq\right)$ …(16)

To balance the hydrogen atom $1$ of the water molecules is added to the right side of chemical equation (16).

$2\stackrel{+7}{\text{Mn}}{\text{O}}_4^{-}\left(aq\right)+\stackrel{+2}{\text{C}}{\text{N}}^{-}\left(aq\right)+2{\text{OH}}^{-}\stackrel{}{\to }2\stackrel{+6}{\text{Mn}}{\text{O}}_4^{-2}\left(aq\right)+\stackrel{+4}{\text{C}}{\text{NO}}^{-}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)$ …(17)

The chemical equation (17) represents the balanced chemical equation for the given reaction.

Interpretation Introduction

Interpretation:

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

Explanation of Solution

The skeleton chemical equation is shown as follows,

$\text{Al}\left(s\right)+{\text{OH}}^{-}\left(aq\right)\stackrel{}{\to }\text{Al}{\left(\text{OH}\right)}_4^{-}\left(aq\right)+{\text{H}}_2\left(g\right)$ …(18)

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

$\stackrel{0}{\text{Al}}\left(s\right)+{\text{OH}}^{-}\left(aq\right)\stackrel{}{\to }\stackrel{3+}{\text{Al}}{\left(\text{OH}\right)}_4^{-}\left(aq\right)+{\stackrel{0}{\text{H}}}_2\left(g\right)$ …(19)

Two electrons are added on the left side on the reduction half-reaction.

${\text{2OH}}^{-}\left(aq\right)+2e^{-}\stackrel{}{\to }{\text{H}}_2\left(g\right)$ …(20)

Three electrons are added on the right side of the oxidation-half of the given reaction.

$\stackrel{0}{\text{Al}}\left(s\right)\stackrel{}{\to }\stackrel{3+}{\text{Al}}{\left(\text{OH}\right)}_4^{-}\left(aq\right)+3e^{-}$ …(21)

To balance the electrons on both sides of the overall reaction, equation (21) is multiplied with $2$ .

$2\stackrel{0}{\text{Al}}\left(s\right)\stackrel{}{\to }2\stackrel{3+}{\text{Al}}{\left(\text{OH}\right)}_4^{-}\left(aq\right)+6e^{-}$ …(22)

To balance the electrons on both sides of the overall reaction, equation (20) is multiplied with $3$ .

${\text{6OH}}^{-}\left(aq\right)+6e^{-}\stackrel{}{\to }{\text{3H}}_2\left(g\right)$ …(23)

Further, equation (23) and equation (22) are added.

$\begin{array}{c}\underset{¯}{\begin{array}{l}2\stackrel{0}{\text{Al}}\left(s\right)\stackrel{}{\to }2\stackrel{3+}{\text{Al}}{\left(\text{OH}\right)}_4^{-}\left(aq\right)+6e^{-}\\ {\text{6OH}}^{-}\left(aq\right)+6e^{-}\stackrel{}{\to }{\text{3H}}_2\left(g\right)\end{array}}\\ \underset{¯}{2\stackrel{0}{\text{Al}}\left(s\right)+{\text{6OH}}^{-}\left(aq\right)\stackrel{}{\to }2\stackrel{3+}{\text{Al}}{\left(\text{OH}\right)}_4^{-}\left(aq\right)+{\text{3H}}_2\left(g\right)}\end{array}$ …(24)

To balance ionic charge on both sides of the chemical reaction, $4$ hydroxyl ions are added on the right side of equation (24).

$2\stackrel{0}{\text{Al}}\left(s\right)+2{\text{OH}}^{-}\left(aq\right)\stackrel{}{\to }2\stackrel{3+}{\text{Al}}{\left(\text{OH}\right)}_4^{-}\left(aq\right)+{\text{3H}}_2\left(g\right)$ …(25)

To balance the hydrogen atom $6$ of the water molecules are added to the left side of chemical equation (24).

$2\stackrel{0}{\text{Al}}\left(s\right)+2{\text{OH}}^{-}\left(aq\right)+6{\text{H}}_{\text{2}}\text{O}\left(l\right)\stackrel{}{\to }2\stackrel{3+}{\text{Al}}{\left(\text{OH}\right)}_4^{-}\left(aq\right)+{\text{3H}}_2\left(g\right)$ …(26)

The chemical equation (25) represents the balanced chemical equation for the given reaction.