Question

Balance the following oxidation-reduction. Which occur in basic solution, using the half reaction method?

1. $\mathbf{Al}\left(s\right)\mathbf{+}{{\mathbf{MnO}}_{\mathbf{4}}}^{\mathbf{-}}\left(\mathrm{aq}\right)\mathbf{\to }{\mathbf{MnO}}_{\mathbf{2}}\left(s\right)\mathbf{+}\mathbf{Al}{{\left(\mathrm{OH}\right)}_{\mathbf{4}}}^{\mathbf{-}}\left(\mathrm{aq}\right)$
2. ${\mathbf{CI}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\to }{\mathbf{Cl}}^{\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{ClO}}^{\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}$
3. ${{\mathbf{NO}}_{\mathbf{2}}}^{\mathbf{-}}\left(\mathrm{aq}\right)\mathbf{+}\mathbf{Al}\left(s\right)\mathbf{\to }{\mathbf{NH}}_{\mathbf{3}}\left(g\right)\mathbf{+}{{\mathbf{AlO}}_{\mathbf{2}}}^{\mathbf{-}}\left(\mathrm{aq}\right)$
4. ${{\mathrm{MnO}}_4}^{-}\left(\mathrm{aq}\right)+{\mathrm{S}}^{2-}\left(\mathrm{aq}\right)\to \mathrm{MnS}\left(\mathrm{s}\right)+\mathrm{S}\left(\mathrm{s}\right)$
5. ${\mathbf{CN}}^{\mathbf{-}}\left(\mathrm{aq}\right)\mathbf{+}{{\mathbf{MnO}}_{\mathbf{4}}}^{\mathbf{-}}\left(\mathrm{aq}\right)\mathbf{\to }{\mathbf{CNO}}^{\mathbf{-}}\left(\mathrm{aq}\right)\mathbf{+}{\mathbf{MnO}}_{\mathbf{2}}\left(s\right)$

Short answer

1. $\mathrm{Al}+{{\mathrm{MnO}}_4}^{-}+{\mathrm{H}}_2\mathrm{O}\to \mathrm{Al}{\left(\mathrm{OH}\right)}_4+{\mathrm{MnO}}_2$
   
2. $3{\mathrm{Cl}}_2+4{\mathrm{OH}}^{-}\to 4{\mathrm{ClO}}^{-}+2{\mathrm{Cl}}^{-}+4{\mathrm{H}}^{+}$
   
3. ${\mathrm{H}}_2\mathrm{O}+2\mathrm{Al}+{{\mathrm{NO}}_2}^{-}+{\mathrm{OH}}^{-}\to 2{\mathrm{AlO}}_2+{\mathrm{NH}}_3$
   
4. $2{{\mathrm{MnO}}_4}^{-}+8{\mathrm{H}}_2\mathrm{O}+7{\mathrm{S}}^{2-}\to 2\mathrm{MnS}+5\mathrm{S}+16{\mathrm{OH}}^{-}$
   
5. $3{\mathrm{CN}}^{-}+2{{\mathrm{MnO}}_4}^{-}+{\mathrm{H}}_2\mathrm{O}\to 3{\mathrm{CNO}}^{-}+2{\mathrm{MnO}}_2+2{\mathrm{OH}}^{-}$
   

Step-by-step solution

Subpart a)

By half-reaction method:

$\mathrm{Al}+{\mathrm{MnO}}_4\to {\mathrm{MnO}}_2+\mathrm{Al}{\left(\mathrm{OH}\right)}_4$

Balancing the equation by adding the electrons and charges:

$$\begin{gathered} \mathrm{Al}+4{\mathrm{H}}_2\mathrm{O}+4\mathrm{OH}\to \mathrm{Al}{\left(\mathrm{OH}\right)}_4+4{\mathrm{H}}_2\mathrm{O}+3{\mathrm{e}}^{-} \\ {\mathrm{MnO}}_4+4{\mathrm{H}}_2\mathrm{O}+3{\mathrm{e}}^{-}\to {\mathrm{MnO}}_2+2{\mathrm{H}}_2\mathrm{O}+2\mathrm{OH} \end{gathered}$$

Add the half-reactions together:

$\mathrm{Al}+{{\mathrm{MnO}}_4}^{-}+8{\mathrm{H}}_2\mathrm{O}+4{\mathrm{OH}}^{-}+3{\mathrm{e}}^{-}\to \mathrm{Al}{{\left(\mathrm{OH}\right)}_4}^{-}+{\mathrm{MnO}}_2+6{\mathrm{H}}_2\mathrm{O}+4\mathrm{OH}+3{\mathrm{e}}^{-}$

The final equation will be,

$\mathrm{Al}+{{\mathrm{MnO}}_4}^{-}+{\mathrm{H}}_2\mathrm{O}\to \mathrm{Al}{\left(\mathrm{OH}\right)}_4+{\mathrm{MnO}}_2$

Subpart b)

${\mathrm{Cl}}_2\left(\mathrm{g}\right)\to {\mathrm{Cl}}^{-}\left(\mathrm{aq}\right)+{\mathrm{ClO}}^{-}\left(\mathrm{aq}\right)$

On oxidation:

${\mathrm{Cl}}_2+2{\mathrm{OH}}^{-}\to 2{\mathrm{Cl}}^{-}+{\mathrm{e}}^{-}+2{\mathrm{H}}^{+}$

On reduction:

${\mathrm{Cl}}_2+2{\mathrm{e}}^{-}\to 2{\mathrm{Cl}}^{-}$

The final equation will be,$3{\mathrm{Cl}}_2+4{\mathrm{OH}}^{-}\to 4{\mathrm{ClO}}^{-}+2{\mathrm{Cl}}^{-}+4{\mathrm{H}}^{+}$

Subpart c)

${{\mathrm{NO}}_2}^{-}\left(\mathrm{aq}\right)+\mathrm{Al}\left(\mathrm{s}\right)\to {\mathrm{NH}}_3\left(\mathrm{g}\right)+{{\mathrm{AlO}}_2}^{-}\left(\mathrm{aq}\right)$

Oxidation half-reaction method:

$\mathrm{Al}\to {\mathrm{AlO}}_2$

By adding atoms and charges/oxygen and hydrogen:

$\mathrm{Al}\to {\mathrm{AlO}}_2+3{\mathrm{e}}^{-}$

Add the equations:

$$\begin{gathered} 2{\mathrm{H}}_2\mathrm{O}+\mathrm{Al}\to {\mathrm{AlO}}_2+3{\mathrm{e}}^{-}+4{\mathrm{H}}^{+} \\ \mathrm{Reduction}\mathrm{half}-\mathrm{reaction}: \\ {\mathrm{NO}}_2\to {\mathrm{NH}}_3 \end{gathered}$$

Neutralize on both sides:

$$\begin{gathered} 2{\mathrm{H}}_2\mathrm{O}+2\mathrm{Al}+{{\mathrm{NO}}_2}^{-}+2{\mathrm{OH}}^{-}\to 2{\mathrm{AlO}}_2+{\mathrm{H}}^{+}+{\mathrm{NH}}_3+{\mathrm{OH}}^{-} \\ 2{\mathrm{H}}_2\mathrm{O}+2\mathrm{Al}+{{\mathrm{NO}}_2}^{-}+2{\mathrm{OH}}^{-}\to 2{\mathrm{AlO}}_2+{\mathrm{NH}}_3+{\mathrm{H}}_2\mathrm{O} \end{gathered}$$

The final equation will be, ${\mathrm{H}}_2\mathrm{O}+2\mathrm{Al}+{{\mathrm{NO}}_2}^{-}+{\mathrm{OH}}^{-}\to 2{\mathrm{AlO}}_2+{\mathrm{NH}}_3$

Subpart d)

${{\mathrm{MnO}}_4}^{-}\left(\mathrm{aq}\right)+{\mathrm{S}}^{2-}\left(\mathrm{aq}\right)\to \mathrm{MnS}\left(\mathrm{s}\right)+\mathrm{S}\left(\mathrm{s}\right)$

Oxidation half-reaction method:

$$\begin{gathered} \mathrm{Oxidation}:{\mathrm{S}}^{2-}\to \mathrm{S} \\ \mathrm{Reduction}:{{\mathrm{MnO}}_4}^{-}\to \mathrm{MnS} \end{gathered}$$

By adding atoms and charges/oxygen and hydrogen:

$$\begin{gathered} {{\mathrm{MnO}}_4}^{-}+5{\mathrm{e}}^{-}\to \mathrm{MnS} \\ {{\mathrm{MnO}}_4}^{-}+5{\mathrm{e}}^{-}+8{\mathrm{H}}^{+}\to \mathrm{MnS}+4{\mathrm{H}}_2\mathrm{O} \end{gathered}$$

Add the equations:${{\mathrm{MnO}}_4}^{-}+16{\mathrm{H}}^{+}+7{\mathrm{S}}^{2-}\to 2\mathrm{MnS}+5\mathrm{S}+8{\mathrm{H}}_2\mathrm{O}$

Neutralize on both sides:${{\mathrm{MnO}}_4}^{-}+16{\mathrm{H}}_2\mathrm{O}-8{\mathrm{H}}_2\mathrm{O}+7{\mathrm{S}}^{2-}\to 2\mathrm{MnS}+5\mathrm{S}+16{\mathrm{OH}}^{-}$

The final equation will be, $2{{\mathrm{MnO}}_4}^{-}+8{\mathrm{H}}_2\mathrm{O}+7{\mathrm{S}}^{2-}\to 2\mathrm{MnS}+5\mathrm{S}+16{\mathrm{OH}}^{-}$

Subpart e)

${\mathrm{CN}}^{-}\left(\mathrm{aq}\right)+{{\mathrm{MnO}}_4}^{-}\left(\mathrm{aq}\right)\to {\mathrm{CNO}}^{-}\left(\mathrm{aq}\right)+{\mathrm{MnO}}_2\left(\mathrm{s}\right)$

Oxidation half-reaction method:

$$\begin{gathered} \mathrm{Oxidation}:{\mathrm{CN}}^{-}\to {\mathrm{CNO}}^{-}+{\mathrm{MnO}}_2 \\ \mathrm{Reduction}:{{\mathrm{MnO}}_4}^{-}\to {\mathrm{MnO}}_2 \end{gathered}$$

By adding atoms and charges/oxygen and hydrogen:

$3{\mathrm{CN}}^{-}+3{\mathrm{H}}_2\mathrm{O}+2{{\mathrm{MnO}}_4}^{-}+2{\mathrm{e}}^{-}+4{\mathrm{H}}^{+}\to 3{\mathrm{CNO}}^{-}+6{\mathrm{e}}^{-}+6{\mathrm{H}}^{+}+2{\mathrm{MnO}}_2+4{\mathrm{H}}_2\mathrm{O}$

The final equation will be, $3{\mathrm{CN}}^{-}+2{{\mathrm{MnO}}_4}^{-}+{\mathrm{H}}_2\mathrm{O}\to 2{\mathrm{MnO}}_2+3{\mathrm{CNO}}^{-}+2{\mathrm{OH}}^{-}$