Question

Balance the following equations by the half-reaction method

1. $\mathbf{Fe}\left(s\right)\mathbf{+}\mathbf{HCl}\left(\mathrm{aq}\right)\mathbf{\to }{\mathbf{HFeCl}}_{\mathbf{4}}\left(\mathrm{aq}\right)\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\left(g\right)$
2. ${{\mathbf{IO}}_{\mathbf{3}}}^{\mathbf{-}}\left(\mathrm{aq}\right)\mathbf{+}{\mathbf{I}}^{\mathbf{-}}\left(\mathrm{aq}\right)\stackrel{\mathbf{Acidic}}{\mathbf{\to }}{{\mathbf{I}}_{\mathbf{3}}}^{\mathbf{-}}\left(\mathrm{aq}\right)$
3. $\mathbf{Cr}{{\left(\mathrm{NCS}\right)}_{\mathbf{6}}}^{\mathbf{4}\mathbf{-}}\left(\mathrm{aq}\right)\mathbf{+}{\mathbf{Ce}}^{\mathbf{4}\mathbf{+}}\left(\mathrm{aq}\right)\mathbf{\to }{\mathbf{Cr}}^{\mathbf{3}\mathbf{+}}\left(\mathrm{aq}\right)\mathbf{+}{\mathbf{Ce}}^{\mathbf{3}\mathbf{+}}\left(\mathrm{aq}\right)\mathbf{+}{\mathbf{NO}}^{\mathbf{-}}\left(\mathrm{aq}\right)\mathbf{+}{\mathbf{CO}}_{\mathbf{2}}\left(g\right)\mathbf{+}{{\mathbf{SO}}_{\mathbf{4}}}^{\mathbf{2}\mathbf{-}}\left(\mathrm{aq}\right)$
4. ${\mathbf{CrI}}_{\mathbf{3}}\left(s\right)\mathbf{+}{\mathbf{CI}}_{\mathbf{2}}\left(g\right)\stackrel{\mathbf{Basic}}{\mathbf{\to }}{{\mathbf{CrO}}_{\mathbf{4}}}^{\mathbf{2}\mathbf{-}}\left(\mathrm{aq}\right)\mathbf{+}{{\mathbf{IO}}_{\mathbf{4}}}^{\mathbf{2}\mathbf{-}}\left(\mathrm{aq}\right)\mathbf{+}{\mathbf{Cl}}^{\mathbf{-}}\left(\mathrm{aq}\right)$
   
5. $\mathbf{Fe}{{\left(\mathrm{CN}\right)}_{\mathbf{6}}}^{\mathbf{4}\mathbf{-}}\left(\mathrm{aq}\right)\mathbf{+}{\mathbf{Ce}}^{\mathbf{4}\mathbf{+}}\left(\mathrm{aq}\right)\mathbf{\to }\mathbf{Ce}{\left(\mathrm{OH}\right)}_{\mathbf{3}}\left(s\right)\mathbf{+}\mathbf{Fe}{\left(\mathrm{OH}\right)}_{\mathbf{3}}\left(s\right)\mathbf{+}{\mathbf{CO}}_{\mathbf{3}}^{\mathbf{2}\mathbf{-}}\left(\mathrm{aq}\right)\mathbf{+}{\mathbf{NO}}_{\mathbf{3}}^{\mathbf{-}}\left(\mathrm{aq}\right)$
   

Short answer

All the given equations are balanced using half-reaction method.

Step-by-step solution

Subpart a)

Split the whole reaction into half-reactions:

$$\begin{gathered} \mathrm{Reduction}:{\mathrm{H}}^{+}\to {\mathrm{H}}_2 \\ \mathrm{Oxidation}:{\mathrm{Cl}}^{-}+\mathrm{Fe}\to {\mathrm{HFeCl}}_4 \end{gathered}$$

In this equation disregarding the atoms

$$\begin{gathered} \mathrm{Reduction}:\mathrm{H}\to {\mathrm{H}}_2 \\ \mathrm{Oxidation}:4{\mathrm{Cl}}^{-}+\mathrm{Fe}\to {\mathrm{HFeCl}}_4 \end{gathered}$$

Balancing of H-atom by adding the H-atom

$$\begin{gathered} \mathrm{Oxidation}:2{\mathrm{e}}^{-}+{\mathrm{H}}^{+}\to {\mathrm{H}}_2 \\ \mathrm{Reduction}:{\mathrm{H}}^{+}+4{\mathrm{Cl}}^{-}+\mathrm{Fe}\to {\mathrm{HFeCl}}_4+3{\mathrm{e}}^{-} \end{gathered}$$

Subpart b)

$\mathrm{Fe}\left(\mathrm{s}\right)+\mathrm{HCl}\left(\mathrm{aq}\right)\to {\mathrm{HFeCl}}_4\left(\mathrm{aq}\right)+{\mathrm{H}}_2\left(\mathrm{g}\right)$

Balancing the equation by adding ${\mathrm{e}}^{-}$:

$$\begin{gathered} \mathrm{Reduction}:2{\mathrm{e}}^{-}+2{\mathrm{H}}^{+}\to {\mathrm{H}}_2 \\ \mathrm{Oxidation}:{\mathrm{H}}^{+}+4{\mathrm{Cl}}^{-}+\mathrm{Fe}\to {\mathrm{HFeCl}}_4+3{\mathrm{e}}^{-} \end{gathered}$$

For removing the electrons, one should multiply the reduction reaction by 3 and the oxidation reaction by 2.

$$\begin{gathered} \mathrm{Reduction}:3\left(2{\mathrm{e}}^{-}+2{\mathrm{H}}^{+}\to {\mathrm{H}}_2\right) \\ 6{\mathrm{e}}^{-}+4{\mathrm{H}}^{+}\to 3{\mathrm{H}}_2 \\ \mathrm{Oxidation}:2\left({\mathrm{H}}^{+}+4{\mathrm{Cl}}^{-}+\mathrm{Fe}\to {\mathrm{HFeCl}}_4+3{\mathrm{e}}^{-}\right) \\ 2{\mathrm{H}}^{+}+8{\mathrm{Cl}}^{-}+2\mathrm{Fe}\to 2{\mathrm{HFeCl}}_4+6{\mathrm{e}}^{-} \end{gathered}$$

Cancelling on both sides of the equation:

$$\begin{gathered} \mathrm{Reduction}:6{\mathrm{H}}^{+}\to 3{\mathrm{H}}_2 \\ \mathrm{Oxidation}:2{\mathrm{H}}^{+}+8{\mathrm{Cl}}^{-}+2\mathrm{Fe}\to 2{\mathrm{HFeCl}}_4 \end{gathered}$$

The final equation will be,

$8\mathrm{HCl}+2\mathrm{Fe}\to 2{\mathrm{HFeCl}}_4+3{\mathrm{H}}_2$

Subpart c)

${{\mathrm{IO}}_3}^{-}\left(\mathrm{aq}\right)+{\mathrm{I}}^{-}\left(\mathrm{aq}\right)\stackrel{\mathrm{Acidic}}{\to }{{\mathrm{I}}_3}^{-}\left(\mathrm{aq}\right)$

$$\begin{gathered} \mathrm{Oxidation}:3{\mathrm{I}}^{-}\to {{\mathrm{I}}_3}^{-}+2{\mathrm{e}}^{-} \\ \mathrm{Reduction}:3{{\mathrm{IO}}_3}^{-}+16{\mathrm{e}}^{-}+18\mathrm{H}\to {{\mathrm{I}}_3}^{-}+9{\mathrm{H}}_2\mathrm{O} \end{gathered}$$

The final equation will be,

${\mathrm{IO}}_3^{-}+8{\mathrm{I}}^{-}+6{\mathrm{H}}^{+}\to 3{\mathrm{I}}_3^{-}+3{\mathrm{H}}_2\mathrm{O}$

Subpart d)

$\mathrm{Cr}{\left(\mathrm{NCS}\right)}_6^{4-}\left(\mathrm{aq}\right)+{\mathrm{Ce}}^{4+}\left(\mathrm{aq}\right)\stackrel{\mathrm{Acidic}}{\to }{\mathrm{Cr}}^{3+}\left(\mathrm{aq}\right)+{\mathrm{Ce}}^{3+}\left(\mathrm{aq}\right)+{\mathrm{NO}}^{-}\left(\mathrm{aq}\right)+{\mathrm{CO}}_2\left(\mathrm{g}\right)+{\mathrm{SO}}_4^{2-}\left(\mathrm{aq}\right)$

$$\begin{gathered} \mathrm{Oxidation}:\mathrm{Cr}{{\left(\mathrm{CNS}\right)}_6}^{4-}+54{\mathrm{H}}_2\mathrm{O}+97{\mathrm{e}}^{-}\to {\mathrm{Cr}}^{3+}+6{\mathrm{NO}}_3^{-}+6{\mathrm{CO}}_2+6{\mathrm{SO}}_4^{2-}+108{\mathrm{H}}^{+} \\ \mathrm{Reduction}:{\mathrm{Ce}}^{4+}+{\mathrm{e}}^{-}\to {\mathrm{Ce}}^{3+} \end{gathered}$$

The final equation will be,

$\mathrm{Cr}{\left(\mathrm{NCS}\right)}_6^{4-}\left(\mathrm{aq}\right)+97{\mathrm{Ce}}^{4+}\left(\mathrm{aq}\right)+54{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\to {\mathrm{Cr}}^{3+}\left(\mathrm{aq}\right)+97{\mathrm{Ce}}^{3+}\left(\mathrm{aq}\right)+6{\mathrm{NO}}_3^{-}\left(\mathrm{aq}\right)+6{\mathrm{CO}}_2\left(\mathrm{g}\right)+6{\mathrm{SO}}_4^{2-}\left(\mathrm{aq}\right)+108{\mathrm{H}}^{+}\left(\mathrm{aq}\right)$

Subpart e)

${\mathrm{CrI}}_3\left(\mathrm{s}\right)+{\mathrm{Cl}}_2\left(\mathrm{g}\right)\stackrel{\mathrm{Basic}}{\to }{\mathrm{CrO}}_4^{2-}\left(\mathrm{aq}\right)+{\mathrm{IO}}_4^{2-}\left(\mathrm{aq}\right)+{\mathrm{Cl}}^{-}\left(\mathrm{aq}\right)$

$$\begin{gathered} \mathrm{Oxidation}:{\mathrm{CrI}}_3+16{\mathrm{H}}_2\mathrm{O}+32{\mathrm{OH}}^{-}\to {\mathrm{CrO}}_4^{2-}+27{\mathrm{e}}^{-}+3{\mathrm{IO}}_4^{-}+32{\mathrm{H}}_2\mathrm{O} \\ \mathrm{Reduction}:{\mathrm{Cl}}_2+2{\mathrm{e}}^{-}\to 2{\mathrm{Cl}}^{-} \end{gathered}$$

The final equation will be,

$2{\mathrm{CrI}}_3+27{\mathrm{Cl}}_2+64{\mathrm{OH}}^{-}\to 2{{\mathrm{CrO}}_4}^{2-}+54{\mathrm{Cl}}^{-}+6{\mathrm{IO}}_4^{-}+32{\mathrm{H}}_2\mathrm{O}$

Subpart d)

$\mathrm{Fe}{{\left(\mathrm{CN}\right)}_6}^{4-}\left(\mathrm{aq}\right)+{\mathrm{Ce}}^{4+}\left(\mathrm{aq}\right)\stackrel{\mathrm{Basic}}{\to }\mathrm{Ce}{\left(\mathrm{OH}\right)}_3\left(\mathrm{s}\right)+\mathrm{Fe}{\left(\mathrm{OH}\right)}_3\left(\mathrm{s}\right)+{\mathrm{CO}}_3^{2-}\left(\mathrm{aq}\right)+{\mathrm{NO}}_3^{-}\left(\mathrm{aq}\right)$

$$\begin{gathered} \mathrm{Oxidation}:\mathrm{Fe}{\left(\mathrm{CN}\right)}_6^{4-}+21{\mathrm{H}}_2\mathrm{O}+39{\mathrm{OH}}^{-}\to \mathrm{Fe}{\left(\mathrm{OH}\right)}_3+6{{\mathrm{CO}}_3}^{2-}+39{\mathrm{H}}_2\mathrm{O}+31{\mathrm{e}}^{-} \\ \mathrm{Reduction}:{\mathrm{Ce}}^{4+}+3{\mathrm{H}}_2\mathrm{O}+3{\mathrm{OH}}^{-}+{\mathrm{e}}^{-}\to \mathrm{Ce}{\left(\mathrm{OH}\right)}_3+3{\mathrm{H}}_2\mathrm{O} \end{gathered}$$

The final equation will be,

$\mathrm{Fe}{\left(\mathrm{CN}\right)}_6^{4-}+31{\mathrm{Ce}}^{4+}+32{\mathrm{OH}}^{-}\to \mathrm{Fe}{\left(\mathrm{OH}\right)}_3+31\mathrm{Ce}{\left(\mathrm{OH}\right)}_3+6{\mathrm{CO}}_3^{2-}+18{\mathrm{H}}_2\mathrm{O}$