Question

The following balanced equations represent reactions that occur in aqueous acid. Break them down into balanced oxidation and reduction half-equations.

(a)$\mathbf{2}{\mathbf{H}}_{\mathbf{3}}{\mathbf{O}}^{\mathbf{+}}\left(\mathbf{aq}\right)\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}{}_{\mathbf{2}}\left(\mathbf{aq}\right)\mathbf{+}\mathbf{2}\mathbf{Fe}{}^{\mathbf{2}\mathbf{+}}\left(\mathbf{aq}\right)\to \mathbf{2}{\mathbf{Fe}}^{\mathbf{3}\mathbf{+}}\left(\mathbf{aq}\right)\mathbf{+}\mathbf{4}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(\mathbf{l}\right)$

(b)${\mathbf{H}}_{\mathbf{3}}{\mathbf{O}}^{\mathbf{+}}\left(\mathbf{aq}\right)\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(\mathbf{l}\right){\mathbf{+}\mathbf{2}\mathbf{MnO}}_{\mathbf{4}}^{\mathbf{-}}\left(\mathbf{aq}\right)\mathbf{+}\mathbf{5}{\mathbf{SO}}_{\mathbf{2}}\left(\mathbf{aq}\right)\to \mathbf{2}{\mathbf{Mn}}^{\mathbf{2}\mathbf{+}}\left(\mathbf{aq}\right){\mathbf{+}\mathbf{5}\mathbf{HSO}}_{\mathbf{4}}^{\mathbf{-}}\left(\mathbf{aq}\right)$

(c)${\mathbf{5}\mathbf{ClO}}_{\mathbf{2}}^{\mathbf{-}}\left(\mathbf{aq}\right)\mathbf{+}\mathbf{4}{\mathbf{H}}_{\mathbf{3}}{\mathbf{O}}^{\mathbf{+}}\left(\mathbf{aq}\right)\to \mathbf{4}{\mathbf{ClO}}_{\mathbf{2}}\left(\mathbf{g}\right)\mathbf{+}{\mathbf{Cl}}^{\mathbf{-}}\left(\mathbf{aq}\right)\mathbf{+}\mathbf{6}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(\mathbf{l}\right)$

Short answer

(a)The oxidation half equation is:

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

The reduction half equation is:

${\text{2H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{2}}\text{+ 2 e}{}^{\text{-}}\to {\text{4H}}_{\text{2}}\text{O}$

(b)Oxidation half equation:

${\text{5 SO}}_{\text{2}}\left(\text{aq}\right){\text{+ 10H}}_{\text{2}}\text{O}\left(\text{l}\right)\to {\text{5 HSO}}_{\text{4}}^{\text{-}}\left(\text{aq}\right){\text{+ 15H}}_{\text{3}}{\text{O}}^{\text{+}}\left(\text{aq}\right){\text{+ 10e}}^{\text{-}}$

Reduction half equation:

${\text{2MnO}}_{\text{4}}^{\text{-}}\left(\text{aq}\right){\text{+ 16H}}_{\text{3}}{\text{O}}^{\text{+}}\left(\text{aq}\right)\text{+ 10e}{}^{-}\to {\text{2Mn}}^{\text{2 +}}\left(\text{aq}\right){\text{+ 8H}}_{2}O\left(l\right)$

(c)The oxidation half reaction is:

${\text{4ClO}}_{\text{2}}^{\text{-}}\left(\text{aq}\right)\to {\text{4ClO}}_{\text{2}}\left(\text{g}\right){\text{+ 4e}}^{\text{-}}$

The reduction half reaction is:

${\text{ClO}}_{\text{2}}^{\text{-}}\left(\text{aq}\right){\text{+ 4H}}_{\text{3}}\text{O}{}^{\text{+}}\left(\text{aq}\right){\text{+ 4e}}^{\text{-}}\to {\text{Cl}}^{\text{-}}\left(\text{aq}\right){\text{+ 6H}}_{\text{2}}\text{O}\left(\text{l}\right)$

Step-by-step solution

Redox reaction

There are several kinds ofredox reactions. Some of them aredecomposition reaction and combination reactions.

Breaking the equation into oxidation and reduction half equations

(a)The given reaction is:

${\text{2H}}_{\text{3}}{\text{O}}^{\text{+}}\left(\text{aq}\right){\text{+H}}_{\text{2}}\text{O}{}_{\text{2}}\left(\text{aq}\right)\text{+ 2 Fe}{}^{\text{2 +}}\left(\text{aq}\right)\to {\text{2 Fe}}^{\text{3 +}}\left(\text{aq}\right){\text{+ 4H}}_{\text{2}}\text{O}\left(\text{l}\right)$

In this equation, localid="1663579804002">Fe2 +is getting oxidized into localid="1663579814754" ${\text{Fe}}^{\text{3 +}}$. The molecule${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$is getting reduced to water.

The oxidation half equation is:

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

The reduction half equation is:

${\text{2H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{2}}\text{+ 2 e}{}^{\text{-}}\to {\text{4H}}_{\text{2}}\text{O}$

(b)The given reaction is:

${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\left(\text{aq}\right){\text{+H}}_{\text{2}}\text{O}\left(\text{l}\right){\text{+ 2MnO}}_{\text{4}}^{\text{-}}\left(\text{aq}\right){\text{+ 5SO}}_{\text{2}}\left(\text{aq}\right)\to {\text{2Mn}}^{\text{2 +}}\left(\text{aq}\right){\text{+ 5HSO}}_{\text{4}}^{\text{-}}\left(\text{aq}\right)$

In this equation${\text{MnO}}_{\text{4}}^{\text{-}}$getting converted to ${\text{Mn}}^{\text{2 +}}$.This process is reduction. In the reduction part ${\text{SO}}_{\text{2}}$, is getting converted into ${\text{HSO}}_{\text{4}}^{\text{-}}$.

The oxidation and reduction half equation can be given as:

Oxidation half equation:

${\text{5 SO}}_{\text{2}}\left(\text{aq}\right){\text{+ 10H}}_{\text{2}}\text{O}\left(\text{l}\right)\to {\text{5 HSO}}_{\text{4}}^{\text{-}}\left(\text{aq}\right){\text{+ 15H}}_{\text{3}}{\text{O}}^{\text{+}}\left(\text{aq}\right){\text{+ 10e}}^{\text{-}}$

Reduction half equation:

${\text{2MnO}}_{\text{4}}^{\text{-}}\left(\text{aq}\right){\text{+ 16H}}_{\text{3}}{\text{O}}^{\text{+}}\left(\text{aq}\right)\text{+ 10e}{}^{-}\to {\text{2Mn}}^{\text{2 +}}\left(\text{aq}\right){\text{+ 8H}}_{2}O\left(l\right)$

(c) The given reaction is:

${\text{5ClO}}_{\text{2}}^{\text{-}}\left(\text{aq}\right){\text{+ 4H}}_{\text{3}}{\text{O}}^{\text{+}}\left(\text{aq}\right)\to {\text{4ClO}}_{\text{2}}\left(\text{g}\right){\text{+ Cl}}^{\text{-}}\left(\text{aq}\right){\text{+ 6H}}_{\text{2}}\text{O}\left(\text{l}\right)$

In the oxidation part,${\text{ClO}}_{\text{2}}^{\text{-}}$is getting converted to ${\text{ClO}}_{\text{2}}$. In the reduction part,${\text{ClO}}_{\text{2}}^{\text{-}}$is getting converted to ${\text{Cl}}^{\text{-}}$.

The oxidation half reaction is:

${\text{4ClO}}_{\text{2}}^{\text{-}}\left(aq\right)\to {\text{4ClO}}_{\text{2}}\left(g\right){\text{+ 4e}}^{\text{-}}$

The reduction half reaction is:

${\text{ClO}}_{\text{2}}^{\text{-}}\left(\text{aq}\right){\text{+ 4H}}_{\text{3}}\text{O}{}^{\text{+}}\left(\text{aq}\right){\text{+ 4e}}^{\text{-}}\to {\text{Cl}}^{\text{-}}\left(\text{aq}\right){\text{+ 6H}}_{\text{2}}\text{O}\left(\text{l}\right)$