Question

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

(a)$\mathbf{4}{\mathbf{PH}}_{\mathbf{3}}\left(g\right)\mathbf{+}\mathbf{4}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(l\right){\mathbf{+}\mathbf{4}\mathbf{CrO}}_{\mathbf{4}}^{\mathbf{2}\mathbf{-}}\left(\mathrm{aq}\right)\mathbf{\to }{\mathbf{P}}_{\mathbf{4}}\left(s\right)\mathbf{+}\mathbf{4}\mathbf{Cr}{\left(\mathrm{OH}\right)}_{\mathbf{4}}^{\mathbf{-}}\left(\mathrm{aq}\right)\mathbf{+}\mathbf{4}{\mathbf{OH}}^{\mathbf{-}}\left(\mathrm{aq}\right)$

(b)${\mathbf{NiO}}_{\mathbf{2}}\left(s\right)\mathbf{+}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(l\right)\mathbf{+}\mathbf{Fe}\left(s\right)\mathbf{\to }\mathbf{Ni}{\left(\mathrm{OH}\right)}_{\mathbf{2}}\left(s\right)\mathbf{+}\mathbf{Fe}{\left(\mathrm{OH}\right)}_{\mathbf{2}}\left(s\right)$

(c)${\mathbf{CO}}_{\mathbf{2}}\left(g\right)\mathbf{+}\mathbf{2}{\mathbf{NH}}_{\mathbf{2}}\mathbf{OH}\left(\mathrm{aq}\right)\mathbf{\to }\mathbf{CO}\left(g\right)\mathbf{+}{\mathbf{N}}_{\mathbf{2}}\left(g\right)\mathbf{+}\mathbf{3}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(g\right)$

Short answer

(a)The oxidation half reaction can be given as:

${\text{4PH}}_{\text{3}}\left(\text{g}\right){\text{+ 4H}}_{\text{2}}\text{O}\left(\text{l}\right)\to {\text{P}}_{\text{4}}\left(\text{s}\right){\text{+ 4OH}}^{\text{-}}\left(\text{aq}\right)$

The reduction half reaction can be given as:

${\text{4CrO}}_{\text{4}}^{\text{2 -}}\left(\text{aq}\right)\to \text{4Cr}{\left(\text{OH}\right)}_{\text{4}}^{\text{-}}\left(\text{aq}\right)$

(b)The oxidation reaction can be given as:

${\text{2H}}_{\text{2}}\text{O}\left(\text{l}\right)\text{+ Fe}\left(\text{s}\right)\to \text{Fe}{\left(\text{OH}\right)}_{\text{2}}\left(\text{s}\right)$

The reduction reaction can be given as:

${\text{NiO}}_{\text{2}}\left(\text{s}\right){\text{+ 2H}}_{\text{2}}\text{O}\left(\text{l}\right)\to \text{Ni}{\left(\text{OH}\right)}_{\text{2}}\left(\text{s}\right)$

(c) The oxidation reaction can be given as:

${\text{2NH}}_2\text{OH}\left(\text{aq}\right)\to {\text{N}}_{\text{2}}\left(\text{g}\right){\text{+ 3H}}_{\text{2}}\text{O}\left(\text{g}\right)$

The reduction reaction can be given as:

${\text{CO}}_{\text{2}}\left(\text{g}\right)\to \text{CO}\left(\text{g}\right)$

Step-by-step solution

Redox reaction

Redox reactions constitute oxidation-reduction reactions where the reactants exhibit a change in the oxidation state. These reactions can be further segregated intooxidation reaction and reduction reaction.

Breaking the balanced equations into oxidation and reduction half reactions

(a)The given reaction is:

${\text{4PH}}_{\text{3}}\left(\text{g}\right){\text{+ 4H}}_{\text{2}}\text{O}\left(\text{l}\right){\text{+ 4CrO}}_{\text{4}}^{\text{2 -}}\left(\text{aq}\right)\to {\text{P}}_{\text{4}}\left(\text{s}\right)\text{+ 4Cr}{\left(\text{OH}\right)}_{\text{4}}^{\text{-}}\left(\text{aq}\right){\text{+ 4OH}}^{\text{-}}\left(\text{aq}\right)$

There is an increase in oxidation state of P from -3 to 0 and oxidation takes place. The oxidation half reaction can be given as:

${\text{4PH}}_{\text{3}}\left(\text{g}\right){\text{+ 4H}}_{\text{2}}\text{O}\left(\text{l}\right)\to {\text{P}}_{\text{4}}\left(\text{s}\right){\text{+ 4OH}}^{\text{-}}\left(\text{aq}\right)$

There is a decrease in oxidation state of Cr from +6 to 0. The reduction half reaction can be given as:

${\text{4CrO}}_{\text{4}}^{\text{2 -}}\left(\text{aq}\right)\to \text{4Cr}{\left(\text{OH}\right)}_{\text{4}}^{\text{-}}\left(\text{aq}\right)$

(b)The given reaction is:

${\text{NiO}}_{\text{2}}\left(\text{s}\right){\text{+ 2H}}_{\text{2}}\text{O}\left(\text{l}\right)\text{+ Fe}\left(\text{s}\right)\to \text{Ni}{\left(\text{OH}\right)}_{\text{2}}\left(\text{s}\right)\text{+ Fe}{\left(\text{OH}\right)}_{\text{2}}\left(\text{s}\right)$

There is an increase in oxidation state of Fe from 0 to +2. The oxidation reaction can be given as:

${\text{2H}}_{\text{2}}\text{O}\left(\text{l}\right)\text{+ Fe}\left(\text{s}\right)\to \text{Fe}{\left(\text{OH}\right)}_{\text{2}}\left(\text{s}\right)$

There is a decrease in oxidation state of Ni from +4 to +2. The reduction reaction can be given as:

${\text{NiO}}_{\text{2}}\left(\text{s}\right){\text{+ 2H}}_{\text{2}}\text{O}\left(\text{l}\right)\to \text{Ni}{\left(\text{OH}\right)}_{\text{2}}\left(\text{s}\right)$

(c)The given reaction is:

${\text{CO}}_{\text{2}}\left(\text{g}\right){\text{+ 2NH}}_2\text{OH}\left(\text{aq}\right)\to \text{CO}\left(\text{g}\right){\text{+N}}_{\text{2}}\left(\text{g}\right){\text{+ 3H}}_{\text{2}}\text{O}\left(\text{g}\right)$

There is an increase in oxidation state of N from -1 to 0. The oxidation reaction can be given as:

${\text{2NH}}_2\text{OH}\left(\text{aq}\right)\to {\text{N}}_{\text{2}}\left(\text{g}\right){\text{+ 3H}}_{\text{2}}\text{O}\left(\text{g}\right)$

There is a decrease in oxidation state of C from +4 to +2. The reduction reaction can be given as:

${\text{CO}}_{\text{2}}\left(\text{g}\right)\to \text{CO}\left(\text{g}\right)$