Question #18329

Apr 23, 2015

a) The dissolution of nickel (II) hydroxide in nitric acid

The balanced chemical equation for this reaction will be

$N i {\left(O H\right)}_{2 \left(s\right)} + 2 H N {O}_{\textrm{3 \left(a q\right]}} \to N i {\left(N {O}_{3}\right)}_{2 \left(a q\right)} + 2 {H}_{2} {O}_{\left(l\right)}$

The net ionic equation will be

$N i {\left(O H\right)}_{2 \left(s\right)} + 2 {H}_{\textrm{\left(a q\right]}}^{+} \to N {i}_{\textrm{\left(a q\right]}}^{2 +} + 2 {H}_{2} {O}_{\left(l\right)}$

b) The oxidation of $C {r}^{3 +}$ to $C {r}^{+ 6}$ in alkaline solution

The idea behind this reaction is that an alkaline solution of hypochlorite ions, $C l {O}^{-}$, will react with cromium (III) hydroxide, $C r {\left(O H\right)}_{3}$, to give chromate, $C r {O}_{4}^{2 -}$ and chloride, $C {l}^{-}$, ions.

The cromium (III) hydroxide will be a solid, while the products will be in aqueous solution.

The unbalanced reaction looks like this

$\stackrel{\textcolor{b l u e}{+ 1}}{C l} {O}_{\left(a q\right)}^{-} + \stackrel{\textcolor{b l u e}{+ 3}}{C r} {\left(O H\right)}_{3 \left(s\right)} \to \stackrel{\textcolor{b l u e}{+ 6}}{C r} {O}_{\textrm{4 \left(a q\right]}}^{2 -} + \stackrel{\textcolor{b l u e}{- 1}}{C {l}_{\left(a q\right)}^{-}}$

As you can see, cromium is being oxidized and chlorine is being reduced. Since you're in basic solution, you can balance oxygen by adding ${H}_{2} O$ and hydrogen by adding ${H}_{2} O$ and $O {H}^{-}$.

The reduction half-reaction is

$\stackrel{\textcolor{b l u e}{+ 1}}{C l} {O}^{-} + 2 {e}^{-} \to \stackrel{\textcolor{b l u e}{- 1}}{C {l}^{-}}$

Balance the oxygen by adding ${H}_{2} O$ on the products' side

$C l {O}^{-} + 2 {e}^{-} \to C {l}^{-} + {H}_{2} O$

Balance the hydrogen by adding two water molecules on the reactants' side and two hydroxide ions on the products' side

$2 {H}_{2} O + C l {O}^{-} + 2 {e}^{-} \to C {l}^{-} + {H}_{2} O + 2 O {H}^{-}$

The oxidation half-reaction is

$\stackrel{\textcolor{b l u e}{+ 3}}{C r} {\left(O H\right)}_{3} \to \stackrel{\textcolor{b l u e}{+ 6}}{C r} {O}_{4}^{2 -} + 3 {e}^{-}$

Balance oxygen by adding 1 water molecule on the reactants' side

${H}_{2} O + C r {\left(O H\right)}_{3} \to C r {O}_{4}^{2 -} + 3 {e}^{-}$

Balance hydrogen by adding 5 water molecules on the products' side and 5 hydroxide ions on the reactants' side

$5 O {H}^{-} + {H}_{2} O + C r {\left(O H\right)}_{3} \to C r {O}_{4}^{2 -} + 3 {e}^{-} + 5 {H}_{2} O$

Multiply the reduction half-reaction by 3 and the oxidation half-reaction by 2 to balance the number of electrons transferred during the reaction, and add the two half-reactions

$6 {H}_{2} O + 3 C l {O}^{-} + \cancel{6 {e}^{-}} + 10 O {H}^{-} + 2 {H}_{2} O + 2 C r {\left(O H\right)}_{3} \to 3 C {l}^{-} + 3 {H}_{2} O + 6 O {H}^{-} + 2 C r {O}_{4}^{2 -} + \cancel{6 {e}^{-}} + 10 {H}_{2} O$

The balanced net ionic equation will be

$3 C l {O}^{-} + 4 O {H}^{-} + 2 C r {\left(O H\right)}_{3} \to 3 C {l}^{-} + 5 {H}_{2} O + 2 C r {O}_{4}^{2 -}$

c) The confirmatory test for $F {e}^{3 +}$

A very common compound used in confirmatory tests for the $F {e}^{3 +}$ ion is potassium ferrocyanide, ${K}_{4} F e {\left(C N\right)}_{6}$. The net ionic equation looks like this

$4 F {e}_{\left(a q\right)}^{3 +} + 3 {\left[F e {\left(C N\right)}_{6}\right]}_{\textrm{\left(a q\right]}}^{4 -} \to {\underbrace{F {e}_{4} {\left[F e {\left(C N\right)}_{6}\right]}_{\textrm{3 \left(s\right]}}}}_{\textrm{P r u s s i a n b l u e}}$

The reaction will form a deep blue precipitate called iron (III) ferrocyanide, or Prussian blue.

Another compound used to confirm $F {e}^{3 +}$ ions is potassium thicyanate, or $K S C N$, which reacts to form a deep red solution.

$F {e}_{\left(a q\right)}^{3 +} + 6 S C {N}_{\left(a q\right)}^{-} \to F e {\left[{\left(S C N\right)}_{6}\right]}_{\textrm{\left(s\right]}}^{3 -}$