Question #18329

!! LONG ANSWER !!

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

The balanced chemical equation for this reaction will be

`Ni(OH)_(2(s)) + 2HNO_text(3(aq]) -> Ni(NO_3)_(2(aq)) + 2H_2O_((l))`

The net ionic equation will be

`Ni(OH)_(2(s)) + 2H_text((aq])^(+) -> Ni_text((aq])^(2+) + 2H_2O_((l))`

b) The oxidation of `Cr^(3+)` to `Cr^(+6)` in alkaline solution

The idea behind this reaction is that an alkaline solution of hypochlorite ions, `ClO^(-)`, will react with cromium (III) hydroxide, `Cr(OH)_3`, to give chromate, `CrO_4^(2-)` and chloride, `Cl^(-)`, ions.

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

The unbalanced reaction looks like this

`stackrel(color(blue)(+1))(Cl)O_((aq))^(-) + stackrel(color(blue)(+3))(Cr)(OH)_(3(s)) -> stackrel(color(blue)(+6))(Cr)O_text(4(aq])^(2-) + stackrel(color(blue)(-1))(Cl_((aq))^(-))`

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_2O` and hydrogen by adding `H_2O` and `OH^(-)`.

The reduction half-reaction is

`stackrel(color(blue)(+1))(Cl)O^(-) + 2e^(-) -> stackrel(color(blue)(-1))(Cl^(-))`

Balance the oxygen by adding `H_2O` on the products' side

`ClO^(-) + 2e^(-) ->Cl^(-) + H_2O`

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

`2H_2O + ClO^(-) + 2e^(-) -> Cl^(-) + H_2O + 2OH^(-)`

The oxidation half-reaction is

`stackrel(color(blue)(+3))(Cr)(OH)_3 -> stackrel(color(blue)(+6))(Cr)O_4^(2-) + 3e^(-)`

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

`H_2O + Cr(OH)_3 -> CrO_4^(2-) + 3e^(-)`

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

`5OH^(-) + H_2O + Cr(OH)_3 -> CrO_4^(2-) + 3e^(-) + 5H_2O`

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

`6H_2O + 3ClO^(-) + cancel(6e^(-)) + 10OH^(-) + 2H_2O + 2Cr(OH)_3 -> 3Cl^(-) + 3H_2O + 6OH^(-) + 2CrO_4^(2-) + cancel(6e^(-)) + 10H_2O`

The balanced net ionic equation will be

`3ClO^(-) + 4OH^(-) + 2Cr(OH)_3 -> 3Cl^(-) + 5H_2O + 2CrO_4^(2-)`

c) The confirmatory test for `Fe^(3+)`

A very common compound used in confirmatory tests for the `Fe^(3+)` ion is potassium ferrocyanide, `K_4Fe(CN)_6`. The net ionic equation looks like this

`4Fe_((aq))^(3+) + 3[Fe(CN)_6]_text((aq])^(4-) -> underbrace(Fe_4[Fe(CN)_6]_text(3(s]))_text(Prussian blue)`

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

Another compound used to confirm `Fe^(3+)` ions is potassium thicyanate, or `KSCN`, which reacts to form a deep red solution.

`Fe_((aq))^(3+) + 6SCN_((aq))^(-) -> Fe[(SCN)_6]_text((s])^(3-)`