!! 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-)#
