!! 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))Ni(OH)2(s)+2HNO3(aq]→Ni(NO3)2(aq)+2H2O(l)
The net ionic equation will be
Ni(OH)_(2(s)) + 2H_text((aq])^(+) -> Ni_text((aq])^(2+) + 2H_2O_((l))Ni(OH)2(s)+2H+(aq]→Ni2+(aq]+2H2O(l)
b) The oxidation of Cr^(3+)Cr3+ to Cr^(+6)Cr+6 in alkaline solution
The idea behind this reaction is that an alkaline solution of hypochlorite ions, ClO^(-)ClO−, will react with cromium (III) hydroxide, Cr(OH)_3Cr(OH)3, to give chromate, CrO_4^(2-)CrO2−4 and chloride, Cl^(-)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))^(-))+1ClO−(aq)++3Cr(OH)3(s)→+6CrO2−4(aq]+−1Cl−(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_2OH2O and hydrogen by adding H_2OH2O and OH^(-)OH−.
The reduction half-reaction is
stackrel(color(blue)(+1))(Cl)O^(-) + 2e^(-) -> stackrel(color(blue)(-1))(Cl^(-))+1ClO−+2e−→−1Cl−
Balance the oxygen by adding H_2OH2O on the products' side
ClO^(-) + 2e^(-) ->Cl^(-) + H_2OClO−+2e−→Cl−+H2O
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^(-)2H2O+ClO−+2e−→Cl−+H2O+2OH−
The oxidation half-reaction is
stackrel(color(blue)(+3))(Cr)(OH)_3 -> stackrel(color(blue)(+6))(Cr)O_4^(2-) + 3e^(-)+3Cr(OH)3→+6CrO2−4+3e−
Balance oxygen by adding 1 water molecule on the reactants' side
H_2O + Cr(OH)_3 -> CrO_4^(2-) + 3e^(-)H2O+Cr(OH)3→CrO2−4+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_2O5OH−+H2O+Cr(OH)3→CrO2−4+3e−+5H2O
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-)
