Question

Q15. Write chemical equations for the formation of following compounds by the method mentioned. i) Iron(II)chloride: Simple displacement ii) Zinc sulphate: Neutralization iii) Sodium hydroxide: Synthesis iv) Copper (II) hydroxide: Precipitation

Answer 1

Answer:

Cu(s) --> [Cu(H2O)6]2+(aq) --> Cu(OH)2(s) --> CuO(s) --> [Cu(H2O)6]2+(aq) --> Cu(s)

Copper metal "dissolves" in nitric acid (HNO3). Actually, the nitrate ion oxidizes the copper metal to copper (II) ion while itself being transformed to NO2 gas in the process; the copper (II) ion then binds to six water molecules. The physical change you should observe is the copper-colored metal vanishing as the solution turns blue (from [Cu(H2O)6]2+, the hexaaquacopper ion) and a brown gas (NO2) is evolved.

Cu (s) + 4 H3O+ (aq) + 2 NO3- (aq) --> [Cu(H2O)6]2+ (aq) + 2 NO2 (g)

Hydroxide ion (OH-) binds to the copper (II) ion even more strongly than does water. As a result, hydroxide ion can displace water from the copper (II) ion, yielding copper hydroxide, Cu(OH)2, a blue precipitate.

[Cu(H2O)6]2+ (aq) + 2 OH- --> Cu(OH)2 (s) + 6 H2O (l)

Heating copper hydroxide produces copper oxide, CuO, a black solid.

Cu(OH)2 (s) --> CuO (s) + H2O (l)

Copper oxide dissolves in acid, regenerating the copper (II) ion, which once again binds to water.

CuO (s) + 2 H3O+ (aq) + 3 H2O (l) --> [Cu(H2O)6]2+ (aq)

Finally, zinc metal reduces the hydrated copper (II) ion back to metallic copper while itself turning being oxidized to zinc (II) ions. We have seen this reaction before in the copper chloride lab).

[Cu(H2O)6]2+ (aq) + Zn (s) --> Cu (s) + Zn2+ (aq) + 6 H2O (aq)

At the same time, some of the zinc metal, which is present in excess, reduces hydronium ions to H2.

Zn (s) + 2 H3O+ (aq) --> Zn2+ (aq) + H2 (g) + 2 H2O (l)

Explanation: