A pale green solution of ferrous sulphate was taken in four separate test tubes marked I, II, III, and IV. Pieces of Cu, Zn, and Al were dropped in test tubes II, III and IV respectively. In which case(s)
(A) the color of ferrous sulphate solution will match with the color in a test tube I? Give reason.
(B) the color of ferrous sulphate solution will fade and black mass will be deposited on the surface of the metal?
Solution
Change in color of green ferrous sulphate solution in each case:
(a) The color of ferrous sulphate solution will match with the color in test tube I because no reaction will take place in test tube I. The test tube I contain a mixture of copper pieces and green ferrous sulphate solution. Since copper is placed below iron in the reactivity series of metals, it cannot displace iron from its solution. Therefore, no reaction takes place in test tube I, and the color of the ferrous sulphate solution remains as it is, green.
•The reaction can be represented as under: FeSO₄ (green) + Cu —>No reaction (green).
(b) the black residue will appear due to the Deposition of iron in tubes III and IV because displacement reaction takes place.
2Al(s) + 3FeSO₄(aq)-------> Al₂(SO₄)₃ (aq) + 2Fe (s) (Black residue).
Zn(s) + FeSO₄(aq—--------> ZnSO₄ (aq) + Fe (s) (Black residue)
Was this answer helpful?
0
0
Class 6
Class 7
Class 8
Class 9
Class 10
Class 11
Class 12
Class 5
College Entrance