Question

Pheli prepared a blue coloured solution of copper sulphate in beaker A and placed an iron nail in it. Boojho prepared a yellowish green solution of ferrous sulphate in beaker B and placed a copper wire in it. What changes will they observe in the two beakers after an hour?

Answer 1

hey friend,

here is the answer tp your question
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In beaker A, copper sulphate is blue in colour. On placing the iron nail in it for an hour, the colour of solution starts changing from blue to greenish blue and finally light green.

In the beaker B, no change is observed because copper being less reactive does not displace iron (more reactive) from ferrous sulphate solution. Hence, it remains in original state.

hope this will help you...

Answer 2

In Beaker A,a reddish brown layer of copper will deposit on the iron nail and blue coloured solution of copper sulphate will become yellowish green due to the formation of iron sulphate.This is because iron is more reactive than copper and it displaces copper from its salt solution {Copper Sulphate Solution}.In Beaker B,No Change is observed because copper being less reactive does not replaces iron {More reactive} from ferrace sulphate solution.