Question

Take 50 ml of water in each of 5 beakere & label them A,B,C,D,E. Dissolve 5 g of each substance as mentioned-

A- Copper Sulphate+ Zinc
B- Copper Sulphate+Iron nail
C- Zinc Sulphate+Copper
D-Iron sulphate+copper
E-Zinc sulphate + iron nail

Elaborate each answer.​

Answer 1

Answer:

in beaker A, blue colour of copper sulphate solution gradually disappears due to the formation of colourless zinc sulphate solution and copper metal deposits as a powder it that mass at the bottom of the beaker it means the zinc being more reactive displaces copper that is less reactive from its copper sulphate solution. The reaction can be represented as follows-

copper sulphate(CuSo4)+Zinc(Zn)= Zinc Sulphate(ZnSo4)+Copper(Cu)

In beaker B, blue colour of copper sulphate changes to greenish due to the formation of iron sulphate and copper metal produced forms a Red Brown coating over the iron nail. It means that iron displaces copper from its copper sulphate solution the reaction can be represented as follows-

copper sulphate(CuSo4) + iron(Fe) = iron sulphate(FeSo4) + copper(Cu)

there will be no change or displacement reaction as copper is less reactive then zinc in beaker C

Same is with case in beaker D as copper is less reactive than iron

In beaker E, there will be no change or displacement reaction to as iron is less reactive than zinc.