Explain a procedure to do the experiment thay magnesium ribbon allows the flow pf current
Answers
◆Procedure:
• Clean two iron nails of sufficient size by rubbing with sand paper so that their colour appears greyish.
• Take sufficient quantity of copper sulphate solution in two test tubes and fix one test tube in each stand.
• Tie one iron nail with a thread and hang it in one test tube so that it is completely immersed in copper sulphate solution. Tie the other end of the thread with the stand.
• Keep the other nail in a petri dish for comparison after the experiment.
Keep the two test tubes undisturbed for about 15 min.
• After 15 min. remove the iron nail immersed in copper sulphate solution and put it in the petri dish.
◆Observations:
There is a brown coating on the iron nail which was dipped in the copper sulphate solution, whereas the iron nail placed in petri dish shows greyish colour of iron.
The colour of the copper sulphate solution in which the iron nail was dipped turns light greenish, whereas the solution of copper sulphate in the other test tube does not change.
◆Inferences:
• The brown coating on the iron nail shows that copper is deposited on the iron nail by displacing iron.
• The greenish colour of the solution in the test tube shows that Fe2+ ions are present in the solution.
• This shows that iron is more reactive than copper as Fe2+ ions have displaced Cu2+ ions from copper sulphate solution and form light greenish coloured ferrous sulphate solution.
• This is a single displacement reaction in which copper has been displaced by iron from copper sulphate solution and a new compound, ferrous sulphate, is formed. So, this reaction is a chemical change.