Question #f3172

1 Answer
May 23, 2015

The color of the solution changes because of the presence of the iron (II) ion, #Fe^(2+)#.

Here's how the balanced chemical equation for this single replacement reaction looks like

#Fe_((s)) + underbrace(CuSO_(4(aq)))_(color(blue)("blue solution")) -> underbrace(FeSO_text(4(aq]))_(color(green)("green solution")) + Cu_text((s])#

Iron is more reactive than copper, so it will displace the copper (II) ions, #Cu^(2+)#, from the solution. As a result, iron (II) ions will pass into solution, and copper (II) ions will be deposited onto the nail.

The color of the initial solution is attributed to the presence of the #Cu^(2+)# ions, which have a blue color when placed in aqueous solution.

On the other hand, the color of the final solution can be attributed to the presence of the #Fe^(2+)# ions, which have a light green color when placed in aqueous solution.

http://www.compoundchem.com/2014/03/05/colours-of-transition-metal-ions-in-aqueous-solution/

You can actually observe two changes that take place with this reaction

  • the color of the solution turns from blue to light green

  • a brown coating is deposited on the nail

http://www.ekshiksha.org.in/eContent-Show.do?documentId=92

The brown coating you see on the nail is actually the copper that was displaced from the solution by the iron (II) ions.

You can think of this as a redox reaction as well. Iron metal is being oxidized and becomes #Fe^(2+)#, while #Cu^(2+)# is being reduced to copper metal.

#Fe_((s)) + Cu_((aq))^(2+) -> Fe_((aq))^(2+) + Cu_((s))#