# Question #f3172

May 23, 2015

The color of the solution changes because of the presence of the iron (II) ion, $F {e}^{2 +}$.

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

$F {e}_{\left(s\right)} + {\underbrace{C u S {O}_{4 \left(a q\right)}}}_{\textcolor{b l u e}{\text{blue solution")) -> underbrace(FeSO_text(4(aq]))_(color(green)("green solution}}} + C {u}_{\textrm{\left(s\right]}}$

Iron is more reactive than copper, so it will displace the copper (II) ions, $C {u}^{2 +}$, 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 $C {u}^{2 +}$ 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 $F {e}^{2 +}$ ions, which have a light green color when placed 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

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 $F {e}^{2 +}$, while $C {u}^{2 +}$ is being reduced to copper metal.

$F {e}_{\left(s\right)} + C {u}_{\left(a q\right)}^{2 +} \to F {e}_{\left(a q\right)}^{2 +} + C {u}_{\left(s\right)}$