Understanding the setup of a chemistry lab about redox In chemistry lab we were looking at oxidation and reduction reactions. To determine the relative activity of several metals we observed reactions between the metals and aqueous solutions of dissolved metals. We mixed metals like $\ce{Fe}$ with solution like $\ce{Cu(NO3)2}$ to see if a reaction occurred. 

I'm wondering why is it not necessary to mix a metal like $\ce{Fe}$ into its own solution $\ce{Fe(NO3)2}$ (like you would do when setting up an electrochemical cell)? We mixed $\ce{Fe}$ with $\ce{Zn(NO3)2}$, $\ce{Cu(NO3)2}$, and $\ce{Ni(NO3)2}$.
I'm wondering why is it not necessary to mix a metal like Fe into its own solution $\ce{Fe(NO3)2}$ (like you would do when setting up an electrochemical cell)?


First remember what the purpose of your experiment was:


  To determine the relative activity of several metals we observed reactions between the metals and aqueous solutions of dissolved metals.


In an electrochemical cell setup, $\ce{Fe}$ would be in two-way equilibrium with $\ce{Fe^2+}$:

$$\ce{Fe + Fe^2+ &lt;=&gt; Fe + Fe^2+}$$

When power was drawn from the cell you then get:

$$\ce{Fe -&gt; 2 e- + Fe^2+}$$

Where as with another salt such as copper which is less reactive, the copper will oxidize the iron in one direction with no reverse reaction.

$$\ce{Fe + Cu^2+ -&gt; Cu + Fe^2+}$$

or with zinc which is more reactive no reaction will occur at all:

$$\require{cancel} \ce{Fe + Zn^2+ \cancel{\ce{&lt;=&gt;}} Zn + Fe^2+ -&gt;Fe + Zn^2+}$$