Ecell practical HELP PLEASE!!!

Hi,
I am doing my AQA chemistry required practical 8 and I am really confused about my predictions. I am using Cu2+ and Fe2+ solutions and changing the concentration of my Fe2+ solution. I did values from 1M to 0.0001M and my results were all similar so I am not even sure what to put down.

When trying to see my predictions I really couldn't understand if increasing concentration would make Ecell more negative or not.

Please help me figure out what my trend for increasing conc of Fe2+ should be.

Thank you!!!!

Let’s go through a worked example, where we look at the equilibrium
Fe^2+ (aq) + e^- <=> Fe (s)
Where under standard conditions, the reduction potential is -0.44 V.
If [Fe^2+] is increased in this instance, the equilibrium goes right (by Le Chatelier) and so more electrons are used up. This means that the reduction potential increases if you increase the concentration of Fe^2+.
Now suppose we set up an electrochemical cell, where we are oxidising Fe (s) and reducing Cu^2+ (aq). For the calculations, we must use
E(cell) = E(reduced species) - E(oxidised species).
The following equilibrium also needs to now be considered:
Cu^2+ (aq) + 2e^- <=> Cu (s)
Where the standard reduction potential is +0.34 V.
Suppose the conditions are standard, E(cell) is calculated as follows:
E(cell) = +0.34 V - (-0.44 V) = +0.78 V
Now, let’s consider the same experiment except instead of using 1.00 mol dm^-3 Fe^2+, you make it more concentrated. As explained earlier, the reduction potential will increase (it will be more than -0.44 V, so let’s say it’s -0.40 V in this case for argument’s sake):
E(cell) = +0.34 V - (-0.40 V) = +0.74 V
Notice that increasing [Fe^2+] but leaving everything else the same decreases E(cell)?