How to calculate the electrical potential and mols of electrons transferred from combined half cell equations? Given 2 half-cell rxn's and their potentials, balance the full cell and solve for ECell:

$$\ce{Fe^{3+} + e^{-} -&gt;Fe^{2+}}; E=0.68V$$

$$\ce{Cr_2O_7^{2-}+14H^+ +6e^- -&gt; 2Cr^{3-} + 7H_2O};E=1.33V$$

So I know that both these half-cells are Gaining electrons (Reduction), and in order to solve for the full cell I need to flip one. I flip Fe Cell to oxidation b/c it has a lower potential and adjust for number of electrons moving. To Calculate ECell: $E(Cr)-E(Fe)$ despite adjustment of mols.

So my full Cell is
$$\ce{6Fe^{2+}-&gt; 6Fe^{3+} + 6e^{-}}$$
$$+$$
$$\ce{Cr_2O_7^{2-} +14H^+ +6e^- -&gt; 2Cr^{3-} +7H_2O}$$

My question is, why is the formula to solve ECell:  $E(Cr)-E(Fe)$ and NOT $E(Cr)-6*E(Fe)$?

I'm pretty sure that more electrons generally = more electricity, so why is it when I multiply the amount of electrons, I don't multiply my half cell E?
TL;DR

This basically boils down to the fact that $\mathrm{E_{cell}}$ is an intensive property, not an extensive property. 

Long Version

Lets define some terms first:

Intensive Property:
A property that doesn't change when the size of the sample changes. For example, density and concentration are examples of intensive properties

Extensive Property:
A property that changes when the size of the sample changes. For example, mass and volume are examples of extensive properties

$\mathrm{E_{cell}}$ is the energy in joules of 1 coulomb of charge (1 volt = 1 joule per coulomb). Hence it is an intensive property. Therefore the number of coulombs of charge or number of electrons doesn't effect the value of $\mathrm{E_{cell}}$. 

Another way of thinking of it is that by doubling the amount of moles, the  amount of coulombs and joules also double, hence $\mathrm{E_{cell}}$ remains the same.