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:


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


$$\ce{Cr\_2O\_7^{2-}+14H^+ +6e^- -> 2Cr^{3-} + 7H\_2O};E=1.33V$$


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  

$$\ce{6Fe^{2+}-> 6Fe^{3+} + 6e^{-}}$$
$$+$$
$$\ce{Cr\_2O\_7^{2-} +14H^+ +6e^- -> 2Cr^{3-} +7H\_2O}$$


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.


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