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?