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?

  • $\begingroup$ In the top two reactions both these half-cells are gaining electrons electrons which is reduction. $\endgroup$ – MaxW Mar 15 '16 at 2:17


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.

  • $\begingroup$ Thanks for answering. So the electrical potential stays the same despite the amount of mols of electrons flowing because of how the units work out. I recall a little bit of physics that electrical potential can be equated to the slope of a hill. So adding more of the hill doesn't nescessarily increase the steepness, but overall there will be more velocity when the ball hits the bottom the the hill. Does the analogy hold in this case? would I be correct to say that 100 mols of ReDox solution has a higher "electric power" than 1 mol of the same ReDox solution? $\endgroup$ – Jess L Mar 15 '16 at 16:52
  • 1
    $\begingroup$ Yes, that is a valid and also a great analogy that also applies to the case of volts. If we were to plot a graph of electrical energy vs number of moles of electrons, the gradient would represent the voltage. Hence increasing the number of moles won't effect the voltage, but gives it greater electrical potential energy. $\endgroup$ – Nanoputian Mar 16 '16 at 4:58

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