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I am having a problem with this equation for redox potentials

$$\Delta G = -nFE_\mathrm{cell}$$

In this equation I never am totally sure about what the value of $n$ should be, for example for the reaction shown below, would the $n$ be 2 electrons or 1 electron? Personally, I think it should be 2 electrons because that is the number of mol of electron under the simplest whole number ratio.

\begin{align} \ce{cyt $c$(Fe^3+) + e- &-> cyt $c$(Fe^2+)}\\ \ce{NADH &-> NAD+ + H+ + 2e- }\\ \end{align}

I think this is a different matter than in the already answered question "Does the relationship equation between standard cell potential and equilibrium constant violate potential's intensive properties?" Because in that Nernst equation the change in $n$ would be balanced by the change in the equilibrium constant. I assume that you are insinuating that there is one of the variables here which will correct for the increase in $n$. I don't see which would, as $E_\mathrm{cell}$ should be independent of quantity and $F$ is a constant.

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  • $\begingroup$ Do you think your question is similar to this one, Matthieu? $\endgroup$ – Nicolau Saker Neto Apr 25 '15 at 20:53
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Personally, I think it should be 2 electrons because that is the number of mol of electron under the simplest whole number ratio.

Yes, it is $\ce{2e-}$.

Assuming that the two reactions provided above are half-reactions, the $n$ has to be $2$.

$E_\mathrm{cell}$ is independent of quantity because it is an intensive property. It does not depend on the amount of matter present. The $n$ accounts for the number of moles of $\ce{e-}$, this is used to calculate $\Delta G$, which in turn depends on the amount of matter present.

Overall, one must understand that $E_\mathrm{cell}$ is just the difference in cell potentials of two half-reactions. Increasing or decreasing the amount of both does not matter.

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  • $\begingroup$ This statement that $n$ "has to be 2" is not correct. By no means does $n$ have to be 2, it can be anything OP so chooses. $\endgroup$ – orthocresol Jul 31 '17 at 10:15
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The accepted answer is not correct, the number $n$ is not equal to 2 simply because 2 is the lowest common multiple of 2 and 1. Furthermore $n$ is not the "number of moles of electrons", which would have units of $\pu{mol}$; $n$ is in fact dimensionless.


In fact the number $n$ can be anything you like, and it depends on how you combine the two half-equations.

To use a different (not very realistic, but I don't care) example consider the half-reactions (data from Wikipedia)

$$\begin{align} \ce{Na+(aq) + e- &-> Na(s)} & E_1 &= \pu{-2.71 V} \tag{1} \\ \ce{Br2 (l) + 2e- &-> 2Br-(aq)} & E_2 &= \pu{+1.07 V} \tag{2} \end{align}$$

We have $E_\mathrm{cell} = E_2 - E_1 = \pu{+3.78 V}$. To construct a full reaction for this cell the "usual" procedure taught is to take (1), multiply it by 2, and subtract it from (2).

$$\ce{Br2(l) + 2Na(s) -> 2Na+(aq) + 2Br-(aq)} \tag{3}$$

In this case, we have $n = 2$ as there are two electrons being transferred in this reaction, from bromine to sodium. Equivalently, you can think of (3) as being the sum of two half-reactions, each featuring the gain or loss of two electrons:

$$\begin{align} \ce{2Na(s) &-> 2Na+(aq) + 2e-} \tag{4} \\ \ce{Br2(l) + 2e- &-> 2Br-(aq)} \tag{5} \end{align}$$

Note that the stoichiometric coefficient of $\ce{e-}$ - the number in front of the electrons - is 2, for both half-equations.


However, you could easily construct a different full reaction, simply differing from (3) by a multiplicative factor. There's no reason why you cannot, for example, divide (2) by 2, then subtract (1) to get:

$$\ce{1/2 Br2(l) + Na(s) -> Na+(aq) + Br-(aq)} \tag{3*}$$

and here you should only have $n = 1$, because there is only one electron being transferred here, from bromine to sodium. Again you can decompose this into its half-reactions:

$$\begin{align} \ce{Na(s) &-> Na+(aq) + e-} \tag{4*} \\ \ce{1/2Br2(l) + e- &-> Br-(aq)} \tag{5*} \end{align}$$

and the stoichiometric coefficient of $\ce{e-}$ here is simply 1.


The bottom line is that

  • $E_\mathrm{cell}$ doesn't change with the stoichiometric coefficients.
  • $n$ does, and hence $\Delta G = -nFE_\mathrm{cell}$ does. Your choice of the full cell reaction determines the value of $n$ you work with.
  • The best definition of $n$ is the stoichiometric coefficient of the electrons in the half-reactions into which the cell reaction can be divided (Atkins & de Paula, Physical Chemistry).
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