How to determine the value for n in the Gibbs free energy and redox potential equation?
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)

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

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).

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

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:

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

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:

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

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:

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

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 &amp; de Paula, Physical Chemistry).