Number of electrons transferred & Gibbs free energy To determine the number of electrons transferred for the following reaction:

$$\ce{CH3OH + \frac{3}{2} O2 -&gt; CO2 + 2 H2O}$$

I found the half reactions to be:

$$\text{Anode: } \ce{CH3OH + H2O -&gt; 6H+ + 6e- + CO2}$$
$$\text{Cathode: } \ce{\frac{3}{2}O2 + 6 H+ + 6 e- -&gt; 3 H2O}$$

The number of electrons transferred is therefore 6. If I was given the following overall reaction:
$$\ce{2 CH3OH + 3 O2 -&gt; 2 CO2 + 4 H2O}$$   

Which is just multiplying everything by 2 for the original reaction, would I say 12 electrons have been transferred or still 6? Since the Gibbs free energy change = nFE where n is the number of electrons transferred should surely be the same?
You should use $12$ electrons. Another way to think of it is when you calculate the $\Delta G$ of the first reaction, that value will be units of energy per mole of reaction. So, since you doubled the moles of reaction, the $\Delta G$ would double.