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To determine the number of electrons transferred for the following reaction:

$$\ce{CH3OH + \frac{3}{2} O2 -> CO2 + 2 H2O}$$

I found the half reactions to be:

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

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

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?

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  • $\begingroup$ On topic video $\endgroup$ – user16347 Mar 31 '17 at 12:07
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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.

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When you go from $\ce {CH3OH}+1.5\ce {O2}$ to $2\ce {CH3OH}+3\ce {O2}$, you double the amount of reaction you're working with. Those ciefficients are not just ratios, they're the number of moles of stuff you put in or get out. So the number of (moles of) electrons and free energy change will scale up accordingly.

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It is $12$. Gibbs Free Energy is an extensive thermodynamic property. In electrochemistry, it depends upon the number of electrons taking part in the reaction.

In the second reaction, you are using twice the number of reactants. This implies that twice the number of electrons are used in the redox reaction, so the Gibbs free energy will be doubled.

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