$$\ce{ 2/3 Al2O3 -> 4/3 Al +O2}$$

As we know, to find the Gibbs free energy, we need to use this formula:

$$ G= -nFE^\circ$$

What would $n$ be in this case? After breaking it down into the half reactions $n = 12,$ but this doesn't seem right.

  • 4
    $\begingroup$ 12 is wrong indeed. But I think I understand how you got there. How do you break this into half-reactions, to begin with? $\endgroup$ – Ivan Neretin Jun 24 at 11:56

I guess the misunderstanding arises from the fact that you treat $n$ as the same entity as the lowest common multiple (LCM) used to balance the net equation using half-reactions for the electrolysis of molten aluminium(III) oxide:

$$ \begin{align} &\text{cathode:} &\ce{Al^3+(melt) + 3 e- &-> Al^0(l)} &|\cdot4\tag{red}\\ &\text{anode:} &\ce{2 O^2-(melt) &-> O2^0(g) + 4 e-} &|\cdot3\tag{ox}\\ \hline & &\ce{4 Al^3+ + 6 O^2- &-> 4 Al^0 + 3 O2^0}\tag{redox} \end{align} $$

or, finally, with the smallest whole-number coefficients:

$$\ce{2 Al2O3(l) -> 4 Al(l) + 3 O2(g)}\label{rxn:R1}\tag{R1}$$

as well as normalized to a single formula unit:

$$\ce{Al2O3(l) -> 2 Al(l) + 3/2 O2(g)}\label{rxn:R2}\tag{R2}$$

In the equation

$$Δ_\mathrm rG = -nFE$$

$n$ is the number that matches the amount of electrons (in mol — since we are using molar convention for Gibbs free energy) transferred in the balanced equation from a reducing agent to an oxidizing agent. From the balanced reaction \eqref{rxn:R2} it is evident that the number of electrons $n$ transferred from $\ce{O^2-}$ to $\ce{Al^3+}$ is $3 \cdot 2 = 6,$ where $3$ is the number of oxygens per formula unit, and $2$ is the number of electrons each oxygen donates.

| improve this answer | |
  • $\begingroup$ But my textbooks says n=4? $\endgroup$ – Leah Jun 25 at 11:23
  • $\begingroup$ @Leah I suspect your textbook is not correct, but could you please provide a reference? What textbook and what page number? It's important to see the context. $\endgroup$ – andselisk Jun 25 at 13:01

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