# What is the number of electrons being transferred in electrolysis of aluminum oxide? [closed]

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

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.

• 12 is wrong indeed. But I think I understand how you got there. How do you break this into half-reactions, to begin with? – Ivan Neretin Jun 24 '20 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.

• But my textbooks says n=4? – ljm Jun 25 '20 at 11:23
• @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. – andselisk Jun 25 '20 at 13:01