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(B) The reaction for the electrolysis of $Al_2O_3$ is the reverse of the formation reaction: $\frac{2}{3} Al_2O_3 	o \frac{4}{3} Al + O_2$.For this r

Vedclass · Accepted Answer

$2.14$

Vedclass · Answer

(B) The reaction for the electrolysis of $Al_2O_3$ is the reverse of the formation reaction: $\frac{2}{3} Al_2O_3 	o \frac{4}{3} Al + O_2$.For this reaction,$\Delta G = +827 \ kJ \ mol^{-1} = 827000 \ J \ mol^{-1}$.The number of electrons involved in the reaction is calculated from the oxidation state change: $Al^{3+} + 3e^- 	o Al$. Since there are $\frac{4}{3}$ moles of $Al$,the total electrons $n = \frac{4}{3} 	imes 3 = 4$.Using the formula $\Delta G = -nFE$,the minimum emf $E$ required is $E = -\frac{\Delta G}{nF}$.$E = -\frac{827000 \ J \ mol^{-1}}{4 	imes 96500 \ C \ mol^{-1}} = -2.14 \ V$.Since we are looking for the minimum external potential to drive the non-spontaneous reaction,the magnitude is $2.14 \ V$.

On the basis of the information available from the reaction $\frac{4}{3} Al + O_2 \to \frac{2}{3} Al_2O_3$,$\Delta G = -827 \ kJ \ mol^{-1}$ of $O_2$,the minimum emf required to carry out an electrolysis of $Al_2O_3$ is $............... \ V$

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