The Gibbs energy for the decomposition of Al_ | vedclass

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(C) The relationship between Gibbs energy and cell potential is given by $\Delta G = -nFE$.For the reaction $\frac{2}{3} Al_2O_3 \rightarrow \frac{4}{

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$2.5$

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(C) The relationship between Gibbs energy and cell potential is given by $\Delta G = -nFE$.For the reaction $\frac{2}{3} Al_2O_3 ightarrow \frac{4}{3} Al + O_2$,the number of electrons transferred $(n)$ is calculated as follows: $Al$ goes from $+3$ to $0$ oxidation state. For $\frac{4}{3}$ moles of $Al$,$n = \frac{4}{3} 	imes 3 = 4$.Using the formula $E = -\frac{\Delta G}{nF}$ (where $\Delta G$ is for the reduction reaction,which is $-966 \ kJ \ mol^{-1}$):$E = -\frac{-966 	imes 10^3 \ J \ mol^{-1}}{4 	imes 96500 \ C \ mol^{-1}}$$E = \frac{966000}{386000} = 2.5 \ V$.Thus,the potential difference needed for the reduction is $2.5 \ V$.

The Gibbs energy for the decomposition of $Al_2O_3$ at $500^o C$ is as follows :
$\frac{2}{3} Al_2O_3 \rightarrow \frac{4}{3} Al + O_2, \Delta G = +966 \ kJ \ mol^{-1}$
The potential difference needed for electrolytic reduction of $Al_2O_3$ at $500^o C$ is at least ...... $V$.

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The Gibbs energy for the decomposition of $Al_2O_3$ at $500^o C$ is as follows :$\frac{2}{3} Al_2O_3 \rightarrow \frac{4}{3} Al + O_2, \Delta G = +966 \ kJ \ mol^{-1}$The potential difference needed for electrolytic reduction of $Al_2O_3$ at $500^o C$ is at least ...... $V$.