For the decomposition of Al_2O_3 at 500^{circ | vedclass

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Question

(D) The relationship between Gibbs free energy change $(\Delta G)$ and cell potential $(E_{cell})$ is given by the equation: $\Delta G = -nFE_{cell}$.

Vedclass · Accepted Answer

$2.5$

Vedclass · Answer

(D) The relationship between Gibbs free energy change $(\Delta G)$ and cell potential $(E_{cell})$ is given by the equation: $\Delta G = -nFE_{cell}$.Here,$\Delta G = +966 \ kJ \ mol^{-1} = 966000 \ J \ mol^{-1}$.For the reaction $\frac{2}{3} Al_2O_3 	o \frac{4}{3} Al + O_2$,the oxidation state of $Al$ changes from $+3$ to $0$. The total number of electrons transferred $(n)$ is calculated as: $n = \frac{4}{3} 	imes 3 = 4$.Using $F = 96500 \ C \ mol^{-1}$,we have:$E_{cell} = -\frac{\Delta G}{nF} = -\frac{966000}{4 	imes 96500}$.Since the decomposition is non-spontaneous,the required external potential is $E_{ext} = -E_{cell} = \frac{966000}{386000} \approx 2.5 \ V$.

For the decomposition of $Al_2O_3$ at $500^{\circ}C$,the free energy change is $\frac{2}{3} Al_2O_3 \to \frac{4}{3} Al + O_2$,$\Delta G = +966 \ kJ \ mol^{-1}$. The potential difference required for the electrolytic reduction of $Al_2O_3$ at $500^{\circ}C$ is ........... $V$.

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