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 $\Delta G = -nFE_{cell}$.For the reaction $\frac{2}{3} Al_2O_3 \rightarrow \f

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$2.5$

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(C) The relationship between Gibbs energy and cell potential is given by $\Delta G = -nFE_{cell}$.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 $0$ (change of $3$ per $Al$ atom).For $\frac{4}{3}$ moles of $Al$,$n = \frac{4}{3} 	imes 3 = 4$.Given $\Delta G = +960 \ kJ \ mol^{-1} = 960000 \ J \ mol^{-1}$.Using $\Delta G = -nFE_{cell}$:$960000 = -4 	imes 96500 	imes E_{cell}$.$E_{cell} = -\frac{960000}{386000} \approx -2.487 \ V$.The magnitude of the potential difference required for the electrolytic reduction is approximately $2.5 \ V$.

$LIST$-$I$ (Cell)	$LIST$-$II$ (Use/Property/Reaction)
$A$. Leclanche cell	$I$. Converts energy of combustion into electrical energy
$B$. Ni-Cd cell	$II$. Does not involve any ion in solution and is used in hearing aids
$C$. Fuel cell	$III$. Rechargeable
$D$. Mercury cell	$IV$. Reaction at anode $Zn \rightarrow Zn^{2+} + 2e^-$

The Gibbs energy for the decomposition of $Al_2O_3$ at $500^oC$ is as follows:
$\frac{2}{3} Al_2O_3 \rightarrow \frac{4}{3} Al + O_2$
$\Delta_rG = +960 \ kJ \ mol^{-1}$
The potential difference needed for the electrolytic reduction of aluminium oxide $(Al_2O_3)$ at $500^oC$ is at least ........ $V$.

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The Gibbs energy for the decomposition of $Al_2O_3$ at $500^oC$ is as follows:$\frac{2}{3} Al_2O_3 \rightarrow \frac{4}{3} Al + O_2$$\Delta_rG = +960 \ kJ \ mol^{-1}$The potential difference needed for the electrolytic reduction of aluminium oxide $(Al_2O_3)$ at $500^oC$ is at least ........ $V$.