Based on the reaction frac{4}{3} , Al + O_2 t | vedclass

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(A) The reaction is $\frac{4}{3} \, Al + O_2 	o \frac{2}{3} \, Al_2O_3$.For the formation of $\frac{2}{3} \, mol$ of $Al_2O_3$,the number of electron

Vedclass · Accepted Answer

$2.14$

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(A) The reaction is $\frac{4}{3} \, Al + O_2 	o \frac{2}{3} \, Al_2O_3$.For the formation of $\frac{2}{3} \, mol$ of $Al_2O_3$,the number of electrons involved $(n)$ is calculated as follows:Each $Al$ atom loses $3$ electrons,so for $\frac{4}{3} \, mol$ of $Al$,$n = \frac{4}{3} 	imes 3 = 4$.Using the relation $\Delta G = -nFE_{cell}$,we have $E_{cell} = -\frac{\Delta G}{nF}$.Given $\Delta G = -827 \, kJ \, mol^{-1} = -827000 \, J \, mol^{-1}$.$E_{cell} = -\frac{-827000}{4 	imes 96500} = \frac{8270}{386} \approx 2.14 \, V$.

Electrolyte	$\Lambda_{m}^{o} \, (S \, cm^{2} \, mol^{-1})$
$NaCl$	$126.5$
$NaBr$	$128.2$
$KCl$	$149.5$

Based on the reaction $\frac{4}{3} \, Al + O_2 \to \frac{2}{3} \, Al_2O_3$,with $\Delta G = -827 \, kJ \, mol^{-1}$ of $Al_2O_3$,what is the minimum $e.m.f.$ required for the electrolysis? $(F = 96500 \, C \, mol^{-1})$

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Electrolyte $\Lambda_{m}^{o} \, (S \, cm^{2} \, mol^{-1})$

$NaCl$ $126.5$

$NaBr$ $128.2$

$KCl$ $149.5$

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