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Electrochemistry — Chemistry STD 12 Science — Question

Bihar BoardEnglish MediumSTD 12 ScienceChemistryElectrochemistry3 Marks
Question
Estimate the minimum potential difference needed to reduce Al2O3 at 500°C. The free energy change for the decomposition reaction,
$\frac{2}{3}\text{Al}_2\text{O}_3\frac{4}{3}\text{Al + O}_2\text{ is }960\text{kJ (F = 96500C mol}-1)$
✓

Answer

$\text{Al}_2\text{O}_3(2\text{Al}^{3+}+3\text{O}^{2-})\xrightarrow{ \ \ \ \ \ }2\text{Al}+\frac{3}{2}\text{O}_2,\text{n}=6\text{e}^-$
$\therefore\frac{2}{3}\text{Al}_2\text{O}_3\xrightarrow{ \ \ \ \ \ \ \ }\frac{4}{3}\text{Al + O}_2,\text{n}=\frac{2}{3}\times6\text{e}^-=4\text{e}^-$
$\Delta_{\text{r}}\text{G}=960\times1000=960000\text{J}$
Now, $\Delta_{\text{r}}\text{G}=-\text{nFE}_{\text{cell}}$
$\Rightarrow\text{E}_{\text{cell}}=-\frac{\Delta_{\text{r}}\text{G}}{\text{nF}}=\frac{-960000}{4\times96500}$
$\Rightarrow\text{E}_{\text{cell}}=-2.487\text{V}$
$\therefore$ Minimum potential difference needed to reduce Al2O3 is -2.487V.

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