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The Gibb's energy for the decomposition of \mathrm{Al_{2}O_{3}} at 500°C is as follows :

\mathrm{\frac{2}{3}Al_{2}O_{3}\rightarrow \frac{4}{3}Al+O_{2},\Delta _{r}G=+966\, kJ\, mol^{-1}.}

The potential difference (in V) needed for electrolytic reduction of \mathrm{Al_{2}O_{3}} at 500°C is at least

Option: 1

5


Option: 2

4.5


Option: 3

3


Option: 4

2.5


Answers (1)

best_answer

In the given balanced reaction, 

\mathrm{\frac{2}{3}Al_{2}O_{3}\rightarrow \frac{4}{3}Al+O_{2},\Delta _{r}G=+966\, kJ\, mol^{-1}.}
The total number of moles of electrons transferred from the Oxide ion \mathrm{(O^{2-})} to aluminium ions \mathrm{(Al^{3+})} is 4

Now, the change in Gibb's Free energy is given as 

\mathrm{\Delta G^0 = -nFE^0 = 966~ kJmol^{-1}}

for the given case 

\mathrm{n=4}

\mathrm{F=96500\ C}

Putting values in the equation, we have 

\mathrm{E^{0} = \frac{-\Delta G^0}{nF}=\frac{-966000}{4\times 96500} \simeq 2.5V}

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HARSH KANKARIA

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