Question

\[\Delta G\] for the reaction, \[\dfrac{4}{3}Al+O_2\to \dfrac{2}{3}Al_2O_3\], is -772 KJ/mol of \[O_2\]. The minimum E.M.F. in volts required to carry out electrolysis of \[Al_2O_3\] is:

Answer 1

Hint: The energy accompanying a chemical reaction that can be used to complete the reaction is called Gibbs free energy. The free energy of a chemical reaction is the sum of its enthalpy (H) and the product of the temperature (Kelvin) and the entropy (S) of the chemical reaction. The equation can be represented as follows.
\[G=H-TS\]
Free energy of reaction (\[\Delta G\]): The change in the enthalpy (H) of the reaction minus the product of the temperature and the change in the entropy (S) of the reaction:
\[\Delta G=\Delta H-T\Delta S\]
If \[\Delta G\] is negative, meaning the chemical reaction is spontaneous, the reaction does not require any energy.

Complete step by step answer:
In the question it is given that \[\dfrac{4}{3}\] moles of aluminum (\[Al\]) reacts with one mole of oxygen (\[{{O}_{2}}\]) and forms \[\dfrac{2}{3}\] moles of alumina (\[Al_2O_3\]).
\[\dfrac{4}{3}Al+O_2\to \dfrac{2}{3}Al_2O_3\]
\[\Delta G\]for the reaction is -772 KJ/mol.
But in the question it is mentioned to calculate the E.M.F in volts required to electrolysis of the same reaction.
Electrolysis of alumina (\[Al_2O_3\]) is as follows.

\[\dfrac{2}{3}Al_2O_3\text{ }\xrightarrow{{}}\dfrac{4}{3}Al+O_2\]

Now, we have to calculate the E.M.F in volts required for electrolysis of the same reaction.
The ionic reactions of the above mentioned chemical reactions are

$\dfrac{2}{3}\times (2A{{l}^{3+}})+4{{e}^{-}}\to \dfrac{4}{3}Al$

$\dfrac{2}{3}\times (2{{O}^{2-}})\to {{O}_{2}}+4{{e}^{-}}$

Generally aluminium exists in +3 oxidation state. To neutralize \[\dfrac{4}{3}\]moles of aluminium it needs 4 electrons, those electrons are going to be donated by oxygen.
From the above ionic reactions we can say that the number of electrons transferred from oxygen to aluminium are, n = 4.
We know that , \[\text{ }\Delta G=-nF{{E}_{o}}\]
Where\[\text{ }\!\!\Delta\!\!\text{ G = gibbs free energy}\], n = number of electrons, F = Faraday, and \[{{E}_{o}}\]= E.M.F of the cell.
Therefore ,
$\text{ }\Delta G=-nF{{E}_{o}}$
$\text{-772}\times \text{1}{{\text{0}}^{\text{3}}}\text{ = }-4\times 96500\times {{E}_{o}}$

${{E}_{o}}=\dfrac{772\times {{10}^{3}}}{4\times 96500}=2.0V$

There the E.M.F of the electrolysis of \[Al_2O_3\] is 2.0V.

Note: Don’t be confused with electrolysis of alumina. Electrolysis of a compound means the compound or a chemical is going to ionize into its respective ions in the given solution by passing some amount of electricity into the solution. Electrolysis of alumina is as follows.
\[\dfrac{2}{3}Al2O3\text{ }\xrightarrow{{}}\dfrac{4}{3}Al+O2\]