Question

Use the relationship $\text{ }\\!\\!\Delta\\!\\!\text{ }{{\text{G}}^{\text{0}}}\text{=-nFE}_{\text{cell}}^{\text{0}}$ to estimate the minimum voltage required to electrolyze $\text{A}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}}$in the Hall-Heroult process.
$\text{ }\\!\\!\Delta\\!\\!\text{ G}_{\text{f}}^{\text{o}}\text{(A}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}}\text{)=-1520kJmo}{{\text{l}}^{\text{-1}}}$
$\text{ }\\!\\!\Delta\\!\\!\text{ G}_{\text{f}}^{\text{o}}\text{(C}{{\text{O}}_{\text{2}}}\text{)=-394kJmo}{{\text{l}}^{\text{-1}}}$
The oxidation of the graphite anode to $\text{C}{{\text{O}}_{\text{2}}}$permits the electrolysis to occur at a low voltage then if the electrolysis reactions are,
$\text{A}{{\text{l}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\to \text{2Al+3}{{\text{O}}_{\text{2}}}$
What is the approximate value of low voltage?
A) 2V
B) 3V
C) 4V

Answer 1

Here graphite which is a form of C is reducing the metal oxide and it acts as the reducing agent.
We are given the value standard Gibbs free energy of formation for aluminium oxide and carbon dioxide.

Complete answer:
So in the question we are asked to determine approximate value of the minimum voltage required to electrolyze $\text{A}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}}$in the Hall-Heroult process. The values for the standard Gibbs free energy of formation value of $\text{A}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}}$and $\text{C}{{\text{O}}_{\text{2}}}$is also given.
We know that here graphite is used as the anode and it is getting oxidized to$\text{C}{{\text{O}}_{\text{2}}}$.
Inorder to start calculating the value of the minimum voltage required, lets first write the complete chemical reaction happening.
The net chemical reaction for the electrolytic process is,
$3C+2A{{l}_{2}}{{O}_{3}}\to 4Al+3C{{O}_{2}}$
$4A{{l}^{3+}}+12{{e}^{-}}\to 4Al$
The reaction happening in cathode is,
$A{{l}^{3+}}+3{{e}^{-}}\to Al(l)$
Here the positively charged aluminium ion is getting reduced to elemental Al by losing its three electrons.
The reaction happening in anode is,
$C(s)+{{O}^{2-}}\to CO(g)+2{{e}^{-}}$
$C(s)+2{{O}^{2-}}\to C{{O}_{2}}(g)+4{{e}^{-}}$
In the anode C is getting oxidized to Carbon dioxide.
Now from the net reaction given above, determine the number of electrons involved i.e. the value of n.
$4A{{l}^{3+}}+12{{e}^{-}}\to 4Al$
Here $12{{e}^{-}}s$ are involved in the overall reaction to convert $A{{l}^{3+}}$to Al.
Now let’s calculate the Gibbs free energy by substituting the values of the standard Gibbs free energy of formation,
$\Delta {{G}^{0}}=3\Delta G_{f}^{0}(C{{O}_{2}})-2\Delta G_{f}^{0}\left( A{{l}_{2}}{{O}_{3}} \right)$
$\Delta {{G}^{0}}=3(-394)-2\left( -1520 \right)$
$\Delta {{G}^{0}}=-1182+3040=1858kJ$
Hence we got the value for Gibbs free energy, now let’s find the potential value.
We know the relation,$\text{ }\\!\\!\Delta\\!\\!\text{ }{{\text{G}}^{\text{0}}}\text{=-nFE}_{\text{cell}}^{\text{0}}$
And by rearranging this equation for$E_{cell}^{0}$, we get,
$-E_{cell}^{0}=\dfrac{\Delta {{G}^{0}}}{nF}$
F = Faraday constant, value of 1F=96500
Substitute all the values in the above equation and we get,
$-E_{cell}^{0}=\dfrac{1858\times 1000}{12\times 96500}$
$-E_{cell}^{0}=1.60V$

The value is almost approximate to the value 2V, which is Option (A) and the Hall-Heroult processes carried out in low voltage.

Note:
The value of Gibbs free energy is converted to Joules in the final equation, and we should always take care of the conversion done and the signs related to the digits.
We should be thorough with the mathematical operations related to the integers and there are many chances of getting wrong during the mathematical operation carried out between a positive integer and negative integer.