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Question: Estimate the minimum potential difference needed to reduce \[A{{l}_{2}}{{O}_{3}}\] at \({{500}^{0}}C...

Estimate the minimum potential difference needed to reduce Al2O3A{{l}_{2}}{{O}_{3}} at 5000C{{500}^{0}}C. The Gibbs energy change for the decomposition reaction,23Al2O3→43Al+O2\dfrac{2}{3}A{{l}_{2}}{{O}_{3}}\to \dfrac{4}{3}Al+{{O}_{2}} is 960 J.

Explanation

Solution

The concept that includes the calculation based on reduction and oxidation that takes place at cathode and anode respectively, where the net number of electrons obtained will be multiplied by the equation and calculating Gibbs free energy.

Complete step by step solution:
We have come across the concepts in physical chemistry that involve thermodynamics and their parameters like calculation of enthalpy, entropy, Gibbs free energy etc.
Now, for the above cell reaction given, let us see which undergoes reduction and which undergoes oxidation.
To start with, we have Al2O3A{{l}_{2}}{{O}_{3}} which undergoes decomposition to yield metal that is aluminium. This reaction at the cell is as shown below,
Al3++3e−→AlA{{l}^{3+}}+3{{e}^{-}}\to Al

This reaction is a reduction reaction that takes place at cathode.
Also, the oxygen splits as:
O2−→O+2e−{{O}^{2-}}\to O+2{{e}^{-}}

This is an oxidation process taking place at anode.
Now, by the concept of balancing the cell reaction we can balance the above electrons in the cell reaction by multiplying the first reaction with two and the second reaction with three.
Thus, this gives the net electrons as six electrons.
Now, in the reaction 23Al2O3→43Al+O2\dfrac{2}{3}A{{l}_{2}}{{O}_{3}}\to \dfrac{4}{3}Al+{{O}_{2}}, basically the net electrons present is 23×6=4e−\dfrac{2}{3}\times 6=4{{e}^{-}}
Therefore, a total of four electrons pass from reactant to product side.
Now, we have the formula for Gibbs energy as, ΔG=nFEcell\Delta G=nF{{E}_{cell}}
Rearranging, Ecell=ΔGnF{{E}_{cell}}=\dfrac{\Delta G}{nF}
From the data, we have ΔG=960J\Delta G=960J
Since the reaction is spontaneous, the value of Gibbs free energy will be negative and thus the equation becomes,
Ecell=−ΔGnF{{E}_{cell}}=-\dfrac{\Delta G}{nF}

By substituting the values,
Ecell=−9604×96500{{E}_{cell}}=-\dfrac{960}{4\times 96500} [since F=96500C/mol]
⇒Ecell=−2.48×10−3V\Rightarrow {{E}_{cell}}=-2.48\times {{10}^{-3}}V
Therefore, the correct answer is the minimum potential energy required to reduceAl2O3A{{l}_{2}}{{O}_{3}}at 5000C{{500}^{0}}C is −2.48×10−3V-2.48\times {{10}^{-3}}V

Note: While solving these reactions, note that reduction always takes place at cathode and oxidation always at anode and for the spontaneous reaction, the Gibbs free energy will always be a negative value.