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Question: The Gibbs energy for the decomposition of \(A{{l}_{2}}{{O}_{3}}\) at 500º C is as follows: \(\dfra...

The Gibbs energy for the decomposition of Al2O3A{{l}_{2}}{{O}_{3}} at 500º C is as follows:
23Al2O343Al+O2\dfrac{2}{3}A{{l}_{2}}{{O}_{3}}\to \dfrac{4}{3}Al+{{O}_{2}}, ΔrG=+966KJmol1{{\Delta }_{r}}G=+966KJmo{{l}^{-1}}
The potential difference needed for electrolytic reduction of Al2O3A{{l}_{2}}{{O}_{3}}at 500º C is at least:
A. 4.5 V
B. 3.0 V
C. 2.5 V
D. 5.0 V

Explanation

Solution

In thermodynamics the Gibbs free energy is a thermodynamic potential which can be used to calculate the maximum reversible work that was performed by a thermodynamic system at a constant pressure and temperature. In SI units it is measured in joules.

Complete Step by step explanation:
The Gibbs free energy can be calculated by the formulaΔG=ΔHTΔS\Delta G=\Delta H-T\Delta S. It is the maximum amount of non-expansion work which can be calculated from a thermodynamically closed system and close systems are those systems which can exchange heat and work with its surroundings but not with the matter. The maximum amount of work can be found only in a completely reversible process.
There is one another formula for calculating Gibbs free energy which is as follows:
ΔGo=nFEo\Delta {{G}^{o}}=-nF{{E}^{o}}
Here F = Faraday constant having value 96500 and ΔGo=+966×103Jmol1\Delta {{G}^{o}}=+966\times {{10}^{3}}Jmo{{l}^{-1}}
23Al2O343Al+O2\dfrac{2}{3}A{{l}_{2}}{{O}_{3}}\to \dfrac{4}{3}Al+{{O}_{2}}
Total number of Al atoms in
Al2O3A{{l}_{2}}{{O}_{3}}= 23×2=43\dfrac{2}{3}\times 2=\dfrac{4}{3}
Al3++3eAlA{{l}^{3+}}+3{{e}^{-}}\to Al
There is 3e3{{e}^{-}}change occurs for each Al atom
n=43×3=4\therefore n=\dfrac{4}{3}\times 3=4
Eo=ΔGonF{{E}^{o}}=\dfrac{\Delta {{G}^{o}}}{nF}=
Eo=966×10004×96500{{E}^{o}}=\dfrac{966\times 1000}{4\times 96500}= 2.50 V

The potential difference needed for electrolytic reduction of Al2O3A{{l}_{2}}{{O}_{3}}at 500º C is at least 2.5 V

Note: The Gibbs energy is also the thermodynamic potential that can be minimized when a system reaches chemical equilibrium at constant temperature and pressure. At the equilibrium point, its derivative with respect to the reaction coordinate of the system vanishes. For a reaction to be spontaneous, reduction is necessary at constant temperature and pressure.