Question

$\delta G = -827 kJ{mol}^{-1}$ of ${O}_{2}$ the minimum e.m.f required to carry out an electrolysis of ${Al}_{2}{O}_{3}$ is:
$(F = 96500 C{mol}^{-1})$
(A) 2.14 V
(B) 4.28 V
(C) 6.42 V
(D) 8.56 V

Answer 1

Hint: Gibbs free energy is the energy associated with a chemical reaction that can be used to do work. The free energy change of a reaction $(\delta G)$ tells us whether a reaction is spontaneous or not.

Complete step by step answer:
$\delta G$ determines the spontaneity of a reaction. There are two factors that affect enthalpy and entropy. Enthalpy is the heat content of a system at constant pressure and entropy is the amount of disorder in the system. When $\delta G < 0$, then the reaction is said to be spontaneous and when $\delta G > 0$, the reaction does not occur spontaneously. When $\delta G = 0$, the reaction is said to be in equilibrium.
Now, $G$ is given by the formula:
$\Delta G= -nF{ E }_{ cell }^{ o }$
-----(1)
where, { E }_{ cell }^{ o } = minimum e.m.f required
n = no. of electrons
The reaction involved is given below:
$\cfrac { 4 }{ 3 } Al + { O }_{ 2 } \longrightarrow \cfrac { 2 }{ 3 } { Al }_{ 2 }{ O }_{ 3 }$
The no. electrons, $n = \cfrac { 2 }{ 3 } \times 6 = 4$
And it is given that $\delta G = -827 kJ{mol}^{-1}$ and $F = 96500 C{mol}^{-1}$.
Now , substituting the values in equation (1), we get
$-827000 = - 4 \times 96500 \times { E }_{ cell }^{ o }$
$\implies { E }_{ cell }^{ o } = \cfrac { 827000 }{ 4 \times 96500 }$
$\implies { E }_{ cell }^{ o } = 2.14 V$

Hence, the correct option is (a).

Note: The no. of electrons in the determination of $\delta G$ is the no. of electrons that are being transferred in the redox reaction. Calculation of n should be done with precision or else you might end up getting the wrong answer.