Chapter 23, Problem 49QP

Interpretation Introduction

Interpretation:

The minimum voltage required to produce one mole of aluminum at the temperature at which aluminum produced by the Hall process and the energy required to produce $1.00\text{kg}$

of the metal is to be calculated.

Concept Introduction:

The Hall–Heroult process is considered an enormous industrial process in which the smelting of aluminum takes place.

The decomposition of chemical process takes place by the electrolytic cell that contains electrical energy.

The number of moles can be calculated as:

$\text{moles}\text{=}\frac{\text{given}\text{mass}}{\text{molar}\text{mass}}$.

The relationship between cell voltage and free energy difference is represented as:

$\Delta G=-nFE$.

Answer to Problem 49QP

Solution:

(a) The minimum voltage required to produce one mole of aluminum is $1.03\text{V}$.

(b) The energy required to produce $1.00\text{kg}$

of the metal is $3.32\times {10}^4\text{ kJ}/\text{mol}$.

Explanation of Solution

Given information: The overall reaction for the electrolytic production of aluminum by the Hall process is represented as:

${\text{AL}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)+3\text{C}\left(s\right)\to 2\text{Al}\left(l\right)+3\text{CO}\left(g\right)$.

At $1000°\text{C}$, the standard free-energy change for this process is $594\text{kJ}/\text{mol}$.

a) The minimum voltage required to produce $\text{1 mol}$ of aluminum at this temperature

The relationship between cell voltage and free energy difference is represented as:

$\Delta G=-nFE$.

Here,

$\Delta G=-594\times {10}^3\text{j}/\text{mol}$,

$n=6$.

$F=96500\text{}\text{j/V}\times \text{mol}$.

All the given values are to be put in the formula as:

$\begin{array}{c}E=\frac{-\Delta G}{nF}\\ =\left(\frac{-594\times {10}^3\cancel{\text{J}}\text{/}\cancel{\text{mol}}}{6\times 96500\cancel{\text{J}}\text{/V}\times \cancel{\text{mol}}}\right)\\ =-1.03\text{V}\text{.}\end{array}$

Aluminum contains two moles in the balanced equation so, the whole given equation is divided by two.

Thus, the new equation is as

$\frac{1}{2}{\text{AL}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)+\frac{3}{2}\text{C}\left(s\right)\to \text{Al}\left(l\right)+\frac{3}{2}\text{CO}\left(g\right)$.

For the new equation,

$n=3$

The Gibbs energy is calculated as

$\begin{array}{c}\Delta G=\left(\frac{-594\times {10}^3\text{ J}/\text{mol}}{2}\right)\\ =297\text{kJ/mol}\text{.}\end{array}$

These values are put in the formula as

$\begin{array}{c}E=\frac{-\Delta G}{nF}\\ =\left(\frac{-297\times {10}^3\cancel{\text{J}}/\cancel{\text{mol}}}{3\times 96500\cancel{\text{J}}\text{/V}\times \cancel{\text{mol}}}\right)\\ =-1.03E\text{V}\text{.}\end{array}$

Hence, the required voltage would be the same, whether one mole or $1000\text{moles}$ of aluminum are produced, but the amount of current used in each case would be different.

b) The energy required to produce 1.00 kg of the metal.

Primarily, $1.00\text{kg }\left(1000\text{g}\right)$ of $\text{Al}$

change into moles takes place.

$\text{moles}\text{=}\frac{\text{given}\text{mass}}{\text{molar}\text{mass}}$

$\left(1.00\times {10}^3\cancel{\text{g Al}}\right)\times \frac{1\text{mol Al}}{26.98\cancel{\text{g}\text{Al}}}=37.1\text{mol Al}$.

The electrical energy can be found by using the same formula taking other voltage as

$\begin{array}{c}\Delta G=-nFE\\ =-\left(37.1\cancel{\text{mol Al}}\right)\left(\frac{3\cancel{{\text{mol e}}^{-}}}{1\cancel{\text{mol Al}}}\times \frac{96500\cancel{\text{C}}}{1\cancel{{\text{mol e}}^{-}}}\right)\left(\frac{-3.09\text{J}}{1\cancel{\text{C}}}\right)\\ =3.32\times {10}^7\text{J/mol}\\ \text{=3}\text{.32}\times {\text{10}}^4\text{kJ/mol}\text{.}\end{array}$

The electrical work can be obtained by the multiplication of the voltage and the amount of charge transported through the circuit as

$\text{Joules}=\text{Volts}\times \text{Coulombs}$