Question

A single Hall-Heroult cell (as shown in Fig. 11.22) produces about 1 ton of aluminum in 24 h. What current must be used to accomplish this?

Short answer

The current need to produce 1 ton Al is $1.12\times {10}^5\mathrm{A}.$

Step-by-step solution

Calculating the moles of electron

The reduction reaction can be denoted as,

$\mathrm{Al}{{\mathrm{F}}_6}^{3-}+3{\mathrm{e}}^{-}\to \mathrm{Al}+6{\mathrm{F}}^{-}$

The mole number of electrons can be denoted as,

$\begin{array}{rcl}\mathrm{Moles} \mathrm{of} \mathrm{electron}& =& (1 \mathrm{ton} \mathrm{Al})\left(\frac{907.185 \mathrm{kg} \mathrm{Al}}{1 \mathrm{ton} \mathrm{Al}}\right)\left(\frac{{10}^3 \mathrm{g} \mathrm{Al}}{1 \mathrm{kg} \mathrm{Al}}\right)\left(\frac{1 \mathrm{mol} \mathrm{Al}}{27 \mathrm{g} \mathrm{Al}}\right)\left(\frac{3 \mathrm{mol}}{1 \mathrm{mol} \mathrm{Al}}\right)\\ & =& 1.01\times {10}^5\mathrm{mol}\end{array}$

The charge needed for the release of the ions can be calculated as,

$\begin{array}{rcl}\mathrm{Charge}& =& (1.01\times {10}^5\mathrm{mol})\left(\frac{96485 \mathrm{C}}{1 \mathrm{mol}}\right)\\ & =& 9.7\times {10}^9\mathrm{C}\end{array}$

Calculating the requirement of current

The current can be denoted as,

$$\begin{gathered} \mathrm{Current}=\frac{\mathrm{Charge}\left(\mathrm{C}\right)}{\mathrm{Time}\left(\mathrm{s}\right)} \\ \mathrm{Current}=\left(\frac{9.7\times {10}^9\mathrm{C}}{24 \mathrm{hr}}\right)\left(\frac{1 \mathrm{hr}}{3600 \sec }\right) \\ \mathrm{Current}=1.12\times {10}^5\mathrm{A} \end{gathered}$$