Question

Question: Estimate the cost of the electrical energy needed to produce$\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{10}}\mathbf{\text{kg}}$(a year's supply for the world) of aluminum from${\mathbf{\text{Al}}}_{\mathbf{2}}{\mathbf{\text{O}}}_{\mathbf{3}}\mathbf{\left(}\mathit{s}\mathbf{\right)}$if electrical energy costs 10 cents per kilowatt-hour$\left(1\text{kWh}=3.6\text{MJ}=3.6\times {10}^6\text{J}\right)$and if the cell potential is 5V.

Short answer

Answer

The estimated cost of electrical energy is $4470.48742\times {10}^8$cents

Step-by-step solution

The Gibbs free energy

In a galvanic cell, the Gibbs free energy is said to be the potential by:$\mathbf{▵}\mathit{G}\mathbf{=}\mathbf{-}\mathit{n}\mathit{F}{\mathit{E}}_{\mathbf{c}\mathbf{e}\mathbf{l}\mathbf{l}}$.If$\mathit{E}\mathbf{°}\mathit{c}\mathit{e}\mathit{l}\mathit{l}$> 0, then the process is spontaneous (galvanic cell). If$\mathit{E}\mathbf{°}\mathit{c}\mathit{e}\mathit{l}\mathit{l}$< 0, then the process is nonspontaneous (electrolytic cell)

Find the number of moles.

Amount of Alumunium that is produced every year$=1.5\times {10}^{10}\text{kg}=1.5\times {10}^{13}\text{g}$

$\text{Molar} \text{mass} \text{of} \text{Al = 26.98} \text{g/mol}$

$$\begin{gathered} \text{Number of moles}=\frac{\text{Weight present}}{\text{gram molar weight}} \\ n=\frac{1.5\times {10}^{13}g}{{\displaystyle \text{26.98} \text{g/mol}}} \\ \text{= 0.0556}\times {\text{10}}^{13} \text{mol} \end{gathered}$$

Reaction at cathode

The cathode reaction is represented in the equation below.

${\text{Al}}^{3+}+3{\text{e}}^{-}\to \text{Al}\left(s\right)$

3 moles of electrons are required to reduce 1 mole of ${\text{Al}}^{3+}$

Number of moles of electron.

Number of moles of electrons for $0.0556\times {10}^{13}\text{mol}$of Al is calculated below.

$\text{n}(\text{e}{)}^{-}=0.0556\times {10}^{13}\text{mol}\times 3=0.1668\times {10}^{13}\text{mol}$

Find the difference in free energy

The difference in free energy is,

$$\begin{gathered} W=\Delta G\backslash \mathrm{hfill} \\ \Delta G=-nF{E}_{cell} \end{gathered}$$

Solve the equation further

$$\begin{gathered} \Delta G=-nF{E}_{\min } \\ =-\left(0.1668\times {10}^{13}\text{mol}\right)\times \left(96485.34{\text{Cmol}}^{-1}\right)\times 5\text{V} \\ =-80468.49\times {10}^{13}\text{J} \\ =-8.04\times {10}^{17}\text{J} \\ \text{W}=-8.04\times {10}^{17}\text{J} \end{gathered}$$

Cost of electrical energy.

The estimated cost of electrical energy is

Number of kilo-watt hour spend is calculated below.

Cost of 1 energy cents

Cost of