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Chapter 18: Problem 129

It takes 15 kWh (kilowatt-hours) of electrical energy to produce 1.0 kg aluminum metal from aluminum oxide by the Hall-Heroult process. Compare this to the amount of energy necessary to melt 1.0 kg aluminum metal. Why is it economically feasible to recycle aluminum cans? [The enthalpy of fusion for aluminum metal is 10.7 $\mathrm{kJ} / \mathrm{mol}(1 \text { watt }=1 \mathrm{J} / \mathrm{s}) . ]$

Short Answer

Expert verified
The energy required to melt 1.0 kg of aluminum metal is significantly less than the 15 kWh of energy needed to produce it from aluminum oxide. This difference in energy consumption makes recycling aluminum cans economically feasible, as less energy is expended in the recycling process compared to producing aluminum from raw materials.

Step by step solution

01

Determine the number of moles in 1 kg of aluminum metal

To find the number of moles in 1 kg of aluminum, we will use the atomic weight of aluminum (26.98g/mol). 1 kg of aluminum = 1000 g of aluminum (as kg needs to be converted to grams) Number of moles = \( \frac{1000 \ \mathrm{g}}{26.98\ \mathrm{g/mol}} \)
02

Calculate the energy required to melt 1 kg of aluminum metal

Using the enthalpy of fusion for aluminum metal that is 10.7 kJ/mol, we'll now calculate the energy required to melt 1.0 kg of aluminum metal. Energy required to melt 1.0 kg aluminum = Number of moles × Enthalpy of fusion Energy required to melt 1.0 kg aluminum = \( \frac{1000 \ \mathrm{g}}{26.98\ \mathrm{g/mol}} \times 10.7\ \mathrm{kJ/mol} \)
03

Compare the energy required for melting aluminum with the energy required to produce it from aluminum oxide

Now that we have calculated the energy required to melt aluminum, we will convert it to kilowatt-hours (kWh) and compare it with the 15 kWh energy required to produce aluminum from aluminum oxide. 1 watt = 1 J/s, therefore 1 kilowatt = 1000 J/s. 1 kilowatt-hour = 1 kW × 1 hour = 1000 J/s × 3600 s = 3.6 × 10^6 J Now, convert the energy required to melt 1.0 kg aluminum from kJ to kWh: Energy_required(kWh) = \( \frac{Energy_required(kJ) \times 10^3}{3.6 \times 10^6} \) Energy_required(kWh) = \( \frac{\left( \frac{1000 \ \mathrm{g}}{26.98\ \mathrm{g/mol}} \times 10.7\ \mathrm{kJ/mol} \right) \times 10^3}{3.6 \times 10^6} \)
04

Discuss the economic feasibility of recycling aluminum cans

Comparing the energy required to melt aluminum metal with the energy required to produce it from aluminum oxide, we can see that: Energy_required_for_melting_aluminum(kWh) << 15 kWh The energy required for melting aluminum is significantly less than the energy needed to produce aluminum from aluminum oxide. This shows that recycling aluminum cans saves a substantial amount of energy, making it economically feasible to recycle them.

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Most popular questions from this chapter

Assign oxidation numbers to all the atoms in each of the following: a. \(\mathrm{HNO}_{3} \quad\) g. \(\mathrm{PbSO}_{4}\) b. \(\mathrm{CuCl}_{2} \quad\) h. \(\mathrm{PbO}_{2}\) c. \(\mathrm{O}_{2} \quad\) i. \(\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) d. \(\mathrm{H}_{2} \mathrm{O}_{2} \quad\) j. \(\mathrm{CO}_{2}\) e. $\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \quad \mathrm{k} .\left(\mathrm{NH}_{0}\right)_{2} \mathrm{Ce}\left(\mathrm{SO}_{4}\right)_{3}$ f. \(\mathrm{Ag} \quad\) l. \(\mathrm{Cr}_{2} \mathrm{O}_{3}\)

The free energy change for a reaction, \(\Delta G,\) is an extensive property. What is an extensive property? Surprisingly, one can calculate \(\Delta G\) from the cell potential, \(\mathscr{E}\), for the reaction. This is surprising because \(\mathscr{E}\) is an intensive property. How can the extensive property \(\Delta G\) be calculated from the intensive property \(\mathscr{E}\) ?

The equation \(\Delta G^{\circ}=-\mathrm{nF} \mathscr{E}^{\circ}\) also can be applied to half-reactions. Use standard reduction potentials to estimate \(\Delta G_{\mathrm{f}}^{\circ}\) for \(\mathrm{Fe}^{2+}(a q)\) and $\mathrm{Fe}^{3+}(a q) .\left(\Delta G_{\mathrm{f}}^{\circ} \text { for } \mathrm{e}^{-}=0 .\right)$

Consider a galvanic cell based on the following half-reactions: $$\begin{array}{ll}{\text {}} & { \mathscr{E}^{\circ}(\mathbf{V}) } \\ \hline {\mathrm{La}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{La}} & {-2.37} \\\ {\mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe}} & {-0.44}\end{array}$$ a. What is the expected cell potential with all components in their standard states? b. What is the oxidizing agent in the overall cell reaction? c. What substances make up the anode compartment? d. In the standard cell, in which direction do the electrons flow? e. How many electrons are transferred per unit of cell reaction? f. If this cell is set up at \(25^{\circ} \mathrm{C}\) with \(\left[\mathrm{Fe}^{2+}\right]=2.00 \times 10^{-4} M\) and \(\left[\mathrm{La}^{3+}\right]=3.00 \times 10^{-3} M,\) what is the expected cell potential?

A galvanic cell is based on the following half-reactions: $$\begin{array}{ll}{\mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe}(s)} & {\mathscr{E}^{\circ}=-0.440 \mathrm{V}} \\ {2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2}(g)} & {\mathscr{E}^{\circ}=0.000 \mathrm{V}}\end{array}$$ where the iron compartment contains an iron electrode and \(\left[\mathrm{Fe}^{2+}\right]=1.00 \times 10^{-3} M\) and the hydrogen compartment contains a platinum electrode, $P_{\mathrm{H}_{2}}=1.00 \mathrm{atm},$ and a weak acid, HA, at an initial concentration of 1.00 \(\mathrm{M}\) . If the observed cell potential is 0.333 \(\mathrm{V}\) at \(25^{\circ} \mathrm{C},\) calculate the \(K_{\mathrm{a}}\) value for the weak acid HA.
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