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Chapter 18: Problem 92

Aluminum is produced commercially by the electrolysis of \(\mathrm{Al}_{2} \mathrm{O}_{3}\) in the presence of a molten salt. If a plant has a continuous capacity of \(1.00\) million A, what mass of aluminum can be produced in \(2.00 \mathrm{~h} ?\)

Short Answer

Expert verified
The mass of Aluminum that can be produced in 2.00 hours using a 1.00 million amperes capacity is approximately 2,017,092.61 g or 2.02 metric tons.

Step by step solution

01

Calculate the total charge passed through the electrolyte

We are given the current is 1.00 million A = 1.00 * 10^6 A and the time is 2.00 hours. We can find the total charge Q (in coulombs) by multiplying the current (in amperes) by time (in seconds). We should firstly convert time from hours to seconds, i.e. 2.00 hours = 2.00 * 3600 seconds. Q = current × time Q = (1.00 * 10^6 A)(2.00 * 3600 s) Q = 7.20 * 10^9 C
02

Calculate the amount of Aluminum (in moles) produced

Now we need to convert the charge (Q) into moles of Aluminum. This will be done using Faraday's law which states that: Moles of substance = (Q * n) / (F * v) Where Q is charge (in C), n is the number of electrons, F is Faraday's constant (96500 C/mol) and v is the valency of the substance. For Aluminum, it has a valency of 3 as it is an electropositive element with 3 valence electrons (losing 3 electrons to become \(\mathrm{Al}^{3+}\)). Therefore, v = 3 and we know F (Faraday's constant) = 96500 C/mol. Moles of Aluminum = (7.20 * 10^9 C *3) / (96500 C/mol *3) Moles of Aluminum = 74758.55 mol
03

Calculate the mass of Aluminum produced

Now, we need to convert the moles of Aluminum into mass using the molar mass. We know that the molar mass of Aluminum is approximately 26.98 g/mol. Mass of Aluminum = Moles of Aluminum × Molar Mass of Aluminum Mass of Aluminum = 74758.55 mol × 26.98 g/mol Mass of Aluminum = 2017092.61 g Hence, the mass of Aluminum that can be produced in 2.00 hours using a 1.00 million amperes capacity is approximately 2,017,092.61 g or 2.02 metric tons.

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Most popular questions from this chapter

Hydrazine is somewhat toxic. Use the half-reactions shown below to explain why household bleach (a highly alkaline solution of sodium hypochlorite) should not be mixed with household ammonia or glass cleansers that contain ammonia. \(\mathrm{ClO}^{-}+\mathrm{H}_{2} \mathrm{O}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{OH}^{-}+\mathrm{Cl}^{-} \quad \mathscr{E}^{\circ}=0.90 \mathrm{~V}\) \(\mathrm{N}_{2} \mathrm{H}_{4}+2 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{NH}_{3}+2 \mathrm{OH}^{-} \quad \mathscr{E}^{\circ}=-0.10 \mathrm{~V}\)

Calculate \(\mathscr{E}^{\circ}\) values for the following cells. Which reactions are spontaneous as written (under standard conditions)? Balance the equations. Standard reduction potentials are found in Table 18.1. a. \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{I}^{-}(a q) \rightleftharpoons \mathrm{I}_{2}(a q)+\mathrm{Mn}^{2+}(a q)\) b. \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons \mathrm{F}_{2}(g)+\mathrm{Mn}^{2+}(a q)\)

The following standard reduction potentials have been determined for the aqueous chemistry of indium: $$\begin{array}{ll}\mathrm{In}^{3+}(a q)+2 \mathrm{e}^{-} \longrightarrow \operatorname{In}^{+}(a q) & \mathscr{E}^{\circ}=-0.444 \mathrm{~V} \\ \mathrm{In}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \operatorname{In}(s) & \mathscr{E}^{\circ}=-0.126 \mathrm{~V}\end{array}$$ a. What is the equilibrium constant for the disproportionation reaction, where a species is both oxidized and reduced, shown below? $$3 \operatorname{In}^{+}(a q) \longrightarrow 2 \operatorname{In}(s)+\operatorname{In}^{3+}(a q)$$ b. What is \(\Delta G_{\mathrm{f}}^{\circ}\) for \(\mathrm{In}^{+}(a q)\) if \(\Delta G_{\mathrm{f}}^{\circ}=-97.9 \mathrm{~kJ} / \mathrm{mol}\) for \(\mathrm{In}^{3+}(a q)\) ?

An aqueous solution of an unknown salt of ruthenium is electrolyzed by a current of \(2.50\) A passing for \(50.0 \mathrm{~min}\). If \(2.618 \mathrm{~g}\) Ru is produced at the cathode, what is the charge on the ruthenium ions in solution?

Consider the following half-reactions: $$\begin{aligned}\mathrm{Pt}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt} & & \mathscr{E}^{\circ}=1.188 \mathrm{~V} \\ \mathrm{PtCl}_{4}^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt}+4 \mathrm{Cl}^{-} & & \mathscr{C}^{\circ}=0.755 \mathrm{~V} \\ \mathrm{NO}_{3}^{-}+4 \mathrm{H}^{+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{NO}+2 \mathrm{H}_{2} \mathrm{O} & & \mathscr{C}^{\circ}=0.96 \mathrm{~V}\end{aligned}$$ Explain why platinum metal will dissolve in aqua regia (a mixture of hydrochloric and nitric acids) but not in either concentrated nitric or concentrated hydrochloric acid individually.
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