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Chapter 18: Problem 96

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 with a continuous capacity of 1.00 million A is approximately 549022.88 grams.

Step by step solution

01

Identify the information given.

In this problem, we have been given the following information: 1. Current (I) = 1.00 million A (1000000 A) 2. Time (t) = 2.00 hours
02

Convert all given values into base units.

Since we will use the SI unit system, convert the given values to that system: 1. Time (t) = 2.00 hours × 3600 s/hour = 7200 s
03

Recall the Faraday's Law of Electrolysis formula.

Faraday's Law of Electrolysis states that: Mass (m) = (Current * Time * Molecular Weight) / (n * Faraday's constant), where - Current (I) = 1000000 A - Time (t) = 7200 s - Molecular Weight of Aluminum (M) = 26.98 g/mol - n = number of electrons involved in the redox reaction (Al³⁺ + 3e⁻ → Al) - Faraday's constant (F) = 96485 C/mol
04

Determine the number of electrons involved in the redox reaction.

For the given problem, the redox reaction in the electrolysis process is: Al³⁺ + 3e⁻ → Al Clearly, 3 electrons (n = 3) are involved in the reaction.
05

Plug the values into the formula and calculate the mass of aluminum produced.

Now, plug in the given values into the Faraday's Law of Electrolysis formula to calculate the mass of aluminum produced: Mass (m) = (1000000 A * 7200 s * 26.98 g/mol) / (3 * 96485 C/mol) Mass (m) = \( \frac{1000000 \times 7200 \times 26.98}{3\times 96485} \) g ≈ 549022.88 g The mass of aluminum that can be produced in 2.00 hours with a continuous capacity of 1.00 million A is approximately 549022.88 grams.

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Most popular questions from this chapter

What is electrochemistry? What are redox reactions? Explain the difference between a galvanic and an electrolytic cell.

What reaction will take place at the cathode and the anode when each of the following is electrolyzed? a. molten KF b. molten \(\mathrm{CuCl}_{2}\) c. molten \(\mathrm{MgI}_{2}\)

Electrolysis of an alkaline earth metal chloride using a current of \(5.00 \mathrm{~A}\) for \(748 \mathrm{~s}\) deposits \(0.471 \mathrm{~g}\) of metal at the cathode. What is the identity of the alkaline earth metal chloride?

Consider the following galvanic cell:Calculate the \(K_{\text {sp }}\) value for \(\mathrm{Ag}_{2} \mathrm{SO}_{4}(s)\). Note that to obtain silver ions in the right compartment (the cathode compartment), excess solid \(\mathrm{Ag}_{2} \mathrm{SO}_{4}\) was added and some of the salt dissolved.

A galvanic cell is based on the following half-reactions: $$ \begin{aligned} \mathrm{Ag}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s) & \mathscr{E}^{\circ}=0.80 \mathrm{~V} \\ \mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s) & \mathscr{E}^{\circ}=0.34 \mathrm{~V} \end{aligned} $$ In this cell, the silver compartment contains a silver electrode and excess \(\operatorname{AgCl}(s)\left(K_{\mathrm{sp}}=1.6 \times 10^{-10}\right)\), and the copper compartment contains a copper electrode and \(\left[\mathrm{Cu}^{2+}\right]=2.0 M\). a. Calculate the potential for this cell at \(25^{\circ} \mathrm{C}\). b. Assuming \(1.0 \mathrm{~L}\) of \(2.0 \mathrm{M} \mathrm{Cu}^{2+}\) in the copper compartment, calculate the moles of \(\mathrm{NH}_{3}\) that would have to be added to give a cell potential of \(0.52 \mathrm{~V}\) at \(25^{\circ} \mathrm{C}\) (assume no volume change on addition of \(\mathrm{NH}_{3}\) ). $$ \begin{aligned} \mathrm{Cu}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons & \\ \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}{ }^{2+}(a q) & K=1.0 \times 10^{13} \end{aligned} $$
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