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Chapter 1: Problem 24

Molten \(\mathrm{AlCl}_{3}\) is electrolyzed with a constant current of 5.00 amperes over a period of 600.0 seconds. Which of the following expressions is equal to the maximum mass of Al(s) that plates out? (1 faraday = 96,500 coulombs) (A) \(\frac{(600)(5.00)}{(96,500)(3)(27.0)}\) grams (B) \(\frac{(600)(5.00)(3)(27.0)}{(96,500)}\) grams (C) \(\frac{(600)(5.00)(27.0)}{(96,500)(3)}\) grams (D) \(\frac{(96,500)(3)(27.0)}{(600)(5.00)}\) grams

Short Answer

Expert verified
The maximum mass of Al that plates out is represented by the expression in option (A).

Step by step solution

01

Stoichiometry of the reaction

In the production of metallic Al from AlCl3, the reaction is: \[Al^{3+} + 3e^- \rightarrow Al(s)\] This indicates that three moles of electrons (or 3 faradays) are needed to produce one mole of Al.
02

Calculate total charge

The total charge (Q) passed in the electrolysis can be calculated from the provided current (I) and the time (t). Use the formula \[Q = It\] where I is given as 5.00 Amperes and t is 600.0 seconds. This gives \[Q = (5.00 A)(600.0 s)\] coulombs.
03

Apply Faraday's Law

Applying Faraday鈥檚 Law, for the deposition of one mole of Al, three faradays or 3*96,500 coulombs are needed. Hence, calculate the mass using the formula \[mass = \frac{Q}{F}*\frac{1}{n}*M\] where F is Faraday鈥檚 constant (96,500 C), n is the number of electrons involved (3 for aluminium), M is the molar mass of Al (27.0 g/mol). Substituting the values the mass becomes: \[mass = \frac{(5.00 A)(600.0 s)}{(96,500 C/mol)(3 mol)(27.0 g/mol)}\]
04

Compare with given options

On close inspection, the correct expression matches with option A. Therefore, the maximum mass of Al that plates out is given by the expression in option A.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Faraday's Constant
Faraday's constant is a fundamental value that represents the charge of one mole of electrons. It is denoted by the symbol \( F \) and has a value of 96,500 coulombs per mole (C/mol). In electrolysis, this constant plays a crucial role as it helps to quantify the amount of substance that can be deposited or released in an electrochemical reaction.
  • A single mole of electrons, which has a charge of 96,500 coulombs, is essential for converting ions into their respective elements.
  • This concept is central to understanding stoichiometric relationships in electrochemical processes.
  • In the context of our problem, 3 faradays are needed to reduce Al3+ ions to metallic aluminum, due to the charge of aluminum ions.
By using Faraday's constant, one can calculate how much of a material can be generated when a certain charge is passed through the circuit.
Molar Mass
Molar mass is the mass of a given substance (chemical element or chemical compound) divided by the amount of substance. Its unit is usually grams per mole (g/mol). When working with electrolysis, molar mass is essential for converting moles into practical measurable mass, such as grams.
  • Molar mass helps us figure out how much of a material we have in a chemical reaction.
  • For aluminum (Al), the molar mass is 27.0 g/mol.
  • This is an important part of calculating how much metal is formed when an electrical current passes through a substance like AlCl3.
In calculations, knowing the molar mass along with the number of moles from Faraday鈥檚 law allows you to determine the actual mass of the metal produced.
Stoichiometry
Stoichiometry refers to the quantitative relationships between reactants and products in a chemical reaction. Understanding stoichiometry is key in electrolysis as it determines how much of each substance is involved and how they react with one another.
  • The stoichiometric equation for the reaction in the exercise is \(Al^{3+} + 3e^- \rightarrow Al(s)\).
  • This indicates that three moles of electrons (3 faradays) are required to produce one mole of Al.
  • Stoichiometry helps to connect the charges passed through the solution to the amount of aluminum deposited during the process.
By using stoichiometry, students can predict the outcomes of reactions and calculate what is expected from the electrochemical equations involved.
Current Calculation
The current in amperes (A) and time in seconds (s) together allow us to determine the total electric charge that has passed through the system, using the formula \(Q = It\).
  • In this exercise, we are given a current of 5.00 A and a time duration of 600 seconds.
  • Multiplying these together gives us the total charge \(Q = 5.00 \times 600 = 3000\) coulombs.
  • This charge is essential for calculating how much aluminum can be deposited, by comparing it against the charge required to deposit a mole of aluminum according to Faraday's laws of electrolysis.
Understanding current calculation allows you to link electrical energy (in terms of charge) to the chemistry occurring in electrolysis.

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Most popular questions from this chapter

In a voltaic cell with a Cu(s) \(| \mathrm{Cu}^{2+}\) cathode and \(\mathrm{a} \mathrm{Pb}^{2+} | \mathrm{Pb}\) (s) anode, increasing the concentration of \(\mathrm{Pb}^{2+}\) causes the voltage to decrease. What is the reason for this? (A) The value for Q will increase, causing the cell to come closer to equilibrium. (B) The solution at the anode becomes more positively charged, leading to a reduced electron flow. (C) The reaction will shift to the right, causing a decrease in favor ability. (D) Cell potential will always decrease anytime the concentration of any aqueous species present increases.

If the solubility of \(\mathrm{BaF}_{2}\), is equal to \(x,\) which of the following expressions is equal to the solubility product, \(K_{s p}\) , for \(\operatorname{BaF}_{2} ?\) (A) \(x^{2}\) (B) 2\(x^{2}\) (C) 2\(x^{3}\) (D) 4\(x^{3}\)

Use the following information to answer questions 25-28. A voltaic cell is created using the following half-cells: \(\begin{array}{ll}{\mathrm{Cr}^{3+}+3 e \rightarrow \mathrm{Cr}(s)} & {E^{\circ}=-0.41 \mathrm{V}} \\ {\mathrm{Pb}^{2+}+2 e \rightarrow \mathrm{Pb}(s)} & {E^{\circ}=-0.12 \mathrm{V}}\end{array}\) The concentrations of the solutions in each half-cell are 1.0 M. Based on the given reduction potentials, which of the following would lead to a reaction? (A) Placing some \(\operatorname{Cr}(s)\) in a solution containing \(\mathrm{Pb}^{2+}\) ions (B) Placing some \(\mathrm{Pb}(s)\) in a solution containing \(\mathrm{Cr}^{3+}\) ions (C) Placing some \(\mathrm{Cr}(s)\) in a solution containing \(\mathrm{Cr}^{3+}\) ions (D) Placing some \(\mathrm{Pb}(s)\) in a solution containing \(\mathrm{Pb}^{2+}\) ions

Questions 45-48 refer to the following. Inside a calorimeter, 100.0 \(\mathrm{mL}\) of 1.0 \(\mathrm{M}\) hydrocyanic acid (HCN), a weak acid, and 100.0 \(\mathrm{mL}\) of 0.50 \(\mathrm{M}\) sodium hydroxide are mixed. The temperature of the mixture rises from \(21.5^{\circ} \mathrm{C}\) to \(28.5^{\circ} \mathrm{C}\) . The specific heat of the mixture is approximately \(4.2 \mathrm{J} / \mathrm{g}^{\circ} \mathrm{C},\) and the density is identical to that of water. As \(\Delta T\) increases, what happens to the equilibrium constant and why? (A) The equilibrium constant increases because more products are created. (B) The equilibrium constant increases because the rate of the forward reaction increases. (C) The equilibrium constant decreases because the equilibrium shifts to the left. (D) The value for the equilibrium constant is unaffected by temperature and will not change.

A 1 -molar solution of a very weak monoprotic acid has a pH of 5. What is the value of \(K_{a}\) for the acid? (A) \(K_{a}=1 \times 10^{-10}\) (B) \(K_{a}=1 \times 10^{-7}\) (C) \(K_{a}=1 \times 10^{-5}\) (D) \(K_{a}=1 \times 10^{-2}\)
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