Domanda del libro di testo
Predict the anode, cathode, and overall cell reactions when an aqueous solution of each of the following salts is electrolyzed in a cell having inert electrodes. (c) LiOH
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Ch.19 - Electrochemistry
McMurry8th EditionChemistryISBN: 9781292336145
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TOC.tableOfContents
- Ch.1 - Chemical Tools: Experimentation & Measurement170
- Ch.2 - Atoms, Molecules & Ions160
- Ch.3 - Mass Relationships in Chemical Reactions169
- Ch.4 - Reactions in Aqueous Solution199
- Ch.5 - Periodicity & Electronic Structure of Atoms102
- Ch.6 - Ionic Compounds: Periodic Trends and Bonding Theory92
- Ch.7 - Covalent Bonding and Electron-Dot Structures61
- Ch.8 - Covalent Compounds: Bonding Theories and Molecular Structure59
- Ch.9 - Thermochemistry: Chemical Energy122
- Ch.10 - Gases: Their Properties & Behavior155
- Ch.11 - Liquids & Phase Changes54
- Ch.12 - Solids and Solid-State Materials73
- Ch.13 - Solutions & Their Properties120
- Ch.14 - Chemical Kinetics98
- Ch.15 - Chemical Equilibrium125
- Ch.16 - Aqueous Equilibria: Acids & Bases110
- Ch.17 - Applications of Aqueous Equilibria147
- Ch.18 - Thermodynamics: Entropy, Free Energy & Equilibrium86
- Ch.19 - Electrochemistry154
- Ch.20 - Nuclear Chemistry96
- Ch.21 - Transition Elements and Coordination Chemistry156
- Ch.22 - The Main Group Elements187
- Ch.23 - Organic and Biological Chemistry51
Capitolo 19, Problema 141
What constant current (in amperes) is required to produce aluminum by the Hall–Héroult process at a rate of 40.0 kg/h?
Guida verificata passo dopo passo1
Identify the chemical reaction involved in the Hall–Héroult process: \(2\text{Al}_2\text{O}_3 + 3\text{C} \rightarrow 4\text{Al} + 3\text{CO}_2\).
Determine the molar mass of aluminum (Al), which is approximately 26.98 g/mol.
Convert the production rate of aluminum from kg/h to moles per second: \(40.0 \text{ kg/h} \rightarrow \text{g/s} \rightarrow \text{mol/s}\).
Use Faraday's laws of electrolysis to relate the moles of aluminum produced to the charge required: \(n = \frac{Q}{zF}\), where \(n\) is moles of electrons, \(Q\) is charge in coulombs, \(z\) is the number of electrons transferred per mole of aluminum (3 for Al), and \(F\) is Faraday's constant (96485 C/mol).
Calculate the current \(I\) using the formula \(I = \frac{Q}{t}\), where \(t\) is time in seconds.
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Hall–Héroult Process
The Hall–Héroult process is an industrial method for extracting aluminum from its ore, bauxite. It involves the electrolysis of aluminum oxide dissolved in molten cryolite, where aluminum ions are reduced to form aluminum metal at the cathode. Understanding this process is crucial for calculating the current required, as it directly relates to the amount of aluminum produced and the efficiency of the electrolysis.
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Percorso guidato
04:20
Spontaneity of Processes
Faraday's Laws of Electrolysis
Faraday's Laws of Electrolysis describe the relationship between the amount of substance produced at an electrode during electrolysis and the quantity of electric charge passed through the electrolyte. The first law states that the mass of a substance altered at an electrode is proportional to the total electric charge passed. This principle is essential for determining the current needed to produce a specific mass of aluminum in the Hall–Héroult process.
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Percorso guidato
01:40Faraday's Constant in Electrochemistry
Current Calculation
To calculate the current required for electrolysis, one must consider the molar mass of aluminum and the charge needed to reduce aluminum ions. The formula I = (n * F) / t relates current (I) to the number of moles (n), Faraday's constant (F), and time (t). This calculation is vital for determining the amperage necessary to achieve the desired production rate of aluminum, ensuring efficient operation of the electrolysis process.
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Percorso guidato
02:28Stoichiometric Rate Calculations
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Predict the anode, cathode, and overall cell reactions when an aqueous solution of each of the following salts is electrolyzed in a cell having inert electrodes.
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Aluminum, titanium, and several other metals can be col-ored by an electrochemical process called anodizing. Anodizing oxidizes a metal anode to yield a porous metal oxide coating that can incorporate dye molecules to give brilliant colors. In the oxidation of aluminum, for instance, the electrode reactions are The thickness of the aluminum oxide coating that forms on the anode can be controlled by varying the current flow during the electrolysis. How many minutes are required to produce a 0.0100-mm thick coating of Al2O3 (density 3.97 g/cm^3) on a square piece of aluminum metal 10.0 cm on an edge if the current passed through the piece is 0.600 A?
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