Textbook Question
(d) Why are active metals such as Al obtained by electrolysis using molten salts rather than aqueous solutions?
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Ch.20 - Electrochemistry
Brown14th EditionChemistry: The Central ScienceISBN: 9780134414232
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Table of contents
- Ch.1 - Introduction: Matter, Energy, and Measurement140
- Ch.2 - Atoms, Molecules, and Ions192
- Ch.3 - Chemical Reactions and Reaction Stoichiometry172
- Ch.4 - Reactions in Aqueous Solution156
- Ch.5 - Thermochemistry151
- Ch.6 - Electronic Structure of Atoms137
- Ch.7 - Periodic Properties of the Elements127
- Ch.8 - Basic Concepts of Chemical Bonding143
- Ch.9 - Molecular Geometry and Bonding Theories167
- Ch.10 - Gases147
- Ch.11 - Liquids and Intermolecular Forces97
- Ch.12 - Solids and Modern Materials129
- Ch.13 - Properties of Solutions126
- Ch.14 - Chemical Kinetics175
- Ch.15 - Chemical Equilibrium107
- Ch.16 - Acid-Base Equilibria127
- Ch.17 - Additional Aspects of Aqueous Equilibria133
- Ch.18 - Chemistry of the Environment88
- Ch.19 - Chemical Thermodynamics140
- Ch.20 - Electrochemistry137
- Ch.21 - Nuclear Chemistry99
- Ch.22 - Chemistry of the Nonmetals66
- Ch.23 - Transition Metals and Coordination Chemistry69
- Ch.24 - The Chemistry of Life: Organic and Biological Chemistry11
Brown14th EditionChemistry: The Central ScienceISBN: 9780134414232Not the one you use?Change textbook
Chapter 20, Problem 91b
(b) What amperage is required to plate out 0.250 mol Cr from a Cr3+ solution in a period of 8.00 h?
Verified step by step guidance1
Identify the relevant electrochemical reaction: The reduction of Cr³⁺ to Cr metal is represented by the half-reaction: Cr³⁺ + 3e⁻ → Cr.
Determine the total charge needed: Since 1 mole of Cr requires 3 moles of electrons, calculate the total moles of electrons needed for 0.250 mol of Cr using the stoichiometry of the reaction.
Convert moles of electrons to charge: Use Faraday's constant (approximately 96485 C/mol) to convert the moles of electrons to total charge in coulombs.
Calculate the current (amperage): Use the formula I = Q/t, where I is the current in amperes, Q is the total charge in coulombs, and t is the time in seconds. Convert the given time from hours to seconds.
Solve for the current: Substitute the values for Q and t into the formula to find the required current in amperes.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Faraday's Laws of Electrolysis
Faraday's Laws of Electrolysis state that the amount of substance deposited during electrolysis is directly proportional to the quantity of electric charge passed through the electrolyte. The first law quantifies this relationship, while the second law relates the mass of the substance to its molar mass and the number of electrons involved in the reaction.
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Amperage and Charge
Amperage, or current (I), is the flow of electric charge measured in amperes (A). The total charge (Q) can be calculated using the formula Q = I × t, where t is the time in seconds. Understanding this relationship is crucial for determining the current needed to achieve a specific electrochemical reaction over a given time.
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01:53Formal Charge
Electrochemical Equivalent
The electrochemical equivalent (ECE) is the mass of a substance that can be deposited or dissolved by the passage of one coulomb of electric charge. For chromium ions (Cr³⁺), the ECE can be calculated using the molar mass and the number of electrons transferred in the reduction process, which is essential for determining the required amperage for plating.
Related Practice
Textbook Question
(d) Why is sodium metal not obtained when an aqueous solution of NaCl undergoes electrolysis?
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Textbook Question
(c) What process occurs at the anode in the electrolysis of molten NaCl?
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Textbook Question
A mixture of copper and gold metals that is subjected toelectrorefining contains tellurium as an impurity. The standard reduction potential between tellurium and its lowestcommon oxidation state, Te4+, isTe4+1aq2 + 4 e- ¡ Te1s2 E°red = 0.57 VGiven this information, describe the probable fate of tellurium impurities during electrorefining. Do the impuritiesfall to the bottom of the refining bath, unchanged, as copper is oxidized, or do they go into solution as ions? If theygo into solution, do they plate out on the cathode?
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