Ch.18 - ElectrochemistryWorksheetSee all chapters
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Ch.1 - Intro to General Chemistry
Ch.2 - Atoms & Elements
Ch.3 - Chemical Reactions
BONUS: Lab Techniques and Procedures
BONUS: Mathematical Operations and Functions
Ch.4 - Chemical Quantities & Aqueous Reactions
Ch.5 - Gases
Ch.6 - Thermochemistry
Ch.7 - Quantum Mechanics
Ch.8 - Periodic Properties of the Elements
Ch.9 - Bonding & Molecular Structure
Ch.10 - Molecular Shapes & Valence Bond Theory
Ch.11 - Liquids, Solids & Intermolecular Forces
Ch.12 - Solutions
Ch.13 - Chemical Kinetics
Ch.14 - Chemical Equilibrium
Ch.15 - Acid and Base Equilibrium
Ch.16 - Aqueous Equilibrium
Ch. 17 - Chemical Thermodynamics
Ch.18 - Electrochemistry
Ch.19 - Nuclear Chemistry
Ch.20 - Organic Chemistry
Ch.22 - Chemistry of the Nonmetals
Ch.23 - Transition Metals and Coordination Compounds

Solution: How many grams of Cu metal (63.55 g/mol) will be plated out of a Cu(NO)2 solution by passing a current of 3.00 amperes for 11 minutes? (1 Faraday = 96,485 Coulombs)1. 2.608 g2. 0.652 g 3. 1.304 g4. 0.0721 g

Solution: How many grams of Cu metal (63.55 g/mol) will be plated out of a Cu(NO3 )2 solution by passing a current of 3.00 amperes for 11 minutes? (1 Faraday = 96,485 Coulombs)1. 2.608 g2. 0.652 g 3. 1.304 g4.

Problem

How many grams of Cu metal (63.55 g/mol) will be plated out of a Cu(NO)2 solution by passing a current of 3.00 amperes for 11 minutes? (1 Faraday = 96,485 Coulombs)

1. 2.608 g

2. 0.652 g 

3. 1.304 g

4. 0.0721 g

Solution

We’re being asked to calculate the mass of Cu metal plated out of a Cu(NO3)2 solution.

Given:

 molar mass Cu = 63.55 g/mol
current = 3.00 A
             Note that amperes (A) = charge/time = C/s
            Current = 3.00 C/s

 t = 11 minutes
            Since the current is in C/s, we will have to convert time from min to s

 Faraday’s constant = 96,485 C/(mol e-)


Let’s first determine how many electrons are involved when Cu is plated out from the solution:

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