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

A current of 3.07 A is passed through a Pb(NO3)solution. How long (in hours) would this current have to be applied to plate out 7.10 g of lead?


We have to calculate the time required to deposit 7.10 g of lead metal from a solution containing Pb(NO3)2.

  • The time required to deposit a substance on an electrode can be determined through the use of Faraday’s constant.
  • Faraday’s constant is defined as 9.648x104 C charge per mol of e-.
  • NO3- is a polyatomic ion with a charge of -1.
  • Pb(NO3)2 has 2 NO3- ions therefore the negative charge is -2.
  • To balance this negative charge, a charge of +2 on Pb is required.

Pb2+ has a +2 charge, so it requires 2 electrons to discharge completely and deposit as a metal.

Pb2+(aq) + 2e- → Pb(s)

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