Ch.18 - ElectrochemistryWorksheetSee all chapters
All Chapters
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: Use data from the table above to calculate E˚cell for the reaction: 2 ClO2(g) + Pb(s) → 2 ClO2–(aq) + Pb2+(aq)NO3–(aq) + 4 H+(aq) + 3 e– →  NO(g) + 2 H2O(l)E˚ = 0.96 VClO2(g) + e– →  ClO2–(aq)E˚= 0.95

Problem

Use data from the table above to calculate E˚cell for the reaction: 2 ClO2(g) + Pb(s) → 2 ClO2(aq) + Pb2+(aq)

NO3(aq) + 4 H+(aq) + 3 e →  NO(g) + 2 H2O(l)E˚ = 0.96 V
ClO2(g) + e →  ClO2(aq)E˚= 0.95 V
Cu2+(aq) +2 e →  Cu(s)E˚= 0.34 V
2 H+(aq) +2 e →  H2(g)E˚= 0.00 V
Pb2+(aq) +2 e  →  Pb(s)E˚ = –0.13 V
Fe2+(aq) +2 e →  Fe(s)E˚ = –0.45 V