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: Given the following half-reactions and associated standard reduction potentials: AuBr4-(aq) + 3e-  →  Au(s) + 4Br-(aq) Eredo= - 0.858V Eu3+(aq) + e-  →  Eu2+(aq) Eredo= - 0.43V

Problem

Given the following half-reactions and associated standard reduction potentials:
AuBr4-(aq) + 3e-  →  Au(s) + 4Br-(aq)
Eredo= - 0.858V
Eu3+(aq) + e-  →  Eu2+(aq)
Eredo= - 0.43V
IO-(aq) + H2O(l) + 2e-  →  I-(aq) + 2OH-(aq)
Eredo= + 0.49V
Sn2+(aq) + 2e-  →  Sn(s)
Eredo= - 0.14V

Write the cell reaction for the combination of these half-cell reactions that leads to the largest positive cell emf.