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: Calculate the equilibrium constant, K, for the following reaction at 25 °C.Fe3+(aq) + B(s) + 6H 2O(l) → Fe(s)+ H3BO3(s) + 3H3O+(aq)The balanced reduction half-reactions for the above equation and thei

Problem

Calculate the equilibrium constant, K, for the following reaction at 25 °C.

Fe3+(aq) + B(s) + 6H 2O(l) → Fe(s)+ H3BO3(s) + 3H3O+(aq)

The balanced reduction half-reactions for the above equation and their respective standard reduction potential values (E degree) are as follows:

Fe3+(aq) + 3e- → Fe(s)                                           E° = - 0.04 V
H3BO3(s) + 3H3O+(aq) + 3e- → B(s) + 6H 2O(l)       E° = - 0.8698 V