Ch.4 - Chemical Quantities & Aqueous ReactionsWorksheetSee 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 reduction half-reactions: Fe3+(aq) + e-  →  Fe2+(aq) Eredo= + 0.77V S2O62-(aq) + 4H+(aq) + 2e-  →  2H2SO3(aq) Eredo= + 0.60V N2O(g) + 2H+(aq) + 2e-  →  N2(g)

Problem

Given the following reduction half-reactions:
Fe3+(aq) + e-  →  Fe2+(aq)
Eredo= + 0.77V
S2O62-(aq) + 4H+(aq) + 2e-  →  2H2SO3(aq)
Eredo= + 0.60V
N2O(g) + 2H+(aq) + 2e-  →  N2(g) + H2O(l)
Eredo= - 1.77V
VO2+(aq) + 2H+(aq) + e-  →  VO2+(aq) + H2O(l)
Eredo= + 1.00V

Write balanced chemical equation for the oxidation of Fe2+(aq) by N2O(g).