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: Break down reaction into its half-reactions.4Fe + 3O2 → 2Fe2O3

Solution: Break down reaction into its half-reactions.4Fe + 3O2 → 2Fe2O3

Problem

Break down reaction into its half-reactions.

4Fe + 3O→ 2Fe2O3

Solution

We’re being asked to break down the given reaction below into its half-reactions.


Given Reaction:      4 Fe + 3 O2 → 2 Fe2O3


We’ll write the half-reactions using the following steps:

Step 1. Determine the charge of the species in the reactant side
Step 2. Determine the charge of the species in the product side
Step 3. Write the unbalanced half-reactions
Step 4. Balance the elements in each half-reaction
Step 5. Balance the charges: add electrons to the more positive side (or less negative side) 
Step 6. Determine the oxidation half-reaction and the reduction half reaction. 



Step 1. Determine the charge of the species in the reactant side

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