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: Balance the following reaction:Cr2O72-(aq) + CH3OH(aq) ? HCO2H(aq) + Cr3+(aq) (acidic solution)

Problem

Balance the following reaction:

Cr2O72-(aq) + CH3OH(aq) ? HCO2H(aq) + Cr3+(aq) (acidic solution)

Solution

We are being asked to balance the given oxidation-reduction reaction. The reaction is under acidic conditions. When balancing redox reactions under basic conditions, we will follow the following steps.

Step 1: Separate the whole reaction into two half-reactions

Step 2: Balance the non-hydrogen and non-oxygen elements first

Step 3: Balance oxygen by adding H2O to the side that needs oxygen (1 O: 1 H2O)

Step 4: Balance hydrogen by adding H+ to the side that needs hydrogen (1 H: 1 H+)

Step 5: Balance the charges: add electrons to the more positive side (or less negative side)

Step 6: Balance electrons on the two half-reactions

Step 7: Get the overall reaction by adding the two reactions.


Balancethe redox reaction under acidic conditions:

Cr2O72-(aq) + CH3OH(aq) → HCO2H(aq) + Cr3+(aq)

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