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 redox reaction by inserting the appropriate coefficients.H+ + CrO42- + NO2- → Cr3+ + H2O +NO3-

Solution: Balance the following redox reaction by inserting the appropriate coefficients.H+ + CrO42- + NO2- → Cr3+ + H2O +NO3-

Problem

Balance the following redox reaction by inserting the appropriate coefficients.

H+ + CrO42- + NO2- → Cr3+ + H2O +NO3-

Solution

We’re being asked to balance the given oxidation-reduction reaction:

H+ + CrO42– + NO2 Cr3+ + H2O + NO3


We assume that the redox is in acidic solution since H+ is present. When balancing redox reactions under acidic 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+)


View the complete written solution...