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: Consider the unbalanced redox reaction:Cr2O72–(aq) + Cu(s) → Cr 3+ (aq) + Cu 2+(aq)Balance the equation and determine the volume of a 0.850 M K2Cr2O7 solution required to completely react with 5.25 g

Solution: Consider the unbalanced redox reaction:Cr2O72–(aq) + Cu(s) → Cr 3+ (aq) + Cu 2+(aq)Balance the equation and determine the volume of a 0.850 M K2Cr2O7 solution required to completely react with 5.25 g

Problem

Consider the unbalanced redox reaction:
Cr2O72–(aq) + Cu(s) → Cr 3+ (aq) + Cu 2+(aq)
Balance the equation and determine the volume of a 0.850 M K2Cr2O7 solution required to completely react with 5.25 g of Cu.

 

Solution

We are being asked to determine the volume of K2Cr2O7 solution to completely react with 5.25 g of Cu. We will calculate the amount of K2Cr2O7 using the balanced redox reaction equation and mol to mol comparison.


Let’s balance the redox reaction first. We will assume that the reaction is under acidic condition unless stated otherwise. 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+)

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