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: The reaction below has a cell potential, Eo , of – 1.18 V. Consider the reaction below: Mn2+ (aq) + 2 e –       →      Mn (s)   What is the new cell potential for this reaction? 3 Mn (s)      →       3 Mn2+ (aq) + 6 e –   a. + 1.18 V b. + 3.54 V c. + 0.39 V d. – 3.54 V

Problem

The reaction below has a cell potential, Eo , of – 1.18 V. Consider the reaction below:

Mn2+ (aq) + 2 e       →      Mn (s)

 

What is the new cell potential for this reaction?

3 Mn (s)      →       3 Mn2+ (aq) + 6 e

 

a. + 1.18 V

b. + 3.54 V

c. + 0.39 V

d. – 3.54 V