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: An electrochemical cell has the following half cell reactions: Cu2+(aq) +2e- → Cu(s)          E°1/2 = +0.34 Zn2+(aq) + 2e- → Zn(s)          E°1/2 = -0.76 If the cell operates with the Cu electrode as the cathode and the Zn as the anode, What is the cell potential when the [Cu2+] = 0.001M and the [Zn2+] = 0.1M? A) 1.159V B) -1.159 V C) +0.479 V D) -0.479 V E) +1.041 V

Problem

An electrochemical cell has the following half cell reactions:
Cu2+(aq) +2e- → Cu(s)          E°1/2 = +0.34

Zn2+(aq) + 2e- → Zn(s)          E°1/2 = -0.76

If the cell operates with the Cu electrode as the cathode and the Zn as the anode, What is the cell potential when the [Cu2+] = 0.001M and the [Zn2+] = 0.1M?


A) 1.159V
B) -1.159 V
C) +0.479 V
D) -0.479 V
E) +1.041 V