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: Using the following standard reduction potentials, Fe3+ (aq) + e–   → Fe2+ (aq)  E°red = +0.77 V Pb2+ (aq) + 2 e– → Pb (s)      E°red = –0.13 V   Calculate the standard emf for the following cell

Problem

Using the following standard reduction potentials,

Fe3+ (aq) + e   → Fe2+ (aq)  E°red = +0.77 V

Pb2+ (aq) + 2 e → Pb (s)      E°red = –0.13 V

 

Calculate the standard emf for the following cell reaction as written and determine whether or not the reaction is spontaneous or non‐spontaneous:

Pb2+ (aq) + 2 Fe2+ (aq) → 2 Fe3+ (aq) + Pb (s)

a) –0.90 V, non‐spontaneous

b) –0.90 V, spontaneous

c) +0.90 V, non‐spontaneous

d) –0.64 V, non‐spontaneous

e) +0.64 V, spontaneous