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: Both Ti and V are reactive enough to displace H 2 from water; of the two metals, Ti is the stronger reducing agent. The difference in their E°half-cell values is 0.43 V. Given V(s) + Cu2+(aq) ⟶ V2+(aq

Problem

Both Ti and V are reactive enough to displace H 2 from water; of the two metals, Ti is the stronger reducing agent. The difference in their E°half-cell values is 0.43 V. Given 

V(s) + Cu2+(aq) ⟶ V2+(aq) + Cu(s)             ΔG° = −298 kJ/mol

use Appendix D to calculate the E°half-cell values for V2+/V and Ti2+/Ti half-cells.