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: Consider this reaction.Sn2+ (aq) + 2 Fe3+ (aq) → Sn4+ (aq) + 2 Fe2+ (aq)     E° = 0.617 VWhat is the value of E when [Sn2+] and {Fe3+] are equal to 0.50 M and [Sn4+] and [Fe2+] are equal to 0.10 M?(A)

Solution: Consider this reaction.Sn2+ (aq) + 2 Fe3+ (aq) → Sn4+ (aq) + 2 Fe2+ (aq)     E° = 0.617 VWhat is the value of E when [Sn2+] and {Fe3+] are equal to 0.50 M and [Sn4+] and [Fe2+] are equal to 0.10 M?(A)

Problem

Consider this reaction.

Sn2+ (aq) + 2 Fe3+ (aq) → Sn4+ (aq) + 2 Fe2+ (aq)     E° = 0.617 V

What is the value of E when [Sn2+] and {Fe3+] are equal to 0.50 M and [Sn4+] and [Fe2+] are equal to 0.10 M?

(A) 0.069V

(B) 0.679 V

(C) 0.658 V

(D) 0.576

Solution

We are asked to find the cell potential of the given reaction. We will use the Nernst Equation to calculate for the cell potential with the given conditions. The Nernst Equation relates the concentrations of compounds and cell potential.

Ecell = cell potential under non-standard conditions
cell = standard cell potential
n = number of e- transferred
Q= reaction quotient = [products]/[reactants] 


Let’s first determine how many electrons were transferred in the reaction by separating the two half-reactions:

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