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: Sketch a voltaic cell for each redox reaction. Label the anode and cathode and indicate the half-reaction that occurs at each electrode and the species present in each solution. Also indicate the dire

Solution: Sketch a voltaic cell for each redox reaction. Label the anode and cathode and indicate the half-reaction that occurs at each electrode and the species present in each solution. Also indicate the dire

Problem

Sketch a voltaic cell for each redox reaction. Label the anode and cathode and indicate the half-reaction that occurs at each electrode and the species present in each solution. Also indicate the direction of electron flow.
c. 2 NO3 (aq) + 8 H+(aq) + 3 Cu(s) → 2 NO(g) + 4 H 2O (/) + 3 Cu2+(aq)

 

Solution

We’re being asked to sketch a voltaic cell for the following reaction:

2 NO3–(aq) + 8 H+(aq) + 3 Cu(s)  2 NO(g) + 4 H2O(l) + 3 Cu2+(aq)


Recall that a voltaic or galvanic cell is an electrochemical cell that is spontaneous and produces electricity


From the reaction, we can see that the oxidation state of Cu went from 0 (Cu) to +2 (Cu2+) which means it was oxidized or lost electrons


The anode (–) half-reaction is:

Cu(s)  Cu2+(aq) + 2 e

0 = +2 + x

x = –2 = 2 e to the product side


The other half-reaction is then reduced or gained electrons


The cathode (+) half-reaction is:

2 NO3–(aq) + 8 H+(aq) + 6 e  2 NO(g) + 4 H2O(l)

2(–1) + 8(+1) + x = 0

x = –6 = 6 e to the reactant side


View the complete written solution...