
Exploring Electrochemistry Concepts: Standard Potentials, Oxidizing Agents, and More
Dive into the world of electrochemistry with topics such as standard cell potentials, oxidizing/reducing agents, reduction potentials, and oxidation of metals by acids. Understand the Nernst Equation, redox titrations, and the behavior of different compounds in electrochemical reactions. Get ready to master the principles essential to understanding electrochemical processes.
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Presentation Transcript
Announcements I Mastering Ch. 18 assignment due 11/12 Lab next week Doing Experiment 9 (thermodynamics lnK vs. 1/T plots) and 10 (electrochemistry) Experiment 8 due Wed./Thurs. Quiz on Exp. 8 and 9 and on electrochemistry
Announcements III Today s Lecture Electrochemistry Standard Cell Potentials Oxidizing and Reducing Agents Acid Oxidation of Metals Charge and Energy in Voltaic Cells Relationship between G and E The Nernst Equation (for non-standard conditions)
Chapter 18 Electrochemistry Example Question An Ag/AgCl electrode is a common reference electrode. What is the standard potential of a cell made up of a Cu2+solution being reduced to Cu(s) and AgCl(s) being reduced to Ag(s)? E (Cu2++ 2e- Cu(s)) = 0.34 V E (AgCl(s) + e- Ag(s) + Cl-(aq)) = 0.22 V What is the balanced reaction and what species must be present at 1 M?
Chapter 18 Electrochemistry Oxidizing/Reducing Agents Compounds with large positive or negative E (standard reduction) values are frequently used in electrochemistry (or in redox titrations) Example: MnO4-- E (MnO4-(aq) + 8H+(aq) + 5e-) = 1.51 V is frequently used in redox titrations Why? Because if E is high, it strongly reduces, which makes it useful for oxidizing a wide variety of compounds (e.g. Cu(s)) Such a compound is called an oxidizing agent (oxidizes other compounds)
Chapter 18 Electrochemistry Oxidizing/Reducing Agents cont. Products of reduction reactions with large negative E values (e.g. Li(s), K(s)) are easily oxidized and can therefore reduce other compounds Example: Al(s) - E (Al3+(aq) + 3e-) = -1.66 V is capable of reducing transition metals (reaction with iron oxide is in thermite reaction)
Chapter 18 Electrochemistry Reduction Potential and Oxidation of Metals by Acids Just as we can see which metals will oxidize or reduce when pairing two metals (Ag/Cu example), we also can see which metals will react in acid to produce H2(g) Metals with E (standard reduction) < 0 will react with H+ Examples: Fe, Pb, Sn, Ni, Cr, Zn, Al Metals with E (standard reduction) > 0 will not react with acid (except with HNO3which is a stronger oxidizing agent) Cu, Ag, Au, Hg
Chapter 18 Electrochemistry Reducing Potential Questions Given the table below, which of the following oxidizing agents is strong enough to oxidize Ag(s) to Ag+(aq) (under standard conditions)? a) H+(aq) b) Co2+(aq) c) Cu2+(aq) d) Co3+(aq) e) Br2(l) Reaction E (V) Ag+(aq) + e- Ag(s) +0.799 Co2+(aq) + 2e- Co(s) -0.277 Cu2+(aq) + 2e- Cu(s) +0.337 Co3+(aq) + e- Co2+(aq) +1.808 Br2(l) + 2e- 2Br-(aq) +1.065
Chapter 18 Electrochemistry Relating Standard Potential to Free Energy Two measures of the usefulness of a battery (voltaic cell) are: potential (voltage) supplied and charge stored The combination of these two give the energy stored for electrical work Stored charge (allows one to calculate lifetime under given current load) = q = nF where n = moles of e-(involved in balanced chemical equation) and F = Faraday s constant F = charge of a mole of electrons = NA*qelectron= (6.02 x 1023 e s/mol e)(1.60 x 10-19C/e) = 96,485 C/mol e
Chapter 18 Electrochemistry Relating Standard Potential to Free Energy cont. Since we know that potential is energy per charge, we can also determine that energy (in terms of maximum electrical work) = -qEcell (negative sign reflects that work is negative if done by the system on the surroundings) So, wmax= -nFEcell In terms of potential energy, wmax= G Thus G = -nFEcell
Chapter 18 Electrochemistry Example problems: A NiCad battery contains 12.0 g of Cd that is oxidized to Cd(OH)2. How long should the battery last if a motor is drawing 0.421 A? Assume 100% efficiency. (Hint: 1 A = 1 C s-1) Calculate the G (in kJ/mol) for the following reaction: 2Ag+(aq) + Cu(s) 2Ag(s) + Cu2+(aq), based on E values given for half reactions on slide 8.
Chapter 18 Electrochemistry Relating Potential to Non-Standard Conditions: The Nernst Equation Starting from our equation for non-standard Equilibrium: Grxn= G rxn+ RTlnQ, we can convert this to a potential equation: -nFEcell= -nFE cell+ RTlnQ or if we divide both sides by nF: Ecell= E cell- (RT/nF)lnQ Ecell= E cell- (0.0592/n)logQ (Nernst Equation valid for T = 25 C)
Chapter 18 Electrochemistry Nernst Equation Application Example: Determine the voltage for a Ag(s)/AgCl(s) electrode when [Cl-] = 0.010 M if E = 0.222 V (at T = 25 C)? Note: this is the same as when this electrode is attached to a SHE.
Chapter 18 Electrochemistry Nernst Equation Application Cont. 2ndExamples: The following cell, Cd(s)|CdC2O4(s)|C2O42-(aq)||Cu2+(aq)1 M|Cu(s) is used to determine [C2O42-]. If E for the Cd reaction is -0.522 V (reduction potential for oxidation reaction) and E for the Cu reaction is +0.337 V, and the measured voltage is 0.647 V, what is [C2O42-]?