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Chem. 1B – 11/5 Lecture. Announcements Mastering Chemistry –Chapter 18 Assignment is due 11/17 Today’s Lecture – Electrochemistry (Ch. 18) –Standard Reduction.

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Presentation on theme: "Chem. 1B – 11/5 Lecture. Announcements Mastering Chemistry –Chapter 18 Assignment is due 11/17 Today’s Lecture – Electrochemistry (Ch. 18) –Standard Reduction."— Presentation transcript:

1 Chem. 1B – 11/5 Lecture

2 Announcements Mastering Chemistry –Chapter 18 Assignment is due 11/17 Today’s Lecture – Electrochemistry (Ch. 18) –Standard Reduction Potential – for half-cell and full cell reactions –Potential Scale and Oxidizing and Reducing Agents –Relating Standard Potential to Free Energy

3 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 Cu 2+ solution being reduced to Cu(s) and AgCl(s) being reduced to Ag(s)? E°(Cu 2+ + 2e - ↔ Cu(s)) = 0.34 V E°(AgCl(s) + e - ↔ Ag(s) + Cl - (aq)) = 0.22 V What compound is reduced? What compound is oxidized? For the reaction to be spontaneous, E cell ° > 0 What is the balanced reaction and what species must be present at 1 M?

4 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: MnO 4 - - E° (MnO 4 - (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)

5 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° (Al 3+ (aq) + 3e - ) = -1.66 V is capable of reducing transition metals (reaction with iron oxide is in thermite reaction)

6 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 H 2 (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 HNO 3 which is a stronger oxidizing agent)

7 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) Co 2+ (aq) c) Cu 2+ (aq) d) Co 3+ (aq) e) Br 2 (l) ReactionEº (V) Ag + (aq) + e - ↔ Ag(s)+0.799 Co 2+ (aq) + 2e - ↔ Co(s)-0.277 Cu 2+ (aq) + 2e - ↔ Cu(s)+0.337 Co 3+ (aq) + e - ↔ Co 2+ (aq)+1.808 Br 2 (l) + 2e - ↔ 2Br - (aq)+1.065

8 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 = N A *q electron = (6.02 x 10 23 e’s/mol e)(1.60 x 10 -19 C/e) = 96,485 C/mol e

9 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) = -qE cell ° (negative sign reflects that work is negative if done by the system on the surroundings) –So, w max = -nFE cell ° –In terms of potential energy, w max =  G° –Thus  G° = -nFE cell °

10 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) + Cu 2+ (aq), based on E° values given for half reactions on slide 7.

11 Chapter 18 Electrochemistry Relating Potential to Non-Standard Conditions: The Nernst Equation –Starting from our equation for non-standard Equilibrium:  G rxn =  Gº rxn + RTlnQ, we can convert this to a potential equation: -nFE cell = -nFEº cell + RTlnQ or if we divide both sides by –nF: E cell = Eº cell - (RT/nF)lnQ E cell = Eº cell - (0.0592/n)logQ (Nernst Equation – valid for T = 25ºC)

12 Chapter 18 Electrochemistry Nernst Equation Application Example: Determine the voltage for a Ag(s)/AgCl(s) electrode when [Cl - ] = 0.010 M if Eº = 0.222 V (at T = 25°C)? Note: this is the same as when this electrode is attached to a SHE.

13 Chapter 18 Electrochemistry Nernst Equation Application – Cont. 2 nd Examples: –The following cell, Cd(s)|CdC 2 O 4 (s)|C 2 O 4 2- (aq)||Cu 2+ (aq)1 M|Cu(s) is used to determine [C 2 O 4 2- ]. If Eº for the Cd reaction is -0.522 V (reduction potential for oxidation reaction) and Eº for the Cu reaction is +0.337 V, and the measured voltage is 0.647 V, what is [C 2 O 4 2- ]?


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