Chem. 1B – 11/1 Lecture

Announcements I Exam #2 - Results –Average = 59.4 –Worst average so far for any Chem 1B exam here –Fraction of students better than 90 was reasonable, but many, many students under 50 –Many questions like quiz or last year’s exam questions Score Range # Students s14 70s14 60s27 50s31 <5040 Solutions for B version posted – will post equivalent C version question and % correct

Announcements II Post Exam 2 Grades –Very few high scores even if average is better than exam 2 –Score is with 380 to 440 points (~40%) –Cut-offs may change slightly, but too early to define % Range # Students s14 70s32 60s50 50s22 <5016 Students in 60s have a chance to improve (with more effort)

Announcements III Today’s Lecture –Electrochemistry (Ch. 18 – Exam 3 material) Redox Reactions – various formats Voltaic (or Galvanic) Cells Definitions Standard Half-Cells and Cells Standard Reduction Potential Standard Cell Potentials

Chapter 18 Electrochemistry Electrochemical Reactions – Different Forms –“Beaker” Reactions Products form along with heat (assuming  H < 0) Little control of reaction Products co-mingled (from reduction and oxidation) Example: nail “rusts” (oxidation of Fe, reduction of O 2 ) –Voltaic (Galvanic) Cells Oxidation and reduction reactions may be divided into different parts (half-cells sometimes physically separated through two reaction cells) Two electrodes are also needed Reaction can be “harnessed” through voltage/power production Examples: batteries, pH measuring electrodes

Chapter 18 Electrochemistry Electrochemical Reactions – Different Forms –Electrolytic Cell In this type of cell, external electrical energy is used to force unfavorable reactions (e.g. 2H 2 O(l) ↔ 2H 2 (g) + O 2 (g)) to occur Also requires two electrodes – but some differences from electrodes of voltaic cells Examples: Production of Cl 2 gas from NaCl(aq), production of H 2 gas from water (above), instruments that measure degree of oxidation/reduction at specific voltages (analogous to spectrometers)

Chapter 18 Electrochemistry Voltaic Cells - Description of how example cell works –Reaction on anode = oxidation –Anode = Zn electrode (as Zn has a greater tendency to oxidize than Ag) –So, reaction on cathode must be reduction and involve Ag –Oxidation produces e -, so anode has ( – ) charge (voltaic cells only); current runs from cathode to anode –Salt bridge allows replenishment of ions as cations migrate to cathode and anions toward anodes Salt Bridge voltmeter Zn(s) ZnSO 4 (aq) Ag(s) AgNO 3 (aq) VOLTAIC CELL Zn(s) → Zn e - Ag + + e - → Ag(s) – +

Chapter 18 Electrochemistry Basic Electrical Quantities –Current: the flow of electrons (although defined where a positive current has electrons moving backwards) –Current units: Amperes (A) with 1 A = 1 C/s and 1 C = 1 Coulomb where 1 electron (elementary charge) has a value of 1.60 x C –Potential or Voltage: The potential energy associated with the movement of charge (e.g. to electrode of opposite sign) –Potential units: Volts (V) = 1 J/C

Chapter 18 Electrochemistry Basic Electrical Quantities – From Voltaic Cells –Current: related to the flow of electrons –Potential: related to the reaction occurring (more energetic means higher potential) –The ability of a metal (or other elements) to reduce can be measured under standard conditions –Example: Zn(s) + 2Ag + (aq) ↔ Zn 2+ (aq) + 2Ag(s) If [Ag + ] and [Zn 2+ ] = 1 M, E cell º = 1.56 V

Chapter 18 Electrochemistry Voltaic Cells Cell notation –Example Cell: Zn(s)|Zn 2+ (aq)||Ag + (aq)|Ag(s) Salt Bridge voltmeter Zn(s) ZnSO 4 (aq) Ag(s) AgNO 3 (aq) Voltaic CELL left side for anode (right side for cathode) “|” means phase boundary “||” means salt bridge

Chapter 18 Electrochemistry Example Questions –Given the following cell, answer the following question: MnO 2 (s)|Mn 2+ (aq)||Cr 3+ (aq)|Cr(s) –What compound is used for the anode? –What compound is used for the cathode? –Write out both half-cell reactions and a net reaction

Chapter 18 Electrochemistry Given the following cell, write the cell notation: Salt Bridge voltmeter – reads V Pt(s) FeSO 4 (aq), Fe 2 (SO 4 ) 3 (aq) Ag(s) NaCl(aq) GALVANIC CELL AgCl(s) + – Note: In this case the Pt(s) is an “inert” electrode (provides electrons but doesn’t react

Chapter 18 Electrochemistry Standard Reduction Potential A cell used to determine the standard reduction potential consists of two half cells One half-cell, the anode, is the standard hydrogen electrode (2H + (aq) + 2e - ↔ H 2 (g)) E anode º = 0 (defined) Other is the test cell (compound being reduced when half-cell is coupled to standard hydrogen electrode (oxidation electrode) Both cells under standard conditions (1 M, 1 atm) E cell º = E cathode º The SHE is not actually used much any more (just a reference for relative potential) Ag(s) AgNO 3 (aq) Pt(s) H + (aq) H 2 (g)

Chapter 18 Electrochemistry Standard Reduction Potential Meaning of Values –Half-cells that exhibit positive values have electrodes with compounds that easily reduce (e.g. Ag + (aq), MnO 4 -, PbO 2 (s)) –Half-cells that exhibit negative values have electrodes that easily oxidize (e.g. alkali metals) –What if we have two half-cells (neither SHE), can we find E cell º? Example: Zn(s)|Zn 2+ (aq)||Ag + (aq)|Ag(s) E cell º = ? E º = 0 Ag + reduction E º = V Zn 2+ reduction E º = V

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 e - ↔ 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: MnO 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)

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 - ) = 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 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)

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) Co 2+ (aq) + 2e - ↔ Co(s) Cu 2+ (aq) + 2e - ↔ Cu(s) Co 3+ (aq) + e - ↔ Co 2+ (aq) Br 2 (l) + 2e - ↔ 2Br - (aq)+1.065