Understanding Electrochemistry: Redox Reactions and Cells

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 90-103 80s 70s 60s 50s <50 # Students 7 14 14 27 31 40 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 90-103 80s 70s 60s 50s <50 # Students 1 14 32 50 22 16 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 O2) 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. 2H2O(l) occur Also requires two electrodes but some differences from electrodes of voltaic cells Examples: Production of Cl2 gas from NaCl(aq), production of H2 gas from water (above), instruments that measure degree of oxidation/reduction at specific voltages (analogous to spectrometers) 2H2(g) + O2(g)) to

Chapter 18 Electrochemistry VOLTAIC CELL 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 voltmeter Ag+ + e- Ag(s) Zn(s) Ag(s) + AgNO3(aq) ZnSO4(aq) Zn(s) Zn2+ + 2e- Salt Bridge

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 10-19 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) If [Ag+] and [Zn2+] = 1 M, Ecell = 1.56 V Zn2+(aq) + 2Ag(s)

Chapter 18 Electrochemistry Voltaic Cells Cell notation Example Cell: Zn(s)|Zn2+(aq)||Ag+ (aq)|Ag(s) Voltaic CELL voltmeter Zn(s) Ag(s) | means phase boundary left side for anode (right side for cathode) || means salt bridge AgNO3(aq) ZnSO4(aq) Salt Bridge

Chapter 18 Electrochemistry Example Questions Given the following cell, answer the following question: MnO2(s)|Mn2+(aq)||Cr3+(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 GALVANIC CELL Given the following cell, write the cell notation: voltmeter reads +0.43 V Pt(s) Ag(s) + Note: In this case the Pt(s) is an inert electrode (provides electrons but doesn t react AgCl(s) NaCl(aq) FeSO4 (aq), Fe2(SO4)3(aq) Salt Bridge

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- Eanode = 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) Ecell = Ecathode The SHE is not actually used much any more (just a reference for relative potential) Pt(s) Ag(s) H2(g)) H2(g) AgNO3(aq) H+(aq)

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), MnO4-, PbO2(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 Ecell ? Example: Zn(s)|Zn2+(aq)||Ag+ (aq)|Ag(s) Ecell = ? Ag+ reduction E = +0.80 V E = 0 Zn2+ reduction E = -0.76 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 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 HNO3 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) 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