Galvanic Cells Electrochemistry: The area of chemistry concerned with the interconversion of chemical and electrical energy Galvanic (Voltaic) Cell: A spontaneous chemical reaction that generates an electric current Electrolytic Cell: An electric current that drives a nonspontaneous reaction 15-אלקטרוכימיה

Galvanic Cells Zn2+(aq) + Cu(s) Zn(s) + Cu2+(aq) Oxidation half-reaction: Zn2+(aq) + 2 e Zn(s) Reduction half-reaction: Cu(s) Cu2+(aq) + 2 e 15-אלקטרוכימיה

Zn-Cu2+ Spontaneous Redox Reaction 15-אלקטרוכימיה

Cu2+(aq) + Zn(s) → Cu(s) + Zn2+(aq) Voltaic Cells Cu2+(aq) + Zn(s) → Cu(s) + Zn2+(aq) 15-אלקטרוכימיה

Zn-Cu2+ Voltaic Cell 15-אלקטרוכימיה

Voltaic Cells Cu2+(aq) + Zn(s) → Cu(s) + Zn2+(aq) Anode Cathode salt bridge Electrodes (anode & cathode) Allow e- to pass in/out of solution. A salt bridge (or porous barrier) is required... e- move across an external conductor. 15-אלקטרוכימיה

Galvanic Cells Anode: The electrode where oxidation occurs The electrode where electrons are produced Is what anions migrate toward Has a negative sign 15-אלקטרוכימיה

Galvanic Cells Cathode: The electrode where reduction occurs The electrode where electrons are consumed Is what cations migrate toward Has a positive sign 15-אלקטרוכימיה

Voltaic Cells Zinc is removed: Zn(s) → Zn2+(aq) + 2 e- Oxidation at anode (vowels!) The anode loses mass Zn supplies e-. Anode has “-” charge. Copper is deposited: Cu2+(aq) + 2 e- → Cu(s) Reduction at cathode (consonants!) The cathode gains mass Cu2+ accepts e-. Cathode has “+” charge. 15-אלקטרוכימיה

Salt Bridge It is an inverted glass U-tube with glass wool plugged at each end Contains a salt paste (e.g. K2SO4,KNO3 ). A porous plug restricts bulk flow Ions pass into/out of the cells Stops charge buildup. K2SO4 SO42- K+ Zn Cu Zn2+ Cu2+ porous plug SO42- Zn2+ SO42- released Zn2+ removed as Zn → Zn2+ 2 K+ released SO42- removed as Cu2+ → Cu 15-אלקטרוכימיה

Voltaic Cells In all voltaic cells: The redox reaction must be product-favored. Oxidation occurs at the anode. Reduction occurs at the cathode. Electrons move from anode to cathode through an external ‘wire’. The electrical circuit is completed by movement of ions into and out of a salt bridge. 15-אלקטרוכימיה

Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s) Shorthand Notation for Galvanic Cells Anode half-reaction: Zn(s) Zn2+(aq) + 2 e Cathode half-reaction: Cu2+(aq) + 2 e Cu(s) Overall cell reaction: Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) Salt bridge Anode half-cell Cathode half-cell Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s) Electron flow Phase boundary Phase boundary 15-אלקטרוכימיה

Show the Cell Notation for the Following Voltaic Cell 15-אלקטרוכימיה

Electrochemical Cell Notations Pt|H2 (P=1 bar)|H+ (aq,1.0M) ||Fe+3 (aq,1.0M) ,Fe2+(aq,1.0M) |Pt Electrodes Typically The anode is made of the metal that is oxidized. The cathode is made of the same metal as is produced by the reduction. If the redox reaction involves the oxidation or reduction of an ion to a different oxidation state, or the oxidation or reduction of a gas, we may use an inert electrode. An inert electrode is one that not does participate in the reaction but just provides a surface on which the transfer of electrons can take place. 15-אלקטרוכימיה

Cell Potentials and Free-Energy Changes for Cell Reactions Electromotive Force (emf): The force or electrical potential that pushes the negatively charged electrons away from the anode ( electrode) and pulls them toward the cathode (+ electrode) It is also called the cell potential (E) or the cell voltage. 15-אלקטרוכימיה

SI unit of electric potential Cell Potentials and Free-Energy Changes for Cell Reactions 1 J = 1 C × 1 V joule (J) SI unit of energy volt (V) SI unit of electric potential coulomb (C) Electric charge 1 coulomb is the amount of charge transferred when a current of 1 ampere (A) flows for 1 second. 15-אלקטרוכימיה

Cell Potentials and Free-Energy Changes for Cell Reactions faraday or Faraday constant The electric charge on 1 mol of electrons and is equal to 96,500 C/mol e DG = nFE or DG° = nFE° Free-energy change Cell potential Number of moles of electrons transferred in the reaction 15-אלקטרוכימיה

Cell Potentials and Free-Energy Changes for Cell Reactions The standard cell potential at 25 °C is 1.10 V for the reaction: Zn2+(aq) + Cu(s) Zn(s) + Cu2+(aq) Calculate the standard free-energy change for this reaction at 25 °C. DG° = nFE° mol e 96,500 C 1 C V 1 J 1000 J 1 kJ = (2 mol e) (1.10 V) DG° = 212 kJ 15-אלקטרוכימיה

Standard Reduction Potentials Anode half-reaction: H2(g) 2 H+(aq) + 2 e Cathode half-reaction: Cu2+(aq) + 2 e Cu(s) Overall cell reaction: H2(g) + Cu2+(aq) 2 H1+(aq) + Cu(s) The standard potential of a cell is the sum of the standard half-cell potentials for oxidation at the anode and reduction at the cathode: E°cell = E°ox + E°red The measured potential for this cell: E°cell = 0.34 V 15-אלקטרוכימיה

Standard Reduction Potentials The standard hydrogen electrode (S.H.E.) has been chosen to be the reference electrode. 15-אלקטרוכימיה

Standard Reduction Potentials The standard hydrogen electrode (S.H.E.) has been chosen to be the reference electrode. H2(g, 1 atm) 2H+(aq, 1 M) + 2e E°ox = 0 V E°red = 0 V 2H+(aq, 1 M) + 2e H2(g, 1 atm) Gas + soln. in contact with Pt Compact notation: H+(aq), 1M| H2(g, 1bar) | Pt The Standard Hydrogen Electrode (SHE) is assigned a value of E° = 0 V: 15-אלקטרוכימיה

Standard Reduction Potentials Anode half-reaction: H2(g) 2 H+(aq) + 2 e Cathode half-reaction: Cu2+(aq) + 2 e Cu(s) Overall cell reaction: H2(g) + Cu2+(aq) 2 H+(aq) + Cu(s) E°cell = E°ox + E°red 0.34 V = 0 V + E°red A standard reduction potential can be defined: Cu(s) Cu2+(aq) + 2 e E° = 0.34 V 15-אלקטרוכימיה

Standard Reduction Potentials 15-אלקטרוכימיה

As a standard reduction potential: Standard Reduction Potentials Anode half-reaction: Zn(s) Zn2+(aq) + 2 e Cathode half-reaction: 2 H+(aq) + 2 e H2(g) Overall cell reaction: Zn(s) + 2 H+(aq) Zn2+(aq) + H2(g) E°cell = E°ox + E°red 0.76 V = E°ox + 0 V Zn2+(aq) + 2 e Zn(s) E° = 0.76 V As a standard reduction potential: Zn(s) Zn2+(aq) + 2 e E° = 0.76 V 15-אלקטרוכימיה

Standard Reduction Potentials 15-אלקטרוכימיה

(If E°cell < 0 the reaction is reactant favored). Cell Potentials & Half Cells A standard potential (E°) occurs when: All [solute] = 1 M (or saturated if solubility < 1 M). All gases have P = 1 bar. All solids are pure. If E°cell is positive the reaction is product favored (If E°cell < 0 the reaction is reactant favored). E° does not depend on nreactant or nproduct (i.e not on the stoichiometric coefficients). 15-אלקטרוכימיה

Using Standard Half-Cell Potentials Potential differences between SHE many other half-cells have been made. Standard reductions are listed. Each half-reaction can occur in either direction More positive E° = easier reduction. Less positive E° = easier oxidation for the reverse reaction. 15-אלקטרוכימיה

Cell Potentials Consider the following table: Identify: Half Reaction Half Cell E° (V) Au3+(aq) + 3 e- → Au(s) Au3+(aq)|Au(s) +1.52 Cl2(g) + 2 e- 2 Cl-(aq) Cl2(g)|Cl−(aq)|Pt +1.358 Al3+(aq) + 3 e- Al(s) Al3+(aq)|Al(s) -1.676 K+(aq) + e- K(s) K+(aq)|K(s) -2.925 Identify: the strongest oxidizing agent. the weakest oxidizing agent. the strongest reducing agent. the weakest reducing agent. = Au3+(aq) = K+(aq) = K(s) = Au(s) 15-אלקטרוכימיה

Cell Potentials Consider the following table: Will: Half Reaction Half Cell E° (V) Au3+(aq) + 3 e- → Au(s) Au3+(aq)|Au(s) +1.52 Cl2(g) + 2 e- 2 Cl-(aq) Cl2(g)|Cl−(aq)|Pt +1.358 Al3+(aq) + 3 e- Al(s) Al3+(aq)|Al(s) -1.676 K+(aq) + e- K(s) K+(aq)|K(s) -2.925 Will: Cl2(g) oxidize Au(s) to Au3+(aq)? K(s) reduce Cl2(g) to Cl-(aq)? and c) What can be reduced by Al(s)? = No = Yes = Cl2 and Au3+ 15-אלקטרוכימיה

Using Standard Reduction Potentials Zn2+(aq) + 2 e Zn(s) E° = (0.76 V) Cu(s) Cu2+(aq) + 2 e E° = 0.34 V E° = 1.10 V Zn2+(aq) + Cu(s) Zn(s) + Cu2+(aq) Ag(s)] 2 × [Ag+(aq) + e E° = 0.80 V Cu2+(aq) + 2 e Cu(s) E° = 0.34 V 2 Ag(s) + Cu2+(aq) 2 Ag+(aq) + Cu(s) E° = 0.46 V Half-cell potentials are intensive properties. 15-אלקטרוכימיה

The Nernst Equation DG = DG° + RT ln Q DG = nFE and DG° = nFE° Using: DG = nFE and DG° = nFE° nF RT Nernst Equation: E = E°  ln Q or nF 2.303RT E = E° log Q 0.0592 V E = E°  log Q in volts, at 25 °C n 15-אלקטרוכימיה

The Nernst Equation Consider a galvanic cell that uses the reaction: Cu(s) + 2 Fe3+(aq) Cu2+(aq) + 2 Fe2+(aq) What is the potential of a cell at 25 °C that has the following ion concentrations? [Fe3+] = 1.0 × 104 M [Cu2+] = 0.25 M [Fe2+] = 0.20 M 15-אלקטרוכימיה

The Nernst Equation 0.0592 V E = E°  log Q n Cu(s) + 2 Fe3+(aq) Cu2+(aq) + 2 Fe2+(aq) Calculate E°: Cu(s) Cu2+(aq) + 2 e E° = 0.34 V Fe3+(aq) + e Fe2+(aq) E° = 0.77 V E°cell = 0.34 V + 0.77 V = 0.43 V 15-אלקטרוכימיה

The Nernst Equation 0.0592 V E = E°  log Q n Cu(s) + 2 Fe3+(aq) Cu2+(aq) + 2 Fe2+(aq) Calculate E: 0.0592 V [Cu2+][Fe2+]2 E = E° – log n [Fe3+]2 0.0592 V (0.25)(0.20)2 = 0.43 V  log 2 (1.0 x 104)2 E = 0.25 V 15-אלקטרוכימיה

Standard Cell Potentials and Equilibrium Constants Using DG° = nFE° and DG° = RT ln K nFE° = RT ln K nF RT 2.303 RT E° = ln K = log K nF 0.0592 V E° = log K in volts, at 25 °C n 15-אלקטרוכימיה

Standard Cell Potentials and Equilibrium Constants 15-אלקטרוכימיה

Standard Cell Potentials and Equilibrium Constants Three methods to determine equilibrium constants: K = [A]a[B]b [C]c[D]d K from concentration data: RT DG° ln K = K from thermochemical data: nF RT nFE° K from electrochemical data: E° = ln K or ln K = RT 0.0592 V E° = log K in volts, at 25 °C n 15-אלקטרוכימיה

Example Consider the following reaction at 25°C: Find K 15-אלקטרוכימיה

Example Solution: Oxidation numbers Half-reactions: n: Multiply the oxidation half-reaction by 3 to cancel out the three electrons in the reduction half-reaction The electrons cancel out: n = 3 +e- 15-אלקטרוכימיה

Example 𝐾=10 𝑛 𝐸 ° 0.0592 =10 3𝑋0.165 0.0592 = 2.3 X108 0.0592 V E° = log K in volts, at 25 °C, n=3,E = 0.165V n 𝐾=10 𝑛 𝐸 ° 0.0592 =10 3𝑋0.165 0.0592 = 2.3 X108 15-אלקטרוכימיה

Concentration Cells Identical cells? E may not be zero… Example Zn|Zn2+(aq,0.01M)||Zn2+(aq,1M)|Zn Left-to-right… Zn(s) → Zn2+ (0.01M) + 2 e- Zn2+ (1M) + 2 e- → Zn(s) Zn2+(1M) → Zn2+(0.01M) (net reaction) E = E° − log E = 0 − 0.0296 log [Zn2+]dil [Zn2+]conc 0.010 1 0.0592 2 Same cell E = 0.0592 V 15-אלקטרוכימיה

Determination of pH Cell potential measurements can allow one to determine the pH of a solution by determining [H+]. According to the Nernst equation, cell potential depends on the H+ concentration of the test solution. To obtain the relationship between Ecell and pH, substitute the following into the Nernst equation: 15-אלקטרוכימיה

Determination of pH On the left, a small, commercial electrode is pictured. On the right is a sketch showing the construction of a glass electrode for measuring hydrogen concentrations. 15-אלקטרוכימיה

Batteries Lead Storage Battery Anode: Pb(s) + HSO4(aq) PbSO4(s) + H+(aq) + 2 e Cathode: PbO2(s) + 3 H+(aq) + HSO4(aq) + 2e PbSO4(s) + 2 H2O(l) Overall: Pb(s) + PbO2(s) + 2 H+(aq) + 2 HSO41(aq) 2 PbSO4(s) + 2 H2O(l) 15-אלקטרוכימיה

Batteries Dry-Cell Batteries Leclanché cell Anode: Zn(s) Zn2+(aq) + 2 e Cathode: 2 MnO2(s) + 2 NH4+(aq) + 2 e Mn2O3(s) + 2 NH3(aq)+ H2O(l) 15-אלקטרוכימיה

Corrosion Corrosion: The oxidative deterioration of a metal 15-אלקטרוכימיה

Corrosion For some metals, oxidation protects the metal (aluminum, chromium, magnesium, titanium, zinc, and others). For other metals, there are two main techniques. 15-אלקטרוכימיה

Prevention of Corrosion Galvanization: The coating of iron with zinc When some of the iron is oxidized (rust), the process is reversed since zinc will reduce Fe2+ to Fe: Fe(s) Fe2+(aq) + 2 e E° = 0.45 V Zn(s) Zn2+(aq) + 2 e E° = 0.76 V 15-אלקטרוכימיה

Prevention of Corrosion Cathodic Protection: Instead of coating the entire surface of the first metal with a second metal, the second metal is placed in electrical contact with the first metal: Anode: Mg(s) Mg2+(aq) + 2 e E° = 2.37 V Cathode: O2(g) + 4 H+(aq) + 4 e 2 H2O(l) E° = 1.23 V Attaching a magnesium stake to iron will corrode the magnesium instead of the iron. Magnesium acts as a sacrificial anode. 15-אלקטרוכימיה

Sacrificial Anode 15-אלקטרוכימיה

Electrolysis and Electrolytic Cells Electrolysis: The process of using an electric current to bring about chemical change 15-אלקטרוכימיה

Electrolysis of Molten Sodium Chloride Electrolysis and Electrolytic Cells Electrolysis of Molten Sodium Chloride Anode: 2 Cl(l) Cl2(g) + 2 e Cathode: 2 Na+(l) + 2 e 2 Na(l) Overall: 2 Na+(l) + 2 Cl(l) 2 Na(l) + Cl2(g) 15-אלקטרוכימיה

Electroplating 15-אלקטרוכימיה

Quantitative Aspects of Electrolysis Charge(C) = Current(A) × Time(s) 1 mol e Moles of e = Charge(C) × 96,500 C Faraday constant 15-אלקטרוכימיה

Counting Electrons What mass of gold will be electroplated from aqueous gold(III) chloride in 5.0 min by a 10.0 A current? Au3+(aq) + 3 e- → Au(s) Charge = (10.0 A)(5.0 min) = 3.0 x 103 C 60 s 1 min 1 mol e- 96,500 C 197.0 g 1 mol Au 3 mol e- 3.0 x 103 C = 2.0 g 15-אלקטרוכימיה

Counting Electrons How long will it take to produce 14 g of Al from Al3+(melt)? Assume a current of 500.0 A. Al3+ + 3 e- → Al 96,500 C 1 mol e- 1 mol Al 26.98 g 3 mol e- 14. g = 1.50 x 105 C charge current Time = = = 3.0 x 102 s 1.50 x 105 C 500.0 A 15-אלקטרוכימיה