Chapter 20 Electrochemistry

Overview Oxidation-Reduction reactions Voltaic Cells Balancing Redox Reactions Half-Reaction method Acidic Solution Basic Solution Voltaic Cells Cell EMF--standard reduction potentials Oxidizing & Reducing reagents Spontaneity of Redox reactions

Effect of Concentration Nernst Equation Equilibrium Constants Commercial Voltaic Cells Electrolysis Quantitative Aspects Electrical Work

Redox Reactions Involve a transfer of electrons Oxidation  loss of one or more electron(s) oxidation state will increase Reduction  gain of one or more electron(s) oxidation state will decrease Must occur simultaneously Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s) Zn  Zn2+(aq) + 2e- oxidation oxidation ½ rxn Cu2+(aq) + 2e-  Cu(s) reduction ½ rxn reduction

You must know oxidation states: (Review: Section 8.10) What are the oxidation states of each atom in the following: H2 CO ClO2- HC2H3O2 H 0 C +2, O -2 Cl +3, O -2 H +1, C1 +3, C2 -3, O -2

Balancing Redox Reactions Mass balance must be observed e--transfer must be balanced Simple reactions: Sn2+ + Fe3+  Sn4+ + Fe2+ Sn2+  Sn4+ + 2e- Fe3+ + e-  Fe2+ oxidation ½ rxn x 2 reduction ½ rxn 2Fe3+ + 2e-  2Fe2+ Sn2+ + 2Fe3+  Sn4+ + 2Fe2+

Reactions involving H & O in acid: MnO4- + C2O42-  Mn2+ + CO2 write both ½ reactions MnO4-  Mn2+ C2O42-  CO2 mass balance (all except H & O) C2O42-  2CO2 add H2O & H+ to balance O & H 8H+ + MnO4-  Mn2+ + 4H2O

check the balance balance charge by adding electrons 5e- + 8H+ + MnO4-  Mn2+ + 4H2O C2O42-  2CO2 + 2e- balance electrons transferred 10e- + 16H+ + 2MnO4-  2Mn2+ + 8H2O 5C2O42-  10CO2 + 10e- add half reactions 16H+ + 2MnO4-+ 5C2O42-  10CO2 + 2Mn2+ + 8H2O check the balance

check balance Reactions in base: MnO4- + CN-  CNO- + MnO2 use exactly the same process CN-  CNO- MnO4-  MnO2 H2O + + 2H+ + 2e- 3e- + 4H+ + + 2H2O since H+ cannot exist in basic solution, add OH- 2OH- + CN-  CNO- + H2O + 2e- 3e- + 2H2O + MnO4-  MnO2 + 4OH- balance electrons transferred & sum 6OH- + 3CN-  3CNO- + 3H2O + 6e- 6e- + 4H2O + 2MnO4-  2MnO2 + 8OH- 3CN- + H2O + 2MnO4-  2MnO2 + 3CNO- +2OH- check balance

Voltaic Cells A spontaneous redox reaction that does work Anode electrode at which oxidation occurs loses mass electrons released, sign is negative Cathode electrode at which reduction occurs gains mass electrons consumed, sign is positive

Cell EMF Difference in potential energy of electrons at the anode and cathode Diff. in potential energy per electrical charge measured in volts 1 V = 1 J C Potential difference = EMF, electromotive force Ecell = cell potential = cell voltage Eºcell = cell potential under std. conditions 1 M, 1 atm, 25 ºC

Standard reduction potentials E ºred in tables E ºcell = E ºred (cathode) - E ºred (anode) Based on “standard hydrogen electrode” 2H+(aq, 1M) + 2e-  H2(g, 1atm) E ºred = 0 V Zn(s) + 2H+(aq)  Zn2+(aq) + H2(g) E ºcell = 0.76 V 0.76 V = 0 V - E ºred (anode) Zn2+(aq, 1M) + 2e-  Zn(s) E ºred (anode) = -0.76 V

Problem: Calculate Eºcell for Anode: 2Al  2Al3+ + 6e- 2Al(s) + 3I2(s)  2Al3+(aq) + 6I-(aq) Anode: 2Al  2Al3+ + 6e- Cathode: 3I2 + 6e-  6I- Eºcell = E ºred (cathode) - E ºred (anode) E ºcell = 0.54 V - (-1.66 V) E ºcell = 2.20 V

The more positive the E ºcell the more driving force for the reaction Note: stoichiometric coefficient does not affect the value of the E ºred (it is an intensive property) E ºox = - E ºred 2Al(s) + 3I2(s)  2Al3+(aq) + 6I-(aq) 2Al  2Al3+ + 6e- E ºox = +1.66 V 3I2 + 6e-  6I- E ºred = +0.54 V E ºcell = E ºox + E ºred = 2.20V The more positive the E ºcell the more driving force for the reaction

Oxidizing/Reducing Agents Oxidizing agents cause oxidation oxidizing agents are reduced the more (+) the E ºred the better the ox. agent Reducing agents cause reduction reducing agents are oxidized the more (-) the E ºred the better the red. agent

Which is the better oxidizing agent? NO3- + 4H+ + 3e-  NO + 2H2O E ºred 0.96 V Ag+ + e-  Ag E ºred 0.80 V Cr2O72- + 14H+ + 6e-  2Cr3+ + H2O E ºred 1.33 V Which is the strongest reducing agent? I2 + 2e-  2I- Eºred +0.54 V Fe2+ + 2e-  Fe Eºred -0.44 V MnO4- + 8H+ + 5e-  Mn2+ + 4H2O Eºred +1.51 V

Spontaneity of Redox Reactions Spontaneous redox rxns have positive potentials Non-spontaneous redox rxns have negative potentials Is this rxn spont. or non-spont.? MnO4- + 8H+ + 5Fe2+  5Fe3+ + Mn2+ + 4H2O Fe2+  Fe3+ + 1e- Eºox = -0.77 v MnO4- + 8H+ + 5e-  Mn2+ + 4H2O E ºred = +1.51 v E ºox + E ºred = + 0.74 v Yes

EMF & Free Energy If both DG & E are a measure of spontaneity, they must be related DG = - nFE F is Faraday’s constant 1 F = 96,500 J/v mol e- remember: 1 C = 1 J/v n = mol e- transferred In the standard state DGº = - nFEº

DG = - (2 mol e-)(-0.083 v)(96,500 J/v mol e-) = + 16 kJ Calculate the standard free energy change for Hg + 2Fe3+  Hg2+ + 2Fe2+ n = 2 mol electrons transferred Hg  Hg2+ + 2e- Eox = - 0.854 v 2Fe3+ +2e-  2Fe2+ Ered= + 0.771 v Ecell = - 0.083 v DG = - (2 mol e-)(-0.083 v)(96,500 J/v mol e-) = + 16 kJ

Concentration & Cell EMF Nernst Equation relationship between DG & concentrations DG = DGº + RT ln Q Q = [prod]x/[react]y substitute -nFE for DG E = Eº - (RT/nF) ln Q or E = Eº - (2.303 RT/nF) log Q 2.303 RT/F = 0.0592 v-mol e- at std. temp. E = Eº - (0.0592/n) log Q

Calculate the emf that the following cell generates when [Mn2+] = 0 Calculate the emf that the following cell generates when [Mn2+] = 0.10 M & [Al3+] = 1.5 M 2Al + 3Mn2+  2Al3+ + 3Mn Eº = + 0.48 v E = (+ 0.48 v) - (0.0592 v/ 6) log [(1.5)2/(0.10)3] E = + 0.45 v when [Mn2+] = 1.5 M & [Al3+] = 0.10 M E = (+ 0.48 v) - (0.0592 v/ 6) log [(0.10)2/(1.5)3] E = + 0.51 v

Equilibrium Constants Remember DG = DGº + RT ln Q, if Q = K, then DG = 0, therefore -nFE = 0 and 0 = Eº - (RT/nF) ln K or 0 = Eº - (0.0592/n) log K K can be calculated from cell potentials log K = nE º/0.0592

Calculate the equilibrium constant, K, for 2IO3- + 5Cu + 12H+  I2 + 5Cu2+ + 6H2O Eº = + 0.858 v n = 10 mol e- transferred log K = nEº/0.0592 log K = 145 K = 1 x 10145

Voltaic Cells Lead storage battery Dry cell PbO2 + SO4-2 + 4H+ + 2e-  PbSO4 + H2O Pb + SO42-  PbSO4 + 2e- Ecell = + 2.041 v Dry cell NH4+ + 2MnO2 + 2e-  Mn2O3 + 2NH3 + H2O Zn  Zn2+ + 2e- In an alkaline cell the NH4Cl is replaced with KOH

Ni-Cd NiO2 + 2H2O + 2e-  Ni(OH)2 + 2OH- Cd + 2OH-  Cd(OH)2 + 2e- Fuel cells 4e- + O2 + 2H2O  4OH- 2H2 + 4OH-  4H2O

Electrolytic Cells Redox reactions that are not spontaneous Must be driven by an outside source of electrical energy Cathode reduction occurs by sign convention, is negative Anode oxidation occurs by sign convention, is positive

Quantitative Aspects Redox reactions occur in stoichiometric relationship to the transfer of electrons Electrons put into a system through electrical energy, can be quantized Coulomb = quantity of charge passing through electrical circuit in 1 s at 1 ampere (A) current Coulomb = (amp) (seconds)

Problem: Calculate the mass of Mg formed upon passage of a current of 60.0 A for a period of 4.00 x 10 3 s. MgCl2  Mg + Cl2 Mg2+ + 2e-  Mg 2Cl-  Cl2 + 2e- we are concerned with the reduction (60.0 A)(4 x 103s)(1C/1 A-s) = 2.4 x 105 C (2.4 x 105 C)(1 mol e-/ 96,500 C) = 2.49 mol e- (2.49 mol e-)(1 mol Mg/2 mol e-) = 1.24 mol Mg (1.24 mol Mg)(24.3 g/mol) = 30.1 Mg

Electrical Work DG = wmax DG = - nFE wmax = - nFE Max work proportional to potential wmax = - n F E J = (mol) (C/mol) (J/C) Electrical work = (watt) (time) 1 watt (W) = 1 J/s or watt-s = J 1 kWh = 3.6 x 106 J