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Electrochemistry

Electrochemistry. Chapter 17. Contents. Galvanic cells Standard reduction potentials Cell potential, electrical work, and free energy Dependence of cell potential on concentration Batteries Corrosion Electrolysis Commercial electrolytic processes. 17.1 Galvanic Cells.

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Electrochemistry

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  1. Electrochemistry Chapter 17

  2. Contents • Galvanic cells • Standard reduction potentials • Cell potential, electrical work, and free energy • Dependence of cell potential on concentration • Batteries • Corrosion • Electrolysis • Commercial electrolytic processes

  3. 17.1 Galvanic Cells Oxidation reduction reactions • Oxidation reduction reactions involve a transfer of electrons. • Oxidation Involves Loss of electrons • Increase in the oxidation number • Reduction Involves Gain electrons • Decrease in the oxidation number

  4. 8H++MnO4-+ 5Fe+2 ® Mn+2 + 5Fe+3 +4H2O Example • If we break the reactions into half reactions. 8H++MnO4-+5e-® Mn+2 +4H2O (Red) 5(Fe+2® Fe+3 + e- ) (Ox) • Electrons are transferred directly. • This process takes place without doing useful work

  5. When the compartments of the two beakers are connected as shown the reaction starts • Current flows for an instant then stops • No flow of electrons in the wire, Why? • Current stops immediately because charge builds up. H+ MnO4- Fe+2

  6. Galvanic Cell Solutions must be connected so ions can flow to keep the net charge in each compartment zero Salt Bridge allows ions to flow without extensive mixing in order to keep net charge zero. Electronsflow through the wire from reductant to oxidant Oxidant reductant H+ MnO4- Fe+2

  7. Porous Disk H+ MnO4- Fe+2

  8. e- e- e- e- Anode Cathode Fe2+ e- e- Reducing Agent Oxidizing Agent MnO4-

  9. _______ __________ _______ __________ Electrochemical Cells Spontaneous redox reaction 19.2

  10. Thus a Galvanic cell is a device in which a chemical energy is changed to electrical energy • The electrochemical reactions occur at the interface between electrode and solution where the electron transfer occurs • Anode: the electrode compartment at which oxidation occurs • Cathode: the electrode compartment at which reduction occurs

  11. Cell Potential • Oxidizing agent pulls the electrons • Reducing agent pushes the electrons • The total push or pull (“driving force”) is called the cell potential, Ecell • Also called the electromotive force (emf) • Unit is the volt(V) = 1 joule of work/coulomb of charge • Measured with a voltmeter

  12. Measuring the cell potential • Can we measure the total cell potential?? • A galvanic cell is made where one of the two electrodes is a reference electrode whose potential is known. • Standard hydrogen electrode (H+ = 1M and the H2 (g) is at 1 atm) is used as a reference electrode and its potential was assigned to be zero at 25 0C.

  13. Standard Hydrogen Electrode • This is the reference all other oxidations are compared to • Eº = 0 • (º) indicates standard states of 25ºC, 1 atm, 1 M solutions. 1 atm H2 H+ Cl- 1 M HCl

  14. 0.76 H2 Cathode Anode H+ Cl- Zn+2 SO4-2 1 M ZnSO4 1 M HCl

  15. 2e- + 2H+ (1 M) 2H2 (1 atm) Zn (s) Zn2+ (1 M) + 2e- Standard Electrode Potentials Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (1 atm) | Pt (s) Anode (oxidation): Cathode (reduction):

  16. 17.2 Standard Reduction Potentials, E • The E values corresponding to reduction half- reactions with all solutes at 1M and all gases at 1 atm. • E can be measured by making a galvanic cell in which one of the two electrodes is the Standard Hydrogen electrode, SHE, whose E = 0 V • The total potential of this cell can be measured experimentally • However, the individual electrode potential can not be measured experimentally. Why?

  17. If the cathode compartment of the cell is SHE, then the half reaction would be 2H+ + 2e H2 (g); Eo = 0V • And the anode compartment is Zn metal in Zn2+, (1 M) then the half reaction would be Zn  Zn2+ + 2e • The total cell potential measured experimentally was found to be + 0.76 V • Thus, +0.76 V was obtained as a result of this calculation: Eº cell = EºZn® Zn2++EºH+® H2 0.76 V 0.76 V 0 V

  18. Standard Reduction Potentials • The E values corresponding to reduction half- reactions with all solutes at 1M and all gases at 1 atm. can be determined by making them half cells where the other half is the SHE. • E0 values for all species were determined as reduction half potentials and tabulated. For example: • Cu2+ + 2e Cu E = 0.34 V • SO42 + 4H+ + 2e H2SO3 + H2O E = 0.20 V • Li+ + e- Li E = -3.05 V

  19. Some Standard Reduction Potentials Li+ + e- ---> Li -3.045 v Zn+2 + 2 e- ---> Zn -0.763v Fe+2 + 2 e- ---> Fe -0.44v 2 H+(aq) + 2 e- ---> H2(g) 0.00v Cu+2 + 2 e- ---> Cu +0.337v O2(g) + 4 H+(aq) + 4 e- ---> 2 H2O(l) +1.229v F2 + 2e- ---> 2 F- +2.87v

  20. Standard Reduction Potentials at 25°C

  21. E0 is for the reaction as written • The more positive E0 the greater the tendency for the substance to be reduced • The more negative E0 the greater the tendency for the substance to be oxidized • Under standard-state conditions, any species on the left of a given half-reaction will react spontaneously with a species that appears on the right of any half-reaction located below it in the table

  22. The half-cell reactions are reversible • The sign of E0changes when the reaction is reversed • Changing the stoichiometric coefficients of a half-cell reaction does not change the value of E0

  23. Can Sn reduce Zn2+ under standard-state conditions? Look up the Eº values in in the table of reduction potentials Zn+2 + 2 e- ---> Zn(s) -0.763v Sn+2 + 2 e- ---> Sn -0.143v • How do we find the answer? • Look up the Eº values in in the table of reduction potentials • \Which reactions in the table will reduce Zn2+(aq)?

  24. Standard cell potential • Zn(s) + Cu+2 (aq)® Zn+2(aq) + Cu(s) • The total standard cell potential is the sum of the potential at each electrode. • Eº cell = EºZn® Zn2++EºCu+2® Cu • We can look up reduction potentials in a table. • One of the reactions must be reversed, in order to change its sign.

  25. Standard Cell Potential • Determine the cell potential for a galvanic cell based on the redox reaction. • Cu(s) + Fe+3(aq)® Cu+2(aq) + Fe+2(aq) • Fe+3(aq)+ e-® Fe+2(aq) Eº = 0.77 V  • Cu+2(aq)+2e- ® Cu(s) Eº = 0.34 V • Cu(s) ® Cu+2(aq)+2e-Eº = -0.34 V • 2Fe+3(aq)+ 2e-® 2Fe+2(aq) Eº = 0.77 V  • Eocell = EoFe3+®EoFe2+ + EoCu®EoCu2+ • Eocell = 0.77 + (-0.34) = o.43 V

  26. The total reaction: Cu(s) ® Cu+2(aq)+2e-Eº = -0.34 V 2Fe+3(aq)+ 2e-® 2Fe+2(aq) Eº = 0.77 V  Cu(s) + 2Fe+3(aq) Cu2+ + 2Fe2+ Eºcell = +0.43 V

  27. Line Notation • Solid½Aqueous½½Aqueous½solid • Anode on the left½½Cathode on the right • Single line different phases. • Double line porous disk or salt bridge. Zn(s)½Zn2+(aq)½½Cu2+½Cu • If all the substances on one side are aqueous, a platinum electrode is indicated. • For the last reaction • Cu(s)½Cu+2(aq)½½Fe+2(aq),Fe+3(aq)½Pt(s)

  28. Complete description of a Galvanic Cell • The reaction always runs spontaneously in the direction that produces a positive cell potential. • Four parameters are needed for a complete description: • Cell Potential • Direction of flow • Designation of anode and cathode • Nature of all the components- electrodes and ions

  29. Exercise • Describe completely the galvanic cell based on the following half-reactions under standard conditions. MnO4- + 8 H+ +5e-® Mn+2 + 4H2O Eº=1.51 Fe+3 +3e-® Fe(s) Eº=0.036V • Write the total cell reaction • Calculate Eo cell • Define the cathode and anode • Draw the line notation for this cell

  30. 17.3 Cell potential, electrical work and free energy • The work accomplished when electrons are transferred through a wire depends on the “push” (thermodynamic driving force) behind the electrons • The driving force (emf) is defined in terms of potential difference (in volts) between two points in the circuit • emf = potential difference (V) = work (J) / Charge(C) =

  31. The work done by the system has a –ve sign • Potential produced as a result of doing a work should have a +ve sign • The cell potential, E, and the work, w, have opposite signs. • Relationship between E and w can be expressed as follows: • E = work done by system / charge ( )

  32. Charge is measured in coulombs. Thus, • -w = qE • Faraday = 96,485 C/mol e- q = nF = moles of e- x charge/mole e- • w = -qE = -nFE = DG • Thus, DG = -nFE and • DGo = -nFEo

  33. Potential, Work, DG and spontaneity • DGº = -nFE º • if E º > 0, then DGº < 0 spontaneous • if E º < 0, then DGº > 0 nonspontaneous • In fact, the reverse process is spontaneous.

  34. DG0 = -nFEcell 0 0 0 0 0 = -nFEcell Ecell Ecell Ecell F = 96,500 J RT V • mol ln K nF (8.314 J/K•mol)(298 K) ln K = 0.0257 V 0.0592 V log K ln K n (96,500 J/V•mol) = n n = = Spontaneity of Redox Reactions DG = -nFEcell n = number of moles of electrons in reaction = 96,500 C/mol DG0 = -RT ln K

  35. Spontaneity of Redox Reactions If you know one, you can calculate the other… If you know K, you can calculate Eº and Gº If you know Eº, you can calculate Gº

  36. Spontaneity of Redox Reactions Relationships among G º, K, and Eºcell

  37. DG0 = -nFEcell DG0 = -nFEcell 0 0 2(3e- + Al3+ Al) E0 = Ered + Eox cell Calculate DG0 for the following reaction at 250C. 2Al3+(aq) + 3Mg(s) 2Al(s) + 3Mg+2(aq) Oxidation: 3 (Mg Mg2+ + 2e-) n = ? Reduction: 0 0 = ___ X (96,500 J/V mol) X ___ V DG0 = _______ kJ/mol

  38. 17.4 Dependence of Cell Potential on Concentration • Qualitatively: we can predict direction of change in E from LeChâtelier pinciple • 2Al(s) + 3Mn+2(aq) 2Al+3(aq) + 3Mn(s); Eocell = 0.48 V • Predict if Ecell will be greater or less than Eºcell for the following cases: • if [Al+3] = 1.5 M and [Mn+2] = 1.0 M • if [Al+3] = 1.0 M and [Mn+2] = 1.5M • An increase in conc. of reactants would favor forward reaction thus increasing the driving force for electrons; i.e. Ecell becomes > Eocell

  39. Concentration Cell: both compartments contain same components but at different concentrations Half cell potential are not identical Because the Ag+ Conc. On both sides arenot same Eright > Eleft • To make them equal, [Ag+] • On both sides should same • Electrons move from left to • right

  40. DG0 = -nFE 0 RT nF E = E0 - ln Q 0 0 E = E = E E 0.0257 V 0.0592 V log Q ln Q n n - - The Nernst Equation Effect of Concentration on Cell Emf DG = DG0 + RT ln Q DG = -nFE -nFE = -nFE0+ RT ln Q Nernst equation At 298K

  41. The Nernst Equation • As reactions proceed concentrations of products increase and reactants decrease. • When equilibrium is reached Q = K ; Ecell = 0 and G = 0 (the cell no longer has the ability to do work)

  42. Predicting spontaneity using Nernst equation • Qualitatively: we can predict the direction of change in E from Lechatelier principle • Find Q • Calculate E • E > 0; the reaction is spontaneous to the right E < 0; the reaction is spontaneous to the left

  43. Cd Cd2+ + 2e- 0 0 2+ E0 = -0.44 + (+0.40) E0 = EFe /Fe + ECd /Cd 2+ E = -0.04 V E0 = -0.04 V 0.010 0.60 0 E = E 0.0257 V 0.0257 V ln ln Q n 2 - - Will the following reaction occur spontaneously at 250C if [Fe2+] = 0.60 M and [Cd2+] = 0.010 M? Fe2+(aq) + Cd (s) Fe (s) + Cd2+(aq) Oxidation: n = 2 Reduction: 2e- + Fe2+ 2Fe E = ____________ E ___ 0 ________________

  44. 0 E = E 0.0592 V log Q n - Exercise- p. 843 • Determine the cell potential at 25oC for the following cell, given that • 2Al(s) + 3Mn+2(aq) ® 2Al+3(aq) + 3Mn(s) [Mn2+] = 0.50 M; [Al3+]=1.50 M; E0 cell = 0.4 • Always we have to figure out n from the balanced equation 2(Al(s)+® Al+3(aq) + 3e-) 3(Mn+2(aq) + 2e-® Mn(s)) n = 6

  45. Calculation of Equilibrium Constants for redox reactions At equilibrium, Ecell = 0 and Q = K. Then, at 25 oC

  46. 2Ag 2Ag+ + 2e- 0 Ecell 2e- + Fe2+ Fe 0 0 2+ + E0 = EFe /Fe+ EAg /Ag E0 = -0.44 –0.80= -1.24 V E0 = -1.24 V E0 cell x n = exp K = 0.0257 V 0.0257 V -1.24 V x2 0.0257 V ln K exp n = What is the equilibrium constant for the following reaction at 250C? Fe2+(aq) + 2Ag (s) Fe (s) + 2Ag+(aq) Oxidation: n = ___ Reduction: K = ________________

  47. 17.5 Batteries Batteries are Galvanic Cells Lead-Storage Battery • A 12 V car battery consists of 6 cathode/anode pairs each producing 2 V. • Cathode: PbO2 on a metal grid in sulfuric acid: PbO2(s) + SO42-(aq) + 4H+(aq) + 2e- PbSO4(s) + 2H2O(l) • Anode: Pb: Pb(s) + SO42-(aq)  PbSO4(s) + 2e-

  48. Lead storage battery PbO2(s) + 4H+(aq) + SO2-(aq) + 2e- PbSO4(s) + 2H2O (l) 4 Pb (s) + PbO2 (s) + 4H+ (aq) + 2SO2- (aq) 2PbSO4 (s) + 2H2O (l) 4 Anode: Pb (s) + SO2- (aq) PbSO4 (s) + 2e- 4 Cathode:

  49. Lead-Storage Battery • The overall electrochemical reaction is PbO2(s) + Pb(s) + 2SO42-(aq) + 4H+(aq)  2PbSO4(s) + 2H2O(l) for which Ecell = Ered(cathode) - Ered(anode) = (+1.685 V) - (-0.356 V) = +2.041 V. • H2SO4 is consumed while the battery is discharging • H2SO4 is 1.28g/ml and must be kept • Water is depleted thus the battery should be topped off always

  50. Zn (s) Zn2+ (aq) + 2e- + 2NH4(aq) + 2MnO2(s) + 2e- Mn2O3(s) + 2NH3(aq) + H2O (l) Zn (s) + 2NH4 (aq) + 2MnO2 (s) Zn2+ (aq) + 2NH3 (aq) + H2O (l) + Mn2O3 (s) Dry cell Batteries Anode: Cathode:

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