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Chapter 14 Notes

Chapter 14 Notes

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Chapter 14 Notes

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  1. Chapter 14 Notes Fundamentals of Electrochemistry

  2. Redox Reactions • Reactions involving the transfer of electrons from one species to another. • LEO says GER • Oxidation – Loss of electrons • Reduction – Gain of electrons

  3. Definitions • A given species is said to be “reduced” when it gains electrons and “oxidized” when it loss electrons. Example • 2Fe3+ + Cu(s) ↔ 2Fe2+ + Cu2+oxidizing reducing reduced oxidized agent agent species species

  4. Electrochemistry - The study of redox chemistry • Electrochemical cells - The reactants are separated from one another, and the reaction is forced to occur via the flow of electrons through an electrical circuit.

  5. But for what purpose? • produce electricity (batteries; chemical energy from spontaneous redox reactions are converted to electrical work). • To force non-spontaneous reactions to occur by supplying an external energy source. • Quantitative analysis of redox active analytes • To study the energetics and kinetics of redox processes

  6. Types of electrochemical experiments • Construction of a battery (ch 14) • Potentiometry (measurement of cell voltages to extract chemical information, ex. pH meter)/ch 15 • Redox titrations/ch 16 • Electrogravimetric analysis (depositing analyte on an electrode/ ch 17 • Coulometry (measuring the number of electrons being transferred at constant cell voltage)/ch 17 • Voltammety (measuring current as a function of cell voltage, quantitative and qualitative info)/ch 17

  7. Remind the students that they need to review the material presented in section 14.1 • Electrochemical Cells • Cell voltage (E or EMF) – A measure of the spontaneity of the redox reaction. • .. DG = -nFE • E is the cell voltage (E = 0 @ equilibrium) • n is the number of electrons transferred • F is Faradays constant – 9.649*104 C/mol. • Note: when the cell voltage is positive the reaction is spontaneous

  8. Galvantic cells : A cell that uses a spontaneous chemical reaction to generate electricity Ex. Fig 14.3 • Cd(s)  Cd2+ + 2e- Oxidation 2AgCl(s) + 2e- ↔ 2Ag(s) + 2Cl- Reduction ___________________________________ Cd(s) + 2AgCl(s) ↔ Cd 2+ + 2Ag(s) + 2Cl- • E is positive, spontaneous reaction

  9. - + e- e- EMF Cd(s) Ag(s) Anode –where oxidation occurs Cd(s)→ Cd2+ + 2e- Cathode –where reduction occurs AgCl +e- →Ag(s) + Cl- Cl- Cd2+ AgCl(s) CdCl2(aq) Cd(s) + 2AgCl(s) ↔ Cd2+ + 2Ag(s) + 2Cl- Note: The two solids are separated and that the transfer of electrons must flow through the external circuit.

  10. Example 2 Cd(s) Cd2+ + 2e- Oxidation 2Ag+ + 2e- ↔ 2Ag(s) Reduction ____________________________________ Cd(s) + 2Ag+ ↔ Cd2+ + 2Ag(s) • If we set up this redox reaction in the same manner as the other one, no current flows through the circuit even though the energetics of the reaction is the same. Why?

  11. Ag+ - + e- e- EMF Cd(s) Ag(s) Anode –where oxidation occurs Cd(s)→ Cd2+ + 2e- Cathode –where reduction occurs Ag+ +e- →Ag(s) Cd2+ Ag+ Cd(NO3)2(aq) AgNO3(aq) Cd(s) + 2Ag+ ↔ Cd2+ + 2Ag(s) The redox reaction can take place directly on the surface of the electrodes, without electron having to flow through circuit.

  12. KNO3 K+ NO3- - + e- e- EMF Ag(s) Cd(s) Cathode –where reduction occurs Ag+ +e- →Ag(s) Anode –where oxidation occurs Cd(s)→ Cd2+ + 2e- Cd2+ Ag+ AgNO3(aq) Cd(NO3)2(aq) Cd(s) + 2Ag+→ Cd2+ + 2Ag(s) Need Salt Bridge to avoid energy barrier to build up of excess charge

  13. Line Notation: anode║cathode Cd(s)│CdCl2(aq) ║ AgNO3(aq)│Ag(s)

  14. Nersnst Equation For the balance redox reaction: aA + bB ↔ cC + dD • The cell voltage is a function of the activities of the reactants and products (in an analogous manner in which DG is related to Q. • E = E - RT/nF ln {ACcADd /AAaABb} at T = 298.15 K (25 C) • E = E - 0.05916 /n log {ACcADd /AAaABb}, where n = the number of moles of electrons transferred in the balanced redox reaction.

  15. Standard Reduction Potentials (E) analogous to DG • The cell voltage when the activities of all reactants and products are unity. • Standard half cell reduction potentials (Table in back of book) – • E for 2H+ + 2e- ↔ H2 is arbitrarily set to 0 V. • MnO4- + 8H+ + 5e- ↔ Mn2++ 4H2O E0 = + 1.507 V Positive voltage means the rxn with H2 is spontaneous at unit activities, implies Ag+ is a strong oxidizing agent. • Cd2+ + 2e- ↔ Cd(s) E0 = -0.402 V Negative voltage means rxn between H+ and Cd(s) is spontaneous at unit activities, implies Cd2+ is a very weak oxidizing agent.

  16. Using the Nernst Equation for example 2 Cd(s) Cd2+ + 2e- Oxidation 2Ag+ + 2e- ↔ 2Ag(s) Reduction __________________________________ Cd(s) + 2Ag+ ↔ Cd2+ + 2Ag(s) Ecell = Ecathode – Eanode

  17. The Nernst equation for a half-cell reaction is always written as a reduction!!!!! Cd(s) + 2Ag+ ↔ Cd2+ + 2Ag(s) EMF = (E+ - E-) - (0.05916/2)log([Cd2+]/[Ag+]2) = Ecell - (0.05916/2)log ([Cd2+] / [Ag+]2) [Cd2+] / [Ag+]2 = Q (the reaction quotient)

  18. Nernst Eq as half reactions EMF = E+ - E- E+ = E+ – (0.05916)log(1/[Ag+]} E- = E- – (0.05916/2)log(1/[Cd2+])} The nernst Eq for both ½ rxn written as reductions Cd2+ + 2e-  Cd(s) anode ½ rxn 2Ag+ + 2e- ↔ 2Ag(s) cathode ½ rxn Gives same answer. The math is equivalent.

  19. KNO3 K+ NO3- - + e- e- EMF Ag(s) Cd(s) Anode –where oxidation occurs Cd(s)→ Cd2+ + 2e- Cathode –where reduction occurs Ag+ +e- →Ag(s) Cd2+ Ag+ 0.100 M Cd(NO3)2(aq) 0.100 M AgNO3(aq) Cd(s) + 2Ag+→ Cd2+ + 2Ag(s)

  20. Calculating Ecell using the Nerst Eq EMF = (Ec - Ea) - (0.05916/2)log([Cd2+]/[Ag+]2) = (0.7993–(-0.402)) – ((0.05916/2)log(0.100/(0.100)2)) = 1.172 V

  21. KNO3 K+ NO3- - + e- e- EMF = 1.172 V Ag(s) Cd(s) Anode –where oxidation occurs Cd(s)→ Cd2+ + 2e- Cathode –where reduction occurs Ag+ +e- →Ag(s) Cd2+ Ag+ 0.100 M Cd(NO3)2(aq) 0.100 M AgNO3(aq) Cd(s) + 2Ag+→ Cd2+ + 2Ag(s)

  22. Determination of the anode and cathode • On paper • When you write an overall redox reaction in a given direction and solve the Nernst Eq accordingly, if the EMF is positive, then the reaction is spontaneous as written. • If it is negative it is spontaneous in the opposite direction. Reduction always occurs at the cathode and oxidation always occurs at the anode.

  23. Determination of the anode and cathode • In the Lab • The voltmeter has a positive and a negative lead. • If you connect the negative lead to the anode and the positive lead to the cathode, the EMF will be positive. • If the EMF reads negative, it means that you have connected the negative lead to the cathode and the positive lead to the anode. • The sign of the EMF on the voltmeter display indicates the directions in which the electrons flow.

  24. KNO3 K+ NO3- The EMF reads positive, Because the negative Terminal is connected to the Anode and the positive Terminal is connected to the cathode. - + e- e- EMF = 1.172 V Cd(s) Ag(s) Anode –where oxidation occurs Cd(s)→ Cd2+ + 2e- Cathode –where reduction occurs Ag+ +e- →Ag(s) Cd2+ Ag+ 0.100 M Cd(NO3)2(aq) 0.100 M AgNO3(aq) Cd(s) + 2Ag+→ Cd2+ + 2Ag(s)

  25. KNO3 K+ NO3- If the leads are switched, the voltage reads negative. - + e- e- EMF = -1.172 V Ag(s) Cd(s) Anode –where oxidation occurs Cd(s)→ Cd2+ + 2e- Cathode –where reduction occurs Ag+ +e- →Ag(s) Cd2+ Ag+ 0.100 M Cd(NO3)2(aq) 0.100 M AgNO3(aq) Cd(s) + 2Ag+→ Cd2+ + 2Ag(s) Need Salt Bridge to avoid energy barrier to build up of excess charge

  26. Prob 14-19 Calculate EMF for the following cell Pb (s)│PbF2(s)│NaF(aq)(0.10 M)║ NaF(aq)(0.10 M)│AgCl(s)│Ag(s) 2 ways to approach it (the first one is less work) PbF2(s) + 2e- → Pb (s) + 2F- (aq) and AgCl(s) + e- → Ag(s) + Cl- (aq) Or Pb (s) → Pb2+ + 2e- and Ag+ + e- → Ag (s)

  27. Anode –where oxidation occurs Pb(s) + 2I- → PbI2(s) + 2e- or Pb(s)→ Pb2+ + 2e- - + e- e- EMF Pb(s) Ag(s) Cathode –where reduction occurs AgCl +e- →Ag(s) + Cl- or Ag+ + e- → Ag(s) Cl- Cd2+ AgCl(s) CdCl2(aq) PbI2(s) 2I- + Pb(s)+ 2AgCl(s) ↔ PbI2(s) + + 2Ag(s) + 2Cl- Pb(s) + 2Ag+ → 2Ag(s) + Pb2+

  28. PbF2(s) + 2e- → Pb (s) + 2F- (aq) E0 = -0.350 V AgCl(s) + e- → Ag(s) + Cl- (aq) E0 = 0.222 V E+ = 0.222 -0.05916 log([Cl-]) = 0.281 V E- = -0.350 -0.05916/2 log([F-]2) = -0.291 V E = 0.281 – (-0.291) = 0.572 V

  29. Second way Pb2+ + 2e- → Pb(s)E0 = -0.126 V Ag+ + e- → Ag(s) E0 = 0.7993 V E+ = 0.7993 -0.05916 log(1/[Ag+]) E- = -0.126 -0.05916/2 log(1/[Pb2+]) [Ag+] = Ksp(AgCl)/[Cl-] = 1.8E-10/0.10 = 1.8E-9 [Pb2+] = Ksp(PbF2)/[F-]2 =3.6E-8/0.01 = 3.6E-6 E = 0.282 – (-0.287) = 0.569 V

  30. Adding half reactions • Adding together two half reactions to obtain a new half reaction. To do this it is best to convert to Gs Problem 14.22 You must determine which reactions you must to add together to obtain the reaction in question. HOBr  Br 2(aq) Br2(aq) 2Br-(aq) HOBr  2Br -(aq)

  31. Balance Redox Rxns • Balance reaction between HOBr and Br- in an acidic solution • balancing the half-cell reaction: • step 1: add H+ to the reactant side and water to the product side HOBr + H+ Br- + H2O • Step 2: Stoichiometrically balance the reaction. • It already is in this case! • Step 3: balance the charge by adding electrons HOBr + 2e- + H+ Br- + H2O • Repeat for the reactions that you must add to obtain the above rxn.

  32. When you add to half-rxns to obtain a third half-rxn, the safe thing to do is to add the DGs. HOBr + H+ + e-1/2Br2 + H2O DG1 = -F(1.584) 1/2Br2 + e- Br-DG2 = -F(1.098) ____________________________________ HOBr + 2e- + H+ Br- + H2O DG3 = -2FE30 DG3 = DG1 + DG2 = -2F(E30) = -F(1.584) + -F(1.098) = -2FE30 E30 = (1.584 + 1.098)/2 = 1.341 V

  33. Problem 14-29 You are forming a half-reaction. It is safest to add together multiply Ks (or add DGs). Pd(OH)2(s) Pd+2 + 2OH- Ksp = 3·10 -28 Pd+2 + 2e- Pd(s) K1 = 10(nE/.05916) = 8.9·1030 ________________________________________________________ Pd(OH)2(s) +2e- Pd (s) + 2OH- K = KspK1 K = 3·103 Log K = -nE/0.05916 E = -(0.05916/2)log(3·103) = 0.103 V