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ELECTROCHEMISTRY

This article explains the fundamental concepts of electrochemistry, including charge (Q), electric current (I), and potential (Ɛ), and how these properties are related. It also discusses the use of standard reduction potentials to predict the spontaneity of reactions and identify strong oxidizing and reducing agents.

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ELECTROCHEMISTRY

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  1. ELECTROCHEMISTRY CHARGE (Q) – A property of matter which causes it to experience the electromagnetic force COULOMB (C) – The quantity of charge equal to 6.241 × 1018 electrons ELECTRIC CURRENT or AMPERAGE (I) – The rate of flow of electric charge AMPERE (A) – Flow rate of one coulomb of electric charge per second ELECTROMOTIVE FORCE or POTENTIAL or VOLTAGE (Ɛ) – The potential difference between 2 substances, causing electrons to flow from one to the other VOLT (V) – One joule of potential energy per coulomb 3D-1 (of 20)

  2. Spontaneous oxidation-reduction reaction: Fe(s) + Cu2+(aq) → Fe2+(aq) + Cu(s) Iron is more reactive than copper,  iron atoms will release their valence electrons to the copper (II) ions Redox reactions can be written as the sum of 2 half-reactions Oxidation: Reduction: Fe(s) → Fe2+(aq) Cu2+ (aq) → Cu(s) + 2e- 2e- + Fe(s) + 2e- + Cu2+ (aq) → Fe2+ (aq) + Cu(s) + 2e- Fe(s) + Cu2+ (aq) → Fe2+ (aq) + Cu(s) If the Fe(s) and Cu2+ (aq) are separated, the electron transfer can go through a wire 3D-2 (of 20)

  3. Cu ANODE – The electrode where oxidation occurs CATHODE – The electrode where reduction occurs 0.78 V is called the cell potential, the cell voltage, or the cell emf 3D-3 (of 20)

  4. GALVANIC CELL – An electrochemical cell that produces electric current from a chemical reaction Shorthand notation: Anode | Anode Solution || Cathode Solution | Cathode Fe | Fe2+ (1 M) || Cu2+ (1 M) | Cu 3D-4 (of 20)

  5. The ΔG of a reaction occurring in a Galvanic cell is related to Ɛ ΔG = -nFƐ n = number of moles of electrons transferred in the redox reaction F= the Faraday constant the charge of 1 mole of electrons, equal to 96,485 C Ɛ = electromotive force or voltage potential difference causing electrons to flow For Galvanic cells with 1 M concentrations ΔGº = -nFƐº J = (mol) (C/mol) (V) (J/C) 3D-5 (of 20)

  6. Calculate ΔGº for the reaction Fe(s) + Cu2+(aq) → Fe2+(aq) + Cu(s) Ɛº = 0.78 V INTENSIVE PROPERTY – One that does not depend on the amount of chemical change ΔGº = -nFƐº = -(2 mol)(96,485 C/mol)(0.78 V) = -(2 mol)(96,485 C/mol)(0.78 J/C) = -150,000 J ΔG < 0 and Ɛ > 0 means a spontaneous process The more negative ΔG, the more spontaneous the process The more positive Ɛ, the more spontaneous the process 3D-6 (of 20)

  7. REDUCTION AND OXIDATION POTENTIALS REDUCTION POTENTIAL (Ɛred) – The electric potential for a reduction half-reaction OXIDATION POTENTIAL (Ɛox) – The electric potential for an oxidation half-reaction Ɛºred and Ɛºoxare for a standard state half-reactions The more positive the Ɛred or Ɛox, the more spontaneous the half-reaction 2e- + F2(g) → 2F-(aq) Ɛºred= 2.87 V 2e- + l2(s) → 2l-(aq) Ɛºred= 0.54 V K(s) → K+ (aq) + e- Ɛºox= 2.92 V Ca(s) → Ca2+ (aq) + 2e- Ɛºox= 2.76 V Cu(s) → Cu2+ (aq) + 2e- Ɛºox= -0.34 V 3D-7 (of 20)

  8. The potential of an overall redox reaction in a Galvanic cell can be measured with a voltmeter Fe(s) → Fe2+ (aq) + 2e- 2e- + Cu2+ (aq) → Cu(s) Ɛºox =0.20 V Ɛºred = 0.58 V Ɛºox =0.30 V Ɛºred = 0.48 V Fe(s) + Cu2+ (aq) → Fe2+ (aq) + Cu(s) Ɛº = 0.78 V The sum of a reduction potential and an oxidation potential must equal the potential for the overall redox reaction Unfortunately, the potential of a half-reaction cannot be measured, so we make one up! 3D-8 (of 20)

  9. H2 (g, 1 atm) → 2H+(aq, 1 M) + 2e- This standard hydrogen half-reaction is assigned a potential of 0.00 V All other standard reduction potentials are measured relative to this one H2(g) → 2H+ (aq) + 2e- 2e- + Cu2+ (aq) → Cu(s) Ɛºox =0.00 V Ɛºred = 0.34 V H2 (g) + Cu2+ (aq) → 2H+ (aq) + Cu(s) Ɛº = 0.34 V 3D-10 (of 20)

  10. H2 (g, 1 atm) → 2H+(aq, 1 M) + 2e- This standard hydrogen half-reaction is assigned a potential of 0.00 V All other standard reduction potentials are measured relative to this one And, Ɛred’s always mean “per one mole of e- change” 3D-11 (of 20)

  11. USES FOR STANDARD REDUCTION POTENTIALS 1) Predicting the spontaneity of a reaction Determine if the following standard state reaction is spontaneous: 3Fe(s) + 2Cr3+(aq) → 3Fe2+ (aq) + 2Cr(s) Find the 2 reduction potentials that can be used to make the reaction 2e- + Fe2+ (aq) → Fe(s) 3e- + Cr3+ (aq) → Cr(s) Ɛºred = -0.44 V Ɛºred = -0.73 V Add a reduction and oxidation half-reaction to make the desired reaction 3D-12 (of 20)

  12. USES FOR STANDARD REDUCTION POTENTIALS 1) Predicting the spontaneity of a reaction Determine if the following standard state reaction is spontaneous: 3Fe(s) + 2Cr3+(aq) → 3Fe2+ (aq) + 2Cr(s) Find the 2 reduction potentials that can be used to make the reaction Fe(s) → Fe2+ (aq) + 2e- 3e- + Cr3+ (aq) → Cr(s) Ɛºox =0.44 V Ɛºred = -0.73 V Ɛº = -0.29 V Ɛº is negative,  the reaction is nonspontaneous 3D-13 (of 20)

  13. USES FOR STANDARD REDUCTION POTENTIALS 2) Predicting strong oxidizing and reducing agents A large, positive reduction potential means the forward reaction is spontaneous (the REACTANThas a strong tendency to be REDUCED) Best oxidizing agent from the list? F2 A large, negative reduction potential means the reverse reaction is spontaneous (the PRODUCT has a strong tendency to be OXIDIZED) Best reducing agent from the list? Li 3D-14 (of 20)

  14. USES FOR STANDARD REDUCTION POTENTIALS 3) Predicting the potential and spontaneous reaction in a Galvanic cell 3D-15 (of 20)

  15. For a Galvanic cell with silver and nickel electrodes in 1 M solutions of Ag+ and Ni2+ respectively, determine the (a) standard cell potential, (b) spontaneous reaction, and (c) anode and cathode (a) Find the 2 reduction potentials to produce the Galvanic cell e-+ Ag+ (aq) → Ag(s) 2e-+ Ni2+ (aq) → Ni(s) Ɛºred = 0.80 V Ɛºred = -0.23 V The largest positive potential is the spontaneous process, and will be the reduction the other must be reversed, and will be the oxidation 3D-16 (of 20)

  16. For a Galvanic cell with silver and nickel electrodes in 1 M solutions of Ag+ and Ni2+ respectively, determine the (a) standard cell potential, (b) spontaneous reaction, and (c) anode and cathode (a) Find the 2 reduction potentials to produce the Galvanic cell e- + Ag+ (aq) → Ag(s) Ni(s) → Ni2+ (aq) + 2e- Ɛºred = 0.80 V Ɛºox = 0.23 V 1.03 V Add the reduction and oxidation potentials to get the cell potential 3D-17 (of 20)

  17. For a Galvanic cell with silver and nickel electrodes in 1 M solutions of Ag+ and Ni2+ respectively, determine the (a) standard cell potential, (b) spontaneous reaction, and (c) anode and cathode (b) Equalize e-s and add the reduction and oxidation half-reactions together ( ) x 2 e- + Ag+ (aq) → Ag(s) Ni(s) → Ni2+ (aq) + 2e- Ɛºred = 0.80 V Ɛºox = 0.23 V 1.03 V 3D-18 (of 20)

  18. For a Galvanic cell with silver and nickel electrodes in 1 M solutions of Ag+ and Ni2+ respectively, determine the (a) standard cell potential, (b) spontaneous reaction, and (c) anode and cathode (b) Equalize e-s and add the reduction and oxidation half-reactions together 2e-+2Ag+ (aq) → 2Ag(s) Ni(s) → Ni2+ (aq) + 2e- Ɛºred = 0.80 V Ɛºox = 0.23 V 2Ag+(aq) + Ni(s) → 2Ag(s) + Ni2+ (aq) 1.03 V (c) Nickel half-reaction is the oxidation, ∴ Ni is the anode Silver half-reaction is the reduction, ∴ Ag is the cathode 3D-19 (of 20)

  19. For a Galvanic cell with cadmium and chromium electrodes in 1 M solutions of Cd2+ and Cr3+ respectively, determine the (a) standard cell potential, (b) spontaneous reaction, and (c) anode and cathode 2e- + Cd2+ (aq) → Cd(s) 3e-+ Cr3+ (aq) → Cr(s) Ɛºred = -0.40 V Ɛºred= -0.73 V 3 x ( ) 2e- + Cd2+ (aq) → Cd(s) Cr(s) → Cr3+ (aq) + 3e- Ɛºred = -0.40 V Ɛºox= 0.73 V 2 x ( ) 0.23 V 6e- + 3Cd2+ (aq) → 3Cd(s) 2Cr(s) → 2Cr3+ (aq) + 6e- 3Cd2+(aq) + 3Cr(s) → 3Cd(s)+ 2Cr3+ (aq) Cd2+ is reduced ∴ Cd is the cathode Cr is oxidized ∴ Cr is the anode 3D-20 (of 20)

  20. NONSTANDARD STATE CELLS For nonstandard cells ΔG = ΔGº + RT ln Q because ΔG = -nFƐ and ΔGº = -nFƐº -nFƐ = -nFƐº + RT ln Q THE NERNST EQUATION Ɛ = Ɛº – RT ln Q ____ nF 3E-1 (of 11)

  21. Calculate the potential for the following cell at 298 K Zn (s) | Zn2+ (0.200 M) || Ag+ (0.100 M) | Ag (s) e- + Ag+ (aq) → Ag(s) 2e- + Zn2+ (aq) → Zn(s) Ɛºred = 0.80 V Ɛºred = -0.76 V e- + Ag+ (aq) → Ag(s) Zn(s) → Zn2+ (aq) + 2e- Ɛºred = 0.80 V Ɛºox = 0.76 V 0.80 V + 0.76 V = 1.56 V 3E-2 (of 11)

  22. Calculate the potential for the following cell at 298 K Zn (s) | Zn2+ (0.200 M) || Ag+ (0.100 M) | Ag (s) ( ) x 2 e- + Ag+ (aq) → Ag(s) Zn(s) → Zn2+ (aq) + 2e- Q = [Zn2+] _________ [Ag+]2 2Ag+(aq) + Zn(s) → 2Ag(s)+ Zn2+ (aq) Ɛ = Ɛº – RT ln Q ____ nF = 1.56 V – (8.314 CV/K)(298 K) ln 0.200 ___________________________ ________ (2 mol)(96,485 C/mol) 0.1002 = 1.56 V – 0.0385 V = 1.52 V 3E-3 (of 11)

  23. For a nickel-cadmium cell with solutions of 0.00100 M nickel (II) sulfate and 0.10 M cadmium sulfate, determine the (a) standard cell potential, (b) spontaneous reaction, (c) anode and cathode (d) cell potential at 298 K 2e-+ Ni2+ (aq) → Ni(s) 2e-+ Cd2+ (aq) → Cd(s) Ɛºred = -0.23 V Ɛºred = -0.40 V 2e-+ Ni2+ (aq) → Ni(s) Cd(s) → Cd2+ (aq) + 2e- Ɛºred = -0.23 V Ɛºox = 0.40 V -0.23 V + 0.40 V = 0.17 V 3E-4 (of 11)

  24. For a nickel-cadmium cell with solutions of 0.00100 M nickel (II) sulfate and 0.10 M cadmium sulfate, determine the (a) standard cell potential, (b) spontaneous reaction, (c) anode and cathode (d) cell potential at 298 K 2e-+ Ni2+ (aq) → Ni(s) Cd(s) → Cd2+ (aq) + 2e- Ni – cathode Cd – anode Ni2+(aq) + Cd(s) → Ni(s)+ Cd2+ (aq) Q = [Cd2+] ________ [Ni2+] Ɛ = Ɛº – RT ln Q ____ nF = 0.17 V – (8.314 CV/K)(298 K) ln 0.10 ___________________________ __________ (2 mol)(96,485 C/mol) 0.00100 = 0.17 V – 0.059 V = 0.11 V 3E-5 (of 11)

  25. For a nickel-cadmium cell with solutions of 0.00100 M nickel (II) sulfate and 0.10 M cadmium sulfate, determine the (e) the cell potential at 298 K once 90.% of the nickel (II) ions have reacted away Ni2+(aq) + Cd(s) → Ni(s)+ Cd2+ (aq) Q = [Cd2+] _________ [Ni2+] Initial M’s Change in M’s Final M’s 0.00100 0.10 - 0.00090 + 0.00090 0.00010 0.10090 Ɛ = Ɛº – RT ln Q ____ nF = 0.17 V – (8.314 CV/K)(298K) ln 0.10090 ___________________________ __________ (2 mol)(96,485 C/mol) 0.00010 = 0.17 V – 0.089 V = 0.08 V 3E-6 (of 11)

  26. For a reaction at equilibrium ΔG = 0 ∴ Ɛ = 0 Ɛ = Ɛº – RT ln Q ____ nF RT ln Keq = Ɛº ____ nF 0 = Ɛº – RT ln Keq ____ nF ln Keq = ƐºnF ______ RT RT ln Keq = Ɛº ____ nF Keq = eƐºnF/RT 3E-7 (of 11)

  27. Find the equilibrium constant at 298 K for Fe(s) + I2(s) → Fe2+ (aq)+ 2I- (aq) 2e-+ I2(s) → 2I-(aq) 2e-+ Fe2+ (aq) → Fe(s) Ɛºred = 0.54 V Ɛºred = -0.44 V 2e-+ I2(s) → 2I-(aq) Fe(s) → Fe2+ (aq) + 2e- Ɛºred = 0.54 V Ɛºox = 0.44 V I2 (s)+ Fe(s) → 2I- (aq)+ Fe2+ (aq) 0.54 V + 0.44 V = 0.98 V 3E-8 (of 11)

  28. Find the equilibrium constant at 298 K for Fe(s) + I2(s) → Fe2+ (aq)+ 2I- (aq) Keq= eƐºnF/RT = e[(0.98 V)(2 mol)(96,485 C/mol)]/[(8.314 CV/K)(298 K)] = 1.3 x 1033 3E-9 (of 11)

  29. Find the solubility product constant for mercury (I) chloride at 298 K Hg2Cl2 (s)⇄ Hg22+ (aq)+ 2Cl- (aq) 2e-+ Hg2Cl2(s) → 2Hg(l) + 2Cl-(aq) 2e- + Hg22+ (aq) → 2Hg(l) Ɛºred =0.33 V Ɛºred = 0.80 V 2e- + Hg2Cl2(s) → 2Hg(l) + 2Cl-(aq) 2Hg(l) → Hg22+(aq) + 2e- Ɛºred = 0.33 V Ɛºox = -0.80 V Hg2Cl2 (s) ⇄ Hg22+ (aq)+ 2Cl- (aq) 0.33 V – 0.80 V = -0.47 V 3E-10 (of 11)

  30. Find the solubility product constant for mercury (I) chloride at 298 K Hg2Cl2 (s)⇄ Hg22+ (aq)+ 2Cl- (aq) Ksp= eƐºnF/RT = e[(-0.47 V)(2 mol)(96,485 C/mol)]/[(8.314 CV/K)(298 K)] = 1.3 x 10-16 3E-11 (of 11)

  31. BATTERIES BATTERY – One or more electrochemical cells that produce electricity from a chemical reaction 3F-1 (of 17)

  32. Dry Cell Graphite rod (cathode) Paste of MnO2, NH4Cl, and H2O Zinc casing (anode) Anode: Cathode: Zn(s) → Zn2+(aq) + 2e- 2e-+ 2MnO2(s) + 8NH4+ (aq) → 2Mn3+(aq) + 4H2O(l) + 8NH3 (aq) Potential or Voltage : 1.5 V Current or Amperage : Depends on battery size 3F-2 (of 17)

  33. Dry Cell Graphite rod (cathode) Paste of MnO2, NH4Cl, and H2O Zinc casing (anode) The acidic content tends to corrode the Zn Fast usage : NH3 insulates the cathode, reducing the voltage With rest : Zn2+ migrates to center, forming Zn(NH3)42+ to bind the NH3 3F-2 (of 17)

  34. Alkaline Battery Graphite rod (cathode) Paste of MnO2, KOH, and H2O Powdered zinc (anode) Anode: Cathode: Zn(s) + OH-(aq)→ ZnO(s) + H2O(l) + 2e- 2e-+ 2MnO2(s) + H2O(l) → 2Mn2O3 (s) + 2OH-(aq) Zn resists corrosion in a basic solution 3F-4 (of 17)

  35. Lithium-Ion Battery Lithium cobalt oxide (anode) Separator Graphite (cathode) Anode: Cathode: LiCoO2 (s) → CoO2 (s) + Li+(org) + e- e-+ Li+ (org) + 6C(s) → LiC6 (s) Because both products stick to the electrodes, by applying an external source of electricity the reverse reaction will occur, reforming the reactants This is called RECHARGING 3F-5 (of 17)

  36. Lead Storage Battery Lead (anode) Lead + lead (IV) oxide (cathode) 4 M sulfuric acid Anode: Cathode: Pb(s) + SO42-(aq) → PbSO4 (s) + 2e- 2e-+ 2PbO2(s) + 4H+ (aq) + SO42-(aq) → PbSO4 (s) + 2H2O(l) 3F-6 (of 17)

  37. Lead Storage Battery Lead (anode) Lead + lead (IV) oxide (cathode) 4 M sulfuric acid Potential or Voltage: 2.1 V x 6 cells = 12.6 V _______ cell Because products stick to the electrodes, this battery is rechargeable 3F-7 (of 17)

  38. Hydrogen Fuel Cell Platinum Catalyst Polymer Electrolyte Membrane Anode: Pt catalyst splits hydrogen atoms into hydrogen ions and electrons Electrolyte: PEM allows hydrogen ions to pass through to the cathode Cathode: Oxygen and electrons combine with hydrogen ions to make water 3F-8 (of 17)

  39. ELECTROLYTIC CELL – An electrochemical cell that uses electricity to produce a chemical reaction Cathode: Reduction H(OH)(l) → H2 (g) (an element’s oxidation number goes down) H+ → H20 2e- + 2 + OH-(aq) 2 H2O +1 -2 Anode: Oxidation H2O(l) → O2 (g) (an element’s oxidation number goes up) O2- → O20 2 + 4H+ (aq) + 4e- 3F-9 (of 17)

  40. Cathode: the cation is reduced to its elemental form e- + M+ → M0 Anode: the anion is oxidized to its elemental form N- → N0 + e- ElectricityCathodeAnode through M+ N- KF(l) NaCl(l) K(s) Na(s) F2 (g) Cl2 (g) 3F-10 (of 17)

  41. ElectricityCathodeAnode through CuBr2(aq) HCl(aq) NaCl(aq) KF (aq) AgNO3(aq) H2SO4(aq) Na2CO3(aq) Cu (s) H2 (g) Na(s) H2 (g) Ag(g) H2(g) H2(g) Br2 (l) Cl2 (g) Cl2(g) F2 (g) O2 (g) O2(g) O2(g) Ɛºred -2.71 V -0.83 V H2 (g) Na+ (aq) + e-→ Na(s) 2H2O(l)+ 2e-→ H2(g) + 2OH- (aq) Exception 1 – Metals that blow up in water are not reduced, H+ in water is Alkali metals, Ca2+, Sr2+, Ba2+, Ra2+ Exception 2 – Polyatomic ions are (usually) not oxidized, O2- in water is 3F-11 (of 17)

  42. Electrolytic cells are used for (1) producing elements (Na, Cl2, etc.) (2) purification of metals from ore (3) electroplating metals (Au, Ag, Pt, etc.) Ag Anode: Oxidation Ag(s)→ Ag+(aq) + e- Ag+ Cathode: Reduction Ag+ (aq) → Ag(s) e- + 3F-12 (of 17)

  43. FARADAY’S LAWS OF ELECTROLYSIS (1) Passing the same quantity of electricity through a cell always leads to the same amount of chemical change (2) It takes 96,485 C of electricity to deposit or liberate 1 mole of a substance that gains or loses 1 mole of e-s during the cell reaction 96,485 C = 1 mole e- = 1 Faraday 3F-13 (of 17)

  44. Calculate the mass of copper deposited by a current of 7.89 amperes flowing for 1.20 x 103seconds through a copper (II) nitrate solution. Cu2+(aq) + 2e-→Cu(s) 7.89 A x 1.20 x 103 s x 1 C ______ 1 As x 1 F ___________ 96,485 C x 1 mol e- __________ 1 F x 1 mol Cu ___________ 2 mol e- x 63.55 g Cu ______________ 1 mol Cu = 3.12 g Cu 3F-14 (of 17)

  45. Calculate the mass of aluminum deposited by a current of 5.00 amperes flowing for 10.0 minutes through an aluminum nitrate solution. Al3+(aq) + 3e-→Al(s) 5.00 A x 10.0 min x 60 s ________ 1 min x 1 C ______ 1 As x 1 F ___________ 96,485 C x 1 mol e- __________ 1 F x 1 mol Al ___________ 3 mol e- x 26.98 g Al _____________ 1 mol Al = 0.280 g Al 3F-15 (of 17)

  46. Calculate the current needed to plate 0.150 grams of zinc onto an electrode in 60.0 seconds from a zinc acetate solution. Zn2+(aq) + 2e-→Zn(s) 0.150 g Zn x mol Zn _____________ 65.38 g Zn x 2 mol e- ___________ 1 mol Zn x 1 F __________ 1 mol e- x 96,485 C ___________ 1 F x 1 As ______ 1 C x 1 ________ 60.0 s = 7.38 A 3F-16 (of 17)

  47. Calculate the time, in minutes, needed to deposit 0.400 grams of chromium from a chromium (III) nitrate solution with a current of 10.0 amperes. Cr3+(aq) + 3e-→Cr(s) 0.400 g Cr x mol Cr _____________ 52.00 g Cr x 3 mol e- ___________ 1 mol Cr x 1 F __________ 1 mol e- x 96,485 C ___________ 1 F x 1 As ______ 1 C x 1 ________ 10.0 A x 1 min _______ 60 s = 3.71 min 3F-17 (of 17)

  48. REVIEW FOR TEST 3 Entropy Predict relative absolute entropies values Predict entropy changes for chemical reactions 1st Law of Thermodynamics 2ndLaw of Thermodynamics 3rd Law of Thermodynamics Free Energy

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