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21 Electrochemistry

21 Electrochemistry Electricity is such a mystery till we understand chemistry. Plain facts and sciences of electrochemistry continue to be important in technologies. . Electricity. Ancient people noticed electricity

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21 Electrochemistry

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  1. 21 Electrochemistry Electricity is such a mystery till we understand chemistry. Plain facts and sciences of electrochemistry continue to be important in technologies. 21 Electrochemistry

  2. Electricity Ancient people noticed electricity 1746 B. Franklin demonstrated lightening as electric effect and performed the kite experiment in 1751. Two people tried to repeat his kite flying experiment were killed by thunder. 1767 L. Galvani inserted two different metals in frog fluid and constructed a electric cell 1800 A. Volta substituted frog fluid; made batteries, consisted of several cells. 1802 G. Romagnosi noticed magnetism related to electricity Michael Faraday 1791-1867 discovered many theories of electricity and magnetism 21 Electrochemistry

  3. Galvani Luigi Galvani (1737-1798): erroneously concluded that the frog's nervous system generated an electrical charge, his work stimulated much research into the electrochemistry. The depiction of his laboratory  21 Electrochemistry

  4. Electrochemistry A. Volta (1745-1827) experimented with different materials, and made voltaic piles (batteries) William Nicholson (1753-1815) observed bubbles forming on the surfaces of metals submerged in water when they are connected to a voltaic pile Humphry Davy (1778-1829) observe electrolysis of water and metal salts. Following that, … Michael Faraday (1791-1860) studied electrolysis, and discovered the relationship between charges and chemical stoichiometry 21 Electrochemistry

  5. Electrons qe = –1.60217733e-19 C F = 96485 C me = 0.00054856 amu = 9.1093897e-31 kgspin = ½ (two state)magnetic moment = 9.284770e–24 J/tesla Voltaic piles (batteries) made the following study possible W. Crookes (1832-1919) observed cathode rays in low-pressure tubes. 1897:J.J. Thomson determined the charge to mass ratio (e– / me) of cathode rays (electrons). 1916R. Millikan (1868-1953) measured the amount of charge of e–. 21 Electrochemistry

  6. Redox reactions and electrons Energy drives chemical reactions. Redox reactions involve the transfer of electrons. Loss of electron (increase oxidation state) is oxidation, (Leo).Gain of electron (decrease oxidation state) is reduction, (ger). Zn  Zn2+ + 2 e–leo Cu2+ + 2 e – Cu gernet: Zn + Cu2+ Cu + Zn2+redox Chemical energy in redox reactions may be convert to electric energy by applying electrochemistry. In the meantime, we should learn to balance the redox reaction equations. 21 Electrochemistry

  7. Galvanic Cell A galvanic cell consists of two different metals inserted into a solution of an electrolyte (salt, acid or base), simulated Representation:Zn | Zn2+ || Cu2+ | Cu 21 Electrochemistry

  8. Assign oxidation states 0 for any element 1 for H in compounds, but –1 for LiH, NaH, etc – 2 for O in compounds, but –1 for H2O2, Na2O2 +1 for alkali metals, +2 for alkaline earth metals The oxidation states of other elements are then assigned to make the algebraic sum of the oxidation states equal to the net charge on the molecule or ion. –3 NH3 –2 N2H4 –1 NH2OH 0 N2 +1 N2O +2 NO +3 NO2– +4 NO2 +5 NO3– –1 Cl– 0 Cl2 +1 ClO– +3 ClO2– +4 ClO2 +5 ClO3– +7 ClO4– 21 Electrochemistry

  9. Half reaction equations Oxidation and reduction can be written as half-reaction equations such as Zn  Zn2+ + 2 e–leo Cu2+ + 2 e –  Cu ger net: Zn + Cu2+ Cu + Zn2+redox Demonstrate how to balance these Fe2+ Fe3+ + __ e– C2O42- 2 CO2 + __e – MnO4 – + __e– Mn2+ Cr2O72– + __e– + __ H+ 2 Cr3+ + __ H2O Steps to balance half reaction Assign oxidation number Figure out what is oxidized or reduced. Add electrons according to oxidation number change Balance charge with H+ (acid) or OH – (base) Balance atoms with H2O 21 Electrochemistry

  10. More half-reaction equations 2 I– I2 + __e– ClO2 + __ OH– ClO3- + __e– + H2O (in basic solution) 2 S2O32–  S4O62- + __e– HS(=S)O3–S + __e- + HSO4– __ H3O+ + __e – H2(g) + __ H2O H2O2 + __e – 2 H2O ClO2 + __e – ClO2– NO3– + __e – NH4+ 21 Electrochemistry

  11. Electrochemical Series An electrochemical series is a list of metals in order of decreasing strength as reductant, or increasing strength as oxidant.An example of an activity series of metals based on the Standard Potentials given would be: K > Ba > Ca > Na > Mg > Al > Mn > Zn > Fe > Ni > Sn > Pb > Cu > Ag In this series the most active metal is potassium (K) and the least active metal is silver (Ag) Reactions and cells are illustrated in 21-1 of Text (PHH). 21 Electrochemistry

  12. Constructing half cells A half cell consists of an oxidizing and its oxidized species Zn | Zn2+Cu | Cu2+Pt | H2 | H+ (Pt as conductor)Pt |Fe2+ , Fe3+Cl– | Cl2 | Pt Student cell set from School Master Science $30 Explain the cell convention and reactions of cells. 21 Electrochemistry

  13. Galvanic cells Using corns for a galvanic cells is illustrated by an Internet site: (schoolnet.ca/general/electric-club/e/page9.html) This picture illustrates a way to make a pact of battery using coins of different metals. Apply the principles you have learned regarding electrochemical series and electrolytes to make such a pile will be an interesting exercise. 21 Electrochemistry

  14. Cell convention Oxidation takes place always at the anode Zn (s) Zn2+ (aq) + __ e– Zn(s) | Zn2+(aq) Fe2+ (aq) Fe3+ (aq) + e– Pt | Fe2+ , Fe3+ H2 (g) 2 H+ (aq) + __ e– Pt | H2(g) | H+(aq) Reduction takes place at the cathode Cu2+ (aq) + __ e– Cu (s) Cu2+ | Cu(s) Cl2 (g) + __ e– Cl– (aq) Cl2(g) | Cl –(aq) | Pt Fe3+ (aq) + __ e – Fe2+ (aq) Fe3+(aq), Fe2+(aq)| Pt 2 H+ (aq) + __ e– H2 (g) H+(aq) | H2(g)| Pt Concentration (1.1 M) and pressure (0.9 atm) are also included in cell notations. 21 Electrochemistry

  15. Electric energy and work Electric energy or electric work = charge * potential difference W = q * V (1 J = 1 Coulomb Volt, C V)compare W = m g h The Faraday constant F is the charge for one mole of electrons,F = 96485 C; F / NA = 96485 C / 6.022e23 = 1.602177e-19 C, charge per e– The maximum chemical energy of the cell that can be converted to electric work is the Gibbs free energy change, GG = – n F E (n F= q is the charge)electromotive force (emf, or V)number of electrons in the reaction equationn F is the charge q See slides in 16 Equilibria 21 Electrochemistry

  16. Standard cell emf’s and electrode potentials The standard cell emf is the emf of a voltaic cell operating under standard conditions (1 M, 1 atm, 25oC etc). E.g. Zn (s) | Zn2+ (aq, 1 M) || Cu2+ (aq, 1 M) | Cu (s) Eo = 1.10 V Mg (s) | Mg2+ (aq, 1 M) || Cu2+ (aq, 1 M) | Cu (s) Eo = 2.90 V The absolute potential of the electrode cannot be determined. Only relative potential can be measured. The standard reduction potential is measured against a SHE, for which Eo = 0.0000 V Zn (s) | Zn2+(aq, 1 M) || H+(aq, 1 M) | H2(1 atm)(g) Eo = 0.76 V Zn = Zn2+ + 2 e–(oxidation, displace H+ possible)Eo = 0.76 V Zn2+ + 2 e– Zn (reduction)Eo = – 0.76 V 21 Electrochemistry

  17. Gibb’s Free Energy in a Cell How much energy is available for the cell Zn | Zn2+ || Ag+ | Agoperating at standard condition when one mole of Zn is consumed ? SolutionEo Zn = Zn2+ + 2 e 0.762 V 2 Ag+ + 2e = 2 Ag 0.799 V (from table) Zn + 2 Ag+ = Zn2+ + 2 Ag Eo = 1.561 V Go = – nFE = – 2 * 96485 C * 1.561 V = 301226 J (1J = 1 C V) = 301.2 kJ How much silver is consumed?How much energy is available if 6.5 g of Zn is consumed? 21 Electrochemistry

  18. Table of standard reduction potential Reaction E o (V) Li+ + e– = Li (s) – 3.04 Na+ + e– = Na (s) – 2.71 Mg2+ + 2 e– = Mg (s) – 2.38 Zn2+ + 2 e– = Zn (s) – 0.76(reference)2 H+ + 2 e– = H2 (g) 0.000(reference)H2 (g) =2 H+ + 2 e– 0.000 Cu2+ + 2 e– = Cu (s) 0.34 Cu+ + e– = Cu (s) 0.52 I2 (s) + 2e– = 2 I– (aq) 0.54 Br2 (l) + 2 e– = 2 Br– (aq) 1.07 Cl2 (g) + 2 e– = 2 Cl– (aq) 1.36 F2 (g) + 2 e– = 2 F– (aq) 2.87 Standard cell potentials Cell E o Li | Li+ || Cu2+ | Cu ____ Mg | Mg2+ || I2 | I– | Pt ____ Zn | Zn2+ || Br2 | Br– | Pt ____ Cu | Cu2+ || Zn2+ | Zn ____ Which is not spontaneous? See 21-2 21 Electrochemistry

  19. Table of standard reduction potential Reaction E o (V) F2 (g) + 2 e– = 2 F– (aq) 2.87 Cl2 (g) + 2 e– = 2 Cl– (aq) 1.36 Br2 (l) + 2 e– = 2 Br– (aq) 1.07 I2 (s) + 2e– = 2 I– (aq) 0.54 Cu+ + e– = Cu (s) 0.52 Cu2+ + 2 e– = Cu (s) 0.34(reference)H2 (g) =2 H+ + 2 e– 0.000(reference)2 H+ + 2 e– = H2 (g) 0.000 Zn2+ + 2 e– = Zn (s) – 0.76 Mg2+ + 2 e– = Mg (s) – 2.38 Na+ + e– = Na (s) – 2.71 Li+ + e– = Li (s) – 3.04 The listing order in the table may be different in different text books. However, the principles and methods of application remain the same. This is the order given on the Exam Data Sheet, that is different from the text. 21 Electrochemistry

  20. Strength of oxidation The ability of a chemical to oxidize is its ability to take electrons from other species, Oxidizing agent + n e  Reduced species Strength of oxidation of an oxidizing agent is measured by its reduction potential. Similarly, strength of reduction of a reducing agent is measured by its oxidation potential. Oxidized species  Reducing agent + n e Be able to order the species according to oxidizing strength. _____ 21 Electrochemistry

  21. Reaction direction and emf What is the emf for the reaction, Zn2+ (aq) + 2Fe2+ (aq) = Zn (s) + 2Fe3+ (aq)? Solution: Know what data to look for Zn2+ + 2 e  Zn Eo = – 0.76 V Fe3+ + e  Fe2+Eo = + 0.77 V +2 Fe2+2 Fe3+ + 2 e Eo = – 0.77 V Zn2+ (aq) + 2 Fe2+ (aq)  Zn (s) + 2Fe3+ (aq) Eo = – 1.53 Vnon-spontaneous Pt | Zn2+ | Zn || Fe2+ | Fe3+ | Pt impractical The reverse reaction is spontaneous, Zn (s) + 2Fe3+ (aq)  Zn2+ (aq) + 2Fe2+ (aq) Eo = + 1.53 V Zn | Zn2+ || Fe3+ | Fe2+ | Pt 21 Electrochemistry

  22. Free energy and emf What is the free-energy change for the cell, Zn | Zn2+ (aq) 1 M || Ag+ (aq) 1 M | Ag? Solution: Reduction potential required, Zn  Zn2+ + 2e Eo = 0.76 Ag+ + e  Ag Eo = 0.80 2 Ag+ + 2e  2Ag Eo = 0.80 Cell reaction 2 Ag+ + Zn  2Ag + Zn2+ Eo = 1.56 V Go = –n F Eo = – 2 * 96485 * 1.56 = – 3.01e5 J or – 302 kJ Negative indicate energy is released. The free energy for the cell is –301 kJ per mole of Zn, what is the emf? Condition for spontaneous reaction is – G or + E. 21 Electrochemistry

  23. General cell emf G o is the standard energy change. G is for non-standard conditions. G = – n F E Similarly, Eo is the standard emf whereas E is general emf. G = Go + R T ln Q E = Eo – R T / n Fln Qreaction quotient When a system is at equilibrium (Q = K), G = 0. Therefore, G = Go + R T lnK= 0 E = Eo – R T / n Fln K = 0equilibrium constant  Go = – R T lnKE o = R T / n Fln K orGo = – ln(10) R T logKE o = 2.303 R T / n FlogK Text uses Ecell instead of E At 298 K 0.0592E o = ———— log K n 21 Electrochemistry

  24. The Nernst equation For a general reaction, a A + b B = c C + d D R T [C]c [D]dE = E o – ——— ln ———— n F [A]a [B]b This Nernst equation correlates cell emf with [ ] or reactivities of reactants and products as well as T Units for [ ]: mol L-1 for aqueous solution, atm (for gas), and constant for solid and liquid. See 21-4 21 Electrochemistry

  25. R T [C]c [D]dE = E o – ——— ln ———— n F [A]a [B]b Evaluating E At 300 K, evaluate the cell emf for Zn | Zn2+ (0.100 M) || H+ (0.200 M) | H2 (1.111 atm) | Pt Solution:Look up: Zn  Zn2+ (aq) + 2e, E o = 0.76 V 2 H+ (aq) + 2 e  H2 (g), E o = 0.00 V (R = 8.314 J mol-1 K-1, F = 96485 C mol-1)E o = 0.76 VThe reaction is Zn(s) + 2 H+ (aq)  Zn2+ (aq) + H2(g) 8.314 J mol-1 K-1 * 300 K (0.100) (1.111) E = 0.76 – —————————— ln —————— 2 * 96485 C mol-1 (0.200)2 = 0.76 – 0.0129 * (1.02) = 0.76 – 0.013 = 0.75 V See example 19.12 21 Electrochemistry

  26. Concentration cell Problem: At 298 K, evaluate the emf of the cellCu | Cu2+ (0.10 M) | | Cu2+ (1.0 M) | Cu Cu(s)  Cu2+(0.1 M) + 2e; Cu2+ (1.0 M) + 2e  Cu(s) Solution:The standard emf (Eo = 0.00)for Cu | Cu2+ || Cu2+ | Cu The reaction is actually Cu (s) + Cu2+ (1.0 M) = Cu2+ (0.1 M) + Cu (s) R T [Cu2+] R T 0.10E = 0.00 – ——- ln ——— = – –––– ln ––––– 2 F [Cu2+] 2 F 1.00 8.3145 * 298 1.0 = +–––––––––– ln –––– = 0.0295 V 2*96485 0.1The voltage is purely due to concentration difference. Solutions in the two compartment try to become equal. When 2 [ ]’s are equal, E = 0 See p. 841 21 Electrochemistry

  27. Equilibrium Constant K and Eocell Calculate the solubility product of AgCl from data of standard cell Solution: Look up desirable data Ag+Cl– (s) + e  Ag0(s) + Cl–E° = 0.2223 V Ag+ (aq) + e  Ag (s) E° = 0.799 V Ag (s)  Ag+ (aq) + e E° = – 0.799 V Get the desirable eq’n AgCl (s)  Ag+ (aq) + Cl– (aq) DE° = – 0.577 V Ksp = [Ag+][Cl– ] log Ksp = – 0.577 / 0.0592 = – 9.75Ksp = 10– 9.75 = 1.8e–10 Show that for this cellAg | Ag+,1 M || Cl–,1 M | AgCl | AgEo = – 0.577 Vbut for this cellAg | AgCl |Cl–, 1 M || Ag+,1 M | AgEo = +0.577 V At 298 K 0.0592Eo = ———— log K n 21 Electrochemistry See p. 837

  28. Evaluate free-energy change Evaluate G o for the reaction Zn (s) + 2 Ag+ (aq) = Zn2+(aq) + 2 Ag (s) Solution: Required to look up: Eo V Ag+ + e = Ag 0.80 Zn2+ + 2 e = Zn – 0.76 2 Ag+ + 2 e  2Ag 0.80 + Zn  Zn2+ + 2 e + 0.76 Zn (s) + 2 Ag+ (aq)  Zn2+(aq) + 2 Ag (s) 1.56 (= E o) G o = –n F E o= – 2 * 96485 C * 1.56 V = – 301000 J = – 301 kJ See slides 17 and 22 Write the cell for this rxn 21 Electrochemistry

  29. Summary of thermodynamics Chemical energyHo, So Stoichiometry n Electric energyGo = – n F Eo Go = Ho – T So Reaction quotient &equilibrium constant G = G o + R T ln QG o = – R T ln K Reaction quotient &equilibrium constant E = E o – R T/n F ln QE o = R T/n F ln K 21 Electrochemistry

  30. insight from cold denaturation and a two-state water structure • By Tsai CJ, Maizel JV Jr, Nussinov R. • …The exposure of non-polar surface reduces the entropy and enthalpy of the system, at low and at high temperatures. At low temperatures the favorable reduction in enthalpy overcomes the unfavorable reduction in entropy, leading to cold denaturation. At high temperatures, folding/unfolding is a two-step process: in the first, the entropy gain leads to hydrophobic collapse, in the second, the reduction in enthalpy due to protein-protein interactions leads to the native state. The different entropy and enthalpy contributions to the Gibbs energy change at each step at high, and at low, temperatures can be conveniently explained by a two-state model of the water structure…. • Biochem Mol Biol. 2002;37(2):55-69 21 Electrochemistry

  31. pH and emf Consider the cell, Zn | Zn2+ (1.00 M) || H+ (x M) | H2 (1.00 atm) | Pt From table data, Zn2+ + 2e = Zn Eo = – 0.76 2 H+ + 2 e = H2Eo = 0.00 Zn = Zn2+ + 2e Eo = 0.76 Zn + 2H+ = Zn2+ + H2Eo = 0.76 R T [Zn2+] PH2E = Eo – —— ln ————2 F [H+]2 At 298 K (pH meters) 0.76 – EpH = ————— 0.0592 = Eo + 0.0592 log [H+] = 0.76 – 0.0592 pH 21 Electrochemistry

  32. pH electrodes pH Range: 0-14Temp. Range: 0-100 CInternal Ref: ROSSJunction: CeramicDimensions: 120 mm x 12 mmSlope: 92 - 102%Temp. Accuracy: 0.5 CCatalog Number: 8202BN (BNC Connector, 1 meter cable) 21 Electrochemistry

  33. Ion selective electrode More research has gone into pH measurements. Nernst started it. • Concentration; • Temperature; • Electrode surface conditions; • Number of charges of ions (8); • Stirring (6); • Suspension (7); • Zwitterionic nature, net charge density; • Anything changing ionic adsorption; • Isoelectric nature of surface material; • The Nernst equation deals only with concentration and temperature 21 Electrochemistry

  34. Battery technology By use: automobile, flash light, radio, computer, camera, watch, emergency equipment, artificial heart machine, pace makers, hearing aids, calculators, … (portable energy) By type: alkaline, dry, wet, storage, rechargeable, etc By chemistry: alkaline, carbon zinc, dry, cell, lithium, lithium ion, lithium polymer, NiCd, etc. By material: anode material, cathode material, electrode, etc. Aluminum for battery manufacture See 21-5 21 Electrochemistry

  35. Lead Storage Battery for Autos Anode – Negative platePb + SO42-  PbSO4 +  2e- Separator Cathode – Positive plate A 12-V battery consists of 6 such cells H2SO4 PbO2 +  4H+  +  SO42- +  2e-   PbSO4 + 2 H2O Net reaction: Pb + PbO2 + 2 H2SO4 = 2 PbSO4 + H2O 21 Electrochemistry

  36. 21 Electrochemistry

  37. A mercury battery. 21 Electrochemistry

  38. Corrosion: Unwanted Voltaic Cells Fe(s)  Fe2+ (aq) + 2e–O2 + H2O (l) + 4e– 4 OH– (aq) 2 Fe(s) + O2 + H2O  2 Fe2+ (aq) + 4 OH– (aq) What are effective corrosion prevention methods? CoatingUse sacrifice electrode 21 Electrochemistry See 21-6

  39. Cathodic protection of an underground pipe. 21 Electrochemistry

  40. Ion displacement reactions (corrosion) Reaction E o (V) Li+ + e– = Li (s) – 3.04 Na+ + e– = Na (s) – 2.71 Mg2+ + 2 e– = Mg (s) – 2.38 Zn2+ + 2 e– = Zn (s) – 0.76(reference)2 H+ + 2 e– = H2 (g) 0.000(reference)H2 (g) =2 H+ + 2 e– 0.000 Cu2+ + 2 e– = Cu (s) 0.34 Cu+ + e– = Cu (s) 0.52 I2 (s) + 2e– = 2 I– (aq) 0.54 Ag+ (aq) + e- = Ag (s) 0.80 Br2 (l) + 2 e– = 2 Br– (aq) 1.07 Cl2 (g) + 2 e– = 2 Cl– (aq) 1.36 F2 (g) + 2 e– = 2 F– (aq) 2.87 What metal will react with certain ions? Zn + 2 Ag+ Zn2+ + 2 Ag Zn + 2 Cu2+ Zn2+ + Cu Zn2+ + 2 Ag  Zn + 2 Ag+ Zn2+ + Cu  Zn + 2 Cu2+ See 21-1 21 Electrochemistry

  41. Electrolysis of molten salts Eo=-2.71V;2Na+ + 2e– 2Na Battery Eo=-1.36 V; 2 Cl– Cl2 + 2e– e e oxidation reduction ANODE CATHODE 805o C Molten salts consists of Na+ and Cl– ions 2 Cl– Cl2 + 2e– 2 Na+ + 2 e– 2 Na Net2 NaCl  2 Na + Cl2 Charges required to produce 1 mole Cl2 and 2 moles Na = 2F Energy required = 2 FE; (E > 4.07 V) 21 Electrochemistry See 21-7

  42. Electrometallurgy of Sodium 21 Electrochemistry

  43. Electrolysis of NaCl solution 2 Cl– = Cl2 + 2e–Cl2 + H2O = HCl + ½ O2 2Na+ + 2e– = 2Na2Na + 2H+ = H2 + 2Na+ Battery e e oxidation reduction ANODE CATHODE Salt solution consists of Na+ and Cl– ions 21 Electrochemistry

  44. Refining Copper by Electrolsis Copper can be purified by electrolysis. Raw copper is oxidized Cu = Cu2+ + 2e and purer copper deposited on to the cathode from a solution containing CuSO4 Cu2+ + 2e = Cu If a current of 2 amperes pass through the cell, how long will it take to deposit 5.00 g of copper on the cathode? (1 ampere = 1 C s–1) Solution 5.00 2*96485 C 1 s------ mol ----------- ------ = work out your answer New65.5 1 mol 2 C _______ 21 Electrochemistry

  45. Aluminum (Al), the third most abundant elements on Earth crust as bauxite or alumina Al2O3, remain unknown to man until 1827, because it is very reactive. By then, Wohler obtained some Al metal by reducing Al2O3 with potassium vapore. In 1886, two young men electrolyzed molten cryolite Na3AlF6 (melting point 1000° C), but did not get aluminum. Production of aluminum Hall and Heroult tried to mix about 5% alumina in their molten cryolite, and obtained Al metal. This is the Hall process. AlF63– + 3 e– Al + 6 F– . . . Cathode 2 Al2OF62– + C(s) + 12 F– + 4 AlF63– + CO2 + 4 e– . . . Anode 2 Al2O3 + 3 C  4 Al + 3 CO2. . . Overall cell reaction Charge required for each mole Al = 3 F Energy required = 3 FE 21 Electrochemistry

  46. Electrometallurgy of Aluminum 21 Electrochemistry

  47. Electrolysis of acid solution Battery H2O = ½ O2 + 2e– + 2 H+ 2H+ + 2e– = H2 e e oxidation reduction ANODE CATHODE Solutions containing H+ and SO42– ions Charges required to produce 1 mole H2 and ½ moles O2 = 2F Energy required = 2 FE 21 Electrochemistry

  48. Electrolysis of H2SO4 solution Pure water is not a good electric conductor. In the presence of electrolytes, water can be decomposed by electrolysis. On the other hand, electrolysis of electrolyte solutions may reduce H+ and oxidize O2– in H2O. In an H2SO4 solution,cathode reductions are 2 H2O (l) + 2 e– = H2 (g) + 2 OH–(same as 2H+ + 2e– = H2) Anode oxidation: 2 H2O (l) = 4 e– + O2 (g) + 4 H+E o = – 1.23 V (observed) 2 SO42– = [SO3O–OSO3]2– + 2 e–E o = – 2.01 V (not observed) 21 Electrochemistry

  49. Electrolysis of H2SO4 solution E o = – 1.23 V 2 H2O (l) = 4 e– + O2 (g) + 4 H+E o = – 2.01 V2 SO42– = [SO3O–OSO3]2– + 2 e– Battery 2H+ + 2e = H2 e reduction e oxidation ANODE CATHODE Solution consists of H+ and SO42– ions 21 Electrochemistry

  50. Electroplating of metals Galvanizing Zn2+ + 2 e– Zn onto metal surface Copper purification Cu2+ + 2 e– Cu onto pure Cu electrode Silver plating Ag+ + e– Ag onto metal surface • Over a half century of extensive and innovative research has made us one of the nation's leading experts in plating on magnesium Since 1971, Cal-Aurum has provided electroplating services to the electronic component industry with the highest standards of quality, performance, and competitive pricing. Miller specializes in plating metals such as magnesium, aluminum, zinc, copper, powdered metals, steel and various other substrates. 21 Electrochemistry

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