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Slide 2 1http:\\asadipour.kmu.ac.ir 76 slides920311 Slide 3 2 2 Electrochemistry http:\\asadipour.kmu.ac.ir 76 slides920311 Slide 4 3 3 Electrochemistry All of Chemical reactins are related to ELECTRONS Redox reactions http:\\asadipour.kmu.ac.ir 76 slides920311 Slide 5 4 Electric power conversion in electrochemistry Chemical Reactions Electric Power Power consumption Power generation Electrolysis Galvanic cells http:\\asadipour.kmu.ac.ir 76 slides920311 Slide 6 5 5 Electrochemistry Conduction 1)Metalic 2)Electrolytic Temprature Motion of ions Resistance -------------------------------- ----- http:\\asadipour.kmu.ac.ir 76 slides920311 Slide 7 6 battery +- power source e-e- e-e- Ions Chemical change (-)(+) Aqueous NaCl Interionic attractions................................ Ions Solvation . Solvent viscosity .. Na + Cl - H2OH2O Electrolytic conduction Ion-Ion Attr. Ion- Solvent Attr. SolventSolvent Attr. Temprature Attractions & Kinetic energy Conduction http:\\asadipour.kmu.ac.ir 76 slides920311 Conduction Ions mobility Slide 8 7 battery +- inert electrodes power source vessel e-e- e-e- conductive medium Electrolytic Cell Construction http:\\asadipour.kmu.ac.ir 76 slides 920311 Slide 9 8 +- battery Na (l) electrode half-cell Molten NaCl Na + Cl - Na + Na + + e - Na2Cl - Cl 2 + 2e - Cl 2 (g) escapes Observe the reactions at the electrodes NaCl (l) (-) Cl - (+) http:\\asadipour.kmu.ac.ir 76 slides920311 Slide 10 9 +- battery e-e- e-e- NaCl (l) (-)(-)(+)(+) cathode anode Molten NaCl Na + Cl - Na + Na + + e - Na 2Cl - Cl 2 + 2e - cations migrate toward (-) electrode anions migrate toward (+) electrode At the microscopic level http:\\asadipour.kmu.ac.ir 76 slides920311 Slide 11 10 Molten NaCl Electrolytic Cell cathode half-cell (-) REDUCTION Na + + e - Na anode half-cell (+) OXIDATION2Cl - Cl 2 + 2e - overall cell reaction 2Na + + 2Cl - 2Na + Cl 2 X 2 Non-spontaneous reaction! http:\\asadipour.kmu.ac.ir 76 slides920311 Slide 12 11 What chemical species would be present in a vessel of aqueous sodium chloride, NaCl (aq)? Na + Cl - H2OH2O Will the half-cell reactions be the same or different? http:\\asadipour.kmu.ac.ir 76 slides920311 Slide 13 Water Complications in Electrolysis In an electrolysis, the most easily oxidized and most easily reduced reaction occurs. When water is present in an electrolysis reaction, then water (H 2 O) can be oxidized or reduced according to the reaction shown. ElectrodeIons...Anode RxnCathode Rxn E Pt (inert)H 2 O H 2 O (l) + 2e- H 2(g) + 2OH - (aq) -0.83 V H 2 O 2 H 2 O (l) 4e - + 4H + (g) + O 2(g) -1.23 V Net Rxn Occurring: 2 H 2 O 2 H 2(g) + O 2 (g) E = - 2.06 V Slide 14 13http:\\asadipour.kmu.ac.ir 76 slides920311 Slide 15 14 battery +- power source e-e- e-e- NaCl (aq) (-)(+) cathode different half-cell Aqueous NaCl anode 2Cl - Cl 2 + 2e - Na + Cl - H2OH2O What could be reduced at the cathode? http:\\asadipour.kmu.ac.ir 76 slides 2H 2 O + 2e - H 2 + 2OH - 920311 Slide 16 15 Aqueous NaCl Electrolysis possible cathode half-cells (-) REDUCTION Na + + e - Na 2H 2 O + 2e - H 2 + 2OH - possible anode half-cells (+) OXIDATION2Cl - Cl 2 + 2e - 2H 2 O O 2 + 4H + + 4e - overall cell reaction 2Cl - + 2H 2 O H 2 + Cl 2 + 2OH - http:\\asadipour.kmu.ac.ir 76 slides920311 Slide 17 16 Aqueous CuCl 2 Electrolysis possible cathode half-cells (-) REDUCTION Cu 2+ + 2e - Cu 2H 2 O + 2e - H 2 + 2OH - possible anode half-cells (+) OXIDATION2Cl - Cl 2 + 2e - 2H 2 O O 2 + 4H + + 4e - overall cell reaction Cu 2+ + 2Cl - Cu (s) + Cl 2(g) http:\\asadipour.kmu.ac.ir 76 slides920311 Slide 18 17 Aqueous Na 2 SO 4 Electrolysis possible cathode half-cells (-) REDUCTION Na + + e - Na [2H 2 O + 2e - H 2 + 2OH - ] possible anode half-cells (+) OXIDATION SO 4 2- S 4 O 8 2_ + 2e - 2H 2 O O 2 + 4H + + 4e - overall cell reaction 6H 2 O 2H 2 + O 2 +4H + + 4OH - http:\\asadipour.kmu.ac.ir 76 slides920311 2 Slide 19 18 Faradays Law The mass deposited or eroded from an electrode depends on the quantity of electricity. Quantity of electricity = coulomb (Q) Q = It coulomb current in amperes (amp) time in seconds http:\\asadipour.kmu.ac.ir 76 slides920311 Slide 20 19 e-e- Ag + Ag For every electron, an atom of silver is plated on the electrode. Ag + + e - Ag Electrical current is expressed in terms of the ampere, which is defined as that strength of current which, when passed thru a solution of AgNO 3 (aq) under standard conditions, will deposit silver at the rate of 0.001118 g Ag/sec Experimentally: 1 amp = 0.001118 g Ag/sec http:\\asadipour.kmu.ac.ir 76 slides920311 1 coulomb = 1 amp-sec = 0.001118 g Ag Slide 21 20 Ag + + e - Ag 1.00 mole e - = 1.00 mole Ag = 107.87 g Ag 107.87 g Ag/mole e - 0.001118 g Ag/coul = 96,485 coul/mole e - 1 Faraday ( F ) mole e - = Q/ F http:\\asadipour.kmu.ac.ir 76 slides920311 1C=1AS /// 1J=1CV Slide 22 21 A series of solutions have 50,000 coulombs passed thru them, if the solutions were Au +3, Zn +2, and Ag +, and Au, Zn, and Ag were plated out respectively, calculate the amount of metal deposited at each anode. battery -+ +++--- 1.0 M Au +3 1.0 M Zn +2 1.0 M Ag + Au +3 + 3e - AuZn +2 + 2e - ZnAg + + e - Ag e-e- e-e- e-e- e-e- http:\\asadipour.kmu.ac.ir 76 slides920311 Slide 23 22 Examples using Faradays Law 1)How many grams of Cu will be deposited in 1L of A)0.1 M CuSO 4 B) 1 M CuSO 4 After 3.00 hours electrolysis by a current of 4.00 amps?(Cu=64) Cu +2 + 2e - Cu 2)The charge on a single electron is 1.6021 x 10 -19 coulomb. Calculate Avogadros number from the fact that 1 F = 96,487 coulombs/mole e -. http:\\asadipour.kmu.ac.ir 76 slides920311 Slide 24 23http:\\asadipour.kmu.ac.ir 76 slides920311 Slide 25 24http:\\asadipour.kmu.ac.ir 76 slides920311 Slide 26 25 21-8 Industrial Electrolysis Processes Slide 25 of 52 http:\\asadipour.kmu.ac.ir 76 slides920311 Slide 27 26http:\\asadipour.kmu.ac.ir 76 slides920311 Slide 28 27 Voltas battery (1800) Alessandro Volta 1745 - 1827 Paper moisturized with NaCl solution Cu Zn http:\\asadipour.kmu.ac.ir 76 slides920311 Slide 29 28 Cu 1.0 M CuSO 4 Zn 1.0 M ZnSO 4 Salt bridge KCl in agar Provides conduction between half-cells Galvanic Cell Construction Observe the electrodes to see what is occurring. http:\\asadipour.kmu.ac.ir 76 slides920311 Slide 30 29 Cu 1.0 M CuSO 4 Zn 1.0 M ZnSO 4 Cu plates out or deposits on electrode Zn electrode erodes or dissolves Cu +2 + 2e - Cu cathode half-cell Zn Zn +2 + 2e - anode half-cell Anod - Cathod + What about half-cell reactions? What about the sign of the electrodes? What happened at each electrode? Why? http:\\asadipour.kmu.ac.ir 76 slides920311 Compare with Electrolytic cells Slide 31 30 +- battery e-e- e-e- NaCl (l) (-)(-)(+)(+) Cathode - Anode + Electrolytic cells sign of the electrodes? Na + Cl - Na + Na + + e - Na 2Cl - Cl 2 + 2e - http:\\asadipour.kmu.ac.ir 76 slides920311 Slide 32 31 Electrodes are passive (not involved in the reaction) Olmsted Williams http:\\asadipour.kmu.ac.ir 76 slides920311 Slide 33 32 H 2 input 1.00 atm inert metal How do we calculate Standard Redox Potentials? We need a standard electrode to make measurements against! The Standard Hydrogen Electrode (SHE) Pt 1.00 M H + 25 o C 1.00 M H + 1.00 atm H 2 Half-cell 2H + + 2e - H 2 E o SHE = 0.0 volts http:\\asadipour.kmu.ac.ir 76 slides920311 Slide 34 33 19.3 E 0 is for the reaction as written E 0 red // E 0 ox The more positive E 0 the greater the tendency for the substance to be reduced The half-cell reactions are reversible The sign of E 0 changes when the reaction is reversed Changing the stoichiometric coefficients of a half-cell reaction does not change the value of E 0 http:\\asadipour.kmu.ac.ir 76 slides920311 Strongest oxidunt Strongest reductant Slide 35 34 Measuring E 0 red Cu 2+ & Zn 2+ Slide 34 of 52 cathode anode http:\\asadipour.kmu.ac.ir 76 slides920311 Cu +2 + 2e - Cu E=E 0 red Zn Zn +2 + 2e - E=E 0 ox -E=E 0 red Slide 36 35 Cu 1.0 M CuSO 4 Zn 1.0 M ZnSO 4 cathode half-cell Cu +2 + 2e - Cu anode half-cell Zn Zn +2 + 2e - - + Measuring E 0 of a cell 1.1 volts http:\\asadipour.kmu.ac.ir 76 slides920311 ? Slide 37 36 What is the standard emf of an electrochemical cell made of a Cd electrode in a 1.0 M Cd(NO 3 ) 2 solution and a Cr electrode in a 1.0 M Cr(NO 3 ) 3 solution? Cd 2+ (aq) + 2e - Cd (s) E 0 = -0.40 V Cr 3+ (aq) + 3e - Cr (s) E 0 = -0.74 V Cd is the stronger oxidizer Cd will oxidize Cr 2e - + Cd 2+ (1 M) Cd (s) Cr (s) Cr 3+ (1 M) + 3e - Anode (oxidation): Cathode (reduction): 2Cr (s) + 3Cd 2+ (1 M) 3Cd (s) + 2Cr 3+ (1 M) x 2 x 3 E 0 cell = -0.40 +0.74=0.34 cell E 0 = 0.34 V cell 19.3 http:\\asadipour.kmu.ac.ir 76 slides E 0 = -0.40 V E 0 = 0.74 V 920311 E 0 cell = ? !! Slide 38 37 Calculating the cell potential, E o cell, at standard conditions Fe +2 + 2e - Fe E o = -0.44 v O 2 (g) + 2H 2 O + 4e - 4 OH - E o = +0.40 v This is spontaneoues corrosion or the oxidation of a metal. Consider a drop of oxygenated water on an iron object Fe H 2 O with O 2 Fe Fe +2 + 2e - -E o = +0.44 v2x 2Fe + O 2 (g) + 2H 2 O 2Fe(OH) 2 (s) E o cell = +0.84 v reverse http:\\asadipour.kmu.ac.ir 76 slides Fe + O 2 (g) + H 2 O Fe(OH) 2 (s) 920311 Which one is oxidunt ? Slide 39 38http:\\asadipour.kmu.ac.ir 76 slides920311 Slide 40 39 G o = -n F E o cell Free Energy and the Cell Potential Cu Cu +2 + 2e - E o = - 0.34 Ag + + e - Ag E o = + 0.80 v 2x Cu + 2Ag + Cu +2 + 2Ag E o cell = +0.46 v where n is the number of electrons for the balanced reaction What is the free energy for the cell? 1 F = 96,500 J/v http:\\asadipour.kmu.ac.ir 76 slides Cu + 2Ag + Cu +2 + 2Ag G o = -2965000.46=-88780 J 920311 Slide 41 40 - E depends on: -Related half reaction - Concentration -kinetic ------------------------------------------------------ 2e - +2H + H 2 E 0 = 0.000 Fe 3e - +Fe 3+ E 0 = 0.036 ------------------------------------------ Fe +H + Fe 3+ +H 2 E 0 = 0.036 Spontaneous redox reaction ????? !!!!!!!No =========================================================================================== http:\\asadipour.kmu.ac.ir 76 slides 0.036 V 920311 - 0.337 V Slide 42 41 - 0.337 V http:\\asadipour.kmu.ac.ir 76 slides e - +Cu + Cu E 0 = 0.521 V Cu + Cu 2+ +e - E 0 = -0.153 V ------------------------------------------- 2Cu + Cu 2+ +Cu E 0 = 0.368V 2Cu + Cu 2+ +Cu Auto redox=Dis proportionation 920311 Slide 43 42http:\\asadipour.kmu.ac.ir 76 slides 0.036 V Auto redox=Dis proportionation ?????? 2e - +Fe 2+ Fe E 0 = -0.440 V Fe 2+ Fe 3+ +e - E 0 = -0.771 V 920311 2 ------------------------------------------- 3Fe 2+ 2Fe 3+ +Fe E 0 = -1.221V NO Slide 44 43http:\\asadipour.kmu.ac.ir 76 slides -0.036 V 920311 ------------------------------------------------------- 3e +Fe 3+ Fe E0=+0.331 ? No e isnt a function state 1) e +Fe 3+ Fe 2+ E0= 0.771 2) 2e +Fe 2+ Fe E0=-0.440 2e - +Fe 2+ Fe E 0 = -0.440 V Fe 2+ Fe 3+ +e - E 0 = -0.771 V ------------------------------------------- 3Fe 2+ 2Fe 3+ +Fe E 0 = -1.221V Slide 45 44 G 0 =-nE 0 f= -3E 0 f http:\\asadipour.kmu.ac.ir 76 slides920311 G 0 =-nE 0 f 2) 2e +Fe 2+ Fe E 0 =-0.440 1) e +Fe 3+ Fe 2+ E 0 = 0.771 G 0 =-1(+0.771) F=-0.771f G 0 =-2(-0.440) F=+0.880f 3e +Fe 3+ Fe G 0 =+0.109f ------------------------------------------------------ =+0.109f 3E 0 =-0.109 E 0 =-0.036 v Slide 46 Free Energy and Chemical Reactions 45 W W q q GG GG HH HH TSTS TSTS Spontaneous reaction Ideal reverse cell Operating cell 920311http:\\asadipour.kmu.ac.ir 76 slides G = H - TS W = H - q Slide 47 46http:\\asadipour.kmu.ac.ir 76 slides Ni (s) | Ni 2+ (XM) || Sn 2+ (YM) | Sn (s) A cell 2 e - + Sn 2+ Sn (s) Ni (s) 2 e - + Ni 2+ Ni (s) + Sn 2+ Ni 2+ + Sn (s) Redox reaction Cathode Anode Representation of a cell 920311 Slide 48 47http:\\asadipour.kmu.ac.ir 76 slides Ni (s) | Ni 2+ (1M) || Sn 2+ (1M) | Sn (s) Ni (s) 2 e - + Ni 2+ E =0.230 V Ni (s) + Sn 2+ (1M) Ni 2+ (1M) + Sn (s) CathodeAnode Emf of a standard cell E =0.230 -0.140 =0.090V 2 e - + Sn 2+ Sn (s) E=-0.140V 920311 ------------------------------------ Slide 49 48 Effect of Concentration on Cell EMF A voltaic cell is functional until E = 0 at which point equilibrium has been reached. The point at which E = 0 is determined by the concentrations of the species involved in the redox reaction. The Nernst Equation http:\\asadipour.kmu.ac.ir 76 slides920311 E = E o RT ln Q n /-nf E = E o - 0.0591 log Q n Slide 50 49 Effect of Concentration on Cell EMF at 25 o C: E = E o - 0.0591 log Ni 2+ / Sn 2+ n Calculate the E red for the hydrogen electrode where 0.50 M H + and 0.95 atm H 2. http:\\asadipour.kmu.ac.ir 76 slides Ni (s) | Ni 2+ (XM) || Sn 2+ (YM) | Sn (s) Ni (s) + Sn 2+ (YM) Ni 2+ (XM) + Sn (s) E= 0.090 V 920311 Q= Ni 2+ / Sn 2+ E=0.090-0.059/2logx/y E=0.000-0.059/2logpH2/[H + ] 2 2H + +2e H 2 Q=X/Y ------------------------------------------------------- Slide 51 50 Ni (s) + Sn 2+ Ni 2+ + Sn (s) E= 0.090 V http:\\asadipour.kmu.ac.ir 76 slides Ni (s) | Ni 2+ (0.600M) || Sn 2+ (0.300M) | Sn (s) Emf of a cell 920311 Slide 52 51 Emf of a cell Sn (s) | Sn 2+ (1.0M) || Pb 2+ (0.0010M) | Pb (s) 2 e - + Pb 2+ Pb (s) E=-0.126 V Sn (s) 2 e - + Sn 2+ E=0.136V http:\\asadipour.kmu.ac.ir 76 slides E=E -0.059/2log[Sn 2+ ]/[Pb 2+ ] E=-0.079 !!!= Reversed cell Sn (s) + Pb 2+ (0.0010M) Sn 2+ (1.0M) + Pb (s) E cell =0.010 V pb (s) | pb 2+ (1.0M) || sn 2+ (0.0010M) | sn (s) 920311 E=+0.079 (Electrolytic cell) (Galvanic cell) Slide 53 52 equilibrium constant of a cell at equilibrium E = 0 Nernst Equation: http:\\asadipour.kmu.ac.ir 76 slides E = E o - 0.0591 log B n A 920311 A BA B Slide 54 53http:\\asadipour.kmu.ac.ir 76 slides Ni (s) | Ni 2+ (0.600M) || Sn 2+ (0.300M) | Sn (s) Ni (s) + Sn 2+ Ni 2+ + Sn (s) E= 0.090 V equilibrium constant of a cell 920311 Slide 55 54http:\\asadipour.kmu.ac.ir 76 slides920311 Slide 56 55 Electrod potential and electrolysis Theoritical emf of a Voltaic cell is maximum voltage. (Practically is less) Theoritical emf of an electrolysis cell is minimum voltage. (Practically is more) Emf is related to: Resistance Concentration Overvoltage http:\\asadipour.kmu.ac.ir 76 slides 920311 Slide 57 56 Electrod potential and electrolysis E = - 0.83 V) 2H 2 O + 2e - H 2 + 2OH - (E = - 0.83 V) In aqueous salts electrolysis [OH - ] =1 10 -7 M http:\\asadipour.kmu.ac.ir 76 slides E = E - log Q n 0.059 V E = E - log [OH - ] 2 pH 2 2 0.059 V E = -0.83 -0.0295 log [1*10 -7 ] 2 *1=-0.417 920311 Slide 58 57 Electrod potential and electrolysis E = - 1.23V) 2H 2 O O 2 + 4H + + 4e - (E = - 1.23V) In aqueous salts electrolysis [H + ] =1 10 -7 M http:\\asadipour.kmu.ac.ir 76 slides E = E - log Q n 0.059 V E = E - log [H + ] 4 pO 2 4 0.059 V E = -1.23 -0.01475 log [1*10 -7 ] 4 *1=-0.817 920311 Slide 59 58 Effect of concentration in aqueous Na 2 SO 4 electrolysis possible cathode half-cells (-) E = - 2.71 V) REDUCTION Na + + e - Na (E = - 2.71 V) E = - 0.83 V) [2H 2 O + 2e - H 2 + 2OH - ] (E = - 0.83 V) E = - 0.417 V) (E = - 0.417 V) possible anode half-cells (+) E = - 2.01 V) OXIDATION SO 4 2- S 4 O 8 2_ + 2e - (E = - 2.01 V) E = - 1.23V) 2H 2 O O 2 + 4H + + 4e - (E = - 1.23V) E = - 0.817 V) (E = - 0.817 V) overall cell reaction 6H 2 O 2H 2 + O 2 + 4H + + 4OH - E =-0.417-0.817=-1.234 http:\\asadipour.kmu.ac.ir 76 slides920311 2 Slide 60 59 Electrod potential and electrolysis Overvoltage(OV): (Because of slow rate of reaction) OV of deposition of metals are low OV of liberation of gases are appreciable (O 2 & H 2 >Cl 2 ) http:\\asadipour.kmu.ac.ir 76 slides 920311 Slide 61 60 Effect of overvoltage & concentration in aqueous NaCl Electrolysis possible cathode half-cells (-) E = - 2.71 V) REDUCTION Na + + e - Na (E = - 2.71 V) E = - 0.83 V) [2H 2 O + 2e - H 2 + 2OH - ] (E = - 0.83 V) E = - 0.417 V) (E = - 0.417 V) possible anode half-cells (+) E = - 1.36 V) OXIDATION2Cl - Cl 2 + 2e - (E = - 1.36 V) E = - 1.23V) 2H 2 O O 2 + 4H + + 4e - (E = - 1.23V) E = - 0.817 V) (E = - 0.817 V) OVERVOLTAGE H2 & O2 > Cl 2 overall cell reaction 2Cl - + 2H 2 O H 2 + Cl 2 + 2OH - http:\\asadipour.kmu.ac.ir 76 slides920311 Slide 62 61 Effect of overvoltage & concentration in aqueous CuCl 2 Electrolysis possible cathode half-cells (-) E = +0.337V) REDUCTION Cu 2+ + 2e - Cu (E = +0.337V) E = - 0.83 V) 2H 2 O + 2e - H 2 + 2OH - (E = - 0.83 V) E = - 0.417 V) (E = - 0.417 V) possible anode half-cells (+) E = - 1.36 V) OXIDATION2Cl - Cl 2 + 2e - (E = - 1.36 V) E = - 1.23V) 2H 2 O O 2 + 4H + + 4e - (E = - 1.23V) E = -0.817V) (E = -0.817V) OVERVOLTAGE H2 & O2 > Cl 2 overall cell reaction Cu 2+ + 2Cl - Cu (s) + Cl 2(g) http:\\asadipour.kmu.ac.ir 76 slides 920311 Slide 63 62 Cu CuSO 4 Cu Cu +2 + 2e - CuCu Cu +2 + 2e - What happened at each electrode? http:\\asadipour.kmu.ac.ir 76 slides battery Impure Cu pure Cu Anode + Cathode - Pure Cu deposit on cathode = (Pure cathodic Cu) What happens if aqueous CuSO 4 electrolyze between 2 Cu electrodes ?=purification of Cu 920311 Slide 64 63 What happens if aqueous CuSO 4 electrolyze between 2 Cu electrodes ?=purification of Cu possible anode half-cells (+) (Impure Cu) E = -0.337 V) OXIDATION Cu Cu 2+ + 2e - (E = -0.337 V) E = - 1.23V) 2H 2 O O 2 + 4H + + 4e - (E = - 1.23V) E = - 0.817 V) (E = - 0.817 V) 2SO 4 2- E = -2.01V ) 2SO 4 2- S 2 O 8 2- + 2e - (E = -2.01V ) possible cathode half-cells (-) (Purified Cu) E = +0.337V) REDUCTION Cu 2+ + 2e - Cu (E = +0.337V) E = - 0.83 V) 2H 2 O + 2e - H 2 + 2OH - (E = - 0.83 V) E = - 0.417 V) (E = - 0.417 V) http:\\asadipour.kmu.ac.ir 76 slides (((Purified cathodic Cu))) overall cell reaction Cu 2+ + Cu (s) Anod Cu 2+ + Cu (s) Cathode 920311 Slide 65 64 Cu 1.0 M CuSO 4 Cu 1.0 M CuSO 4 A cell with the similar electrods and electrolytes 0.0 http:\\asadipour.kmu.ac.ir 76 slides volts 920311 Slide 66 65 Cu 1.0M CuSO 4 Cu 0.1 M CuSO 4 A cell with the similar electrods but different concentration electrolytes http:\\asadipour.kmu.ac.ir 76 slides volts 920311 Slide 67 66 Electrolysis of Copper Concentration Cells A concentration cell based on the Cu/Cu2+ half-reaction. A, Even though the half-reactions involve the same components, the cell operates because the half-cell concentrations are different. B, The cell operates spontaneously until the half-cell concentrations are equal. Note the change in electrodes (exaggerated here for clarity) and the equal color of solutions. http:\\asadipour.kmu.ac.ir 76 slides920311 Slide 68 http:\\asadipour.kmu.ac.ir 76 slides67 Cu Cu 2+ (0.1M) Cu 2+ (1.0 M) Cu Anod cathod E=E 0 -0.059/2Log(0.1/1.0) =+0.0296 Concentration Cells Cu+Cu 2+ (1.0 M) Cu 2+ (0.1M)+Cu 920311 Slide 69 68 pH meter, A concentration Cell 920311http:\\asadipour.kmu.ac.ir 76 slides Slide 68 of 52 2 H + (1 M) 2 H + (x M) Pt | H 2 (1 atm)|H + (x M) ||H + (1.0 M) |H 2 (1 atm) | Pt(s) 2 H + (1 M) + 2 e - H 2 (g, 1 atm) H 2 (g, 1 atm) 2 H + (x M) + 2 e - H 2 (g, 1 atm) +2 H + (1 M) 2 H + (x M) + H 2 (g, 1 atm) Slide 70 69 Slide 69 of 52 E cell = E cell - log n 0.059 V x2x2 1212 E cell = 0 - log 2 0.059 V x2x2 1 E cell = - 0.059 V log x E cell = (0.059 V) pH 2 H + (1 M) 2 H + (x M) E cell = E cell - log Q n 0.059 V http:\\asadipour.kmu.ac.ir 76 slides pH = E cell /(0.059) 920311 Slide 71 70 The pH Meter In practice, a special pH electrode is much more convenient than using platinum electrodes and a tank of hydrogen gas! A stable reference electrode and a glass-membrane electrode are contained within a combination pH electrode. The electrode is merely dipped into a solution, and the potential difference between the electrodes is displayed as pH. http:\\asadipour.kmu.ac.ir 76 slides920311 Slide 72 71 Corrosion of Fe: Unwanted Voltaic Cells O2+2H 2 O+4e - 4OH - Rust formation: Fe 2+ Fe 3+ +e E 0 =-0.771 V O 2 (g) + 4 H + (aq) + 4 e - 4 H 2 O (aq) E 0 = 1.229 V ---------------------------------------------------------------- 4Fe 2+ (aq) + O 2 (g) + 4H + (aq) 4Fe 3+ (aq) + 2H 2 O(l) E 0 =0.458V 2Fe 3+ (aq) + 4H 2 O(l) Fe 2 O 3 H 2 O(s) + 6H + (aq) http:\\asadipour.kmu.ac.ir 76 slides E0=0.440 VE0=1.229 V E 0 =0.401 V 920311 Slide 73 72 Prevention of Corrosion Cover the Fe surface with a protective coating Paint Tin (Tin plate) Zn (Galvanized iron) http:\\asadipour.kmu.ac.ir 76 slides920311 Slide 74 73 Corrosion Protection Slide 73 of 52 http:\\asadipour.kmu.ac.ir 76 slides Fe Fe 2+ +2e E 0 =0.440 V Cu Cu 2+ +2e E 0 =0.337 V Fe Fe 2+ +2e E 0 =0.440 V Zn Zn 2+ +2e E 0 =0.763 V 920311 Slide 75 74 Corrrosion Protection (cathode) (electrolyte) (anode) http:\\asadipour.kmu.ac.ir 76 slides Fe Fe 2+ +2e E 0 =0.440 V Mg Mg 2+ +2e E 0 =2.363 V Steel pipe dont rust 920311 Fe Fe 2+ +2e E 0 =0.440 V E 0 =0.401 V Slide 76 75 galvanicelectrolytic need power source two electrodes produces electrical current anode (-) cathode (+) anode (+) cathode (-) salt bridge vessel conductive medium Comparison of Electrochemical Cells G < 0 G > 0 http:\\asadipour.kmu.ac.ir 76 slides920311 Slide 77 76 Cathodic Protection In cathodic protection, an iron object to be protected is connected to a chunk of an active metal. The iron serves as the reduction electrode and remains metallic. The active metal is oxidized. Water heaters often employ a magnesium anode for cathodic protection. http:\\asadipour.kmu.ac.ir 76 slides920311