Electrochemistry Exercises: Calculate Oxygen Electrode Potential, Reverse Polarity, and pH Determination
Explore electrochemistry exercises including calculating oxygen electrode potential at pH 14, reversing polarity in a cell reaction, and determining pH in a zinc-hydrogen cell. Dive into the world of materials engineering with Dr. Lubna Ghalib's guidance and solve challenging problems related to electrode reactions and cell potentials.
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Presentation Transcript
Materials Engineering Dr. Lubna Ghalib
Exercise 1: Calculate the potential of oxygen electrode at pH=14.0. Solution : Oxygen electrode reaction: in basic and neutral environment is expressed as: O2 + 2H2O + 4e- 4(OH- ) Electrode potential equation for the electrode basic solution), and at 25 oC. ? = ?? ?? ??ln ? = 0.401 0.0591 4 ??? = 1 4 ?? log?(?? )= 0 Eo= + 0.401 V ????????? ?????????? 4 ?(?? ??2? ) l?? 2 ??2
Exercise 1: O2 + 2H2O + 4e- 4(OH- ) Eo= + 0.401 V 4 ?(?? ??2? (0.0)(?? ) ? = 0.401 0.0591 l?? 2 4 ??2 4 ) ? = 0.401 0.0591 l?? 4 1 E= 0.401 V
Exercise 2: In the cell reaction given below, what is the ratio of the activities of ionic species required to make the polarity reverse? Fe2+ + Sn Sn2+ + Fe Solution: Fe2+ + 2e- Fe Sn Sn2+ + 2e- Final Cell reaction Fe2+ + Sn Eocell= (-0.440+0.136)= - 0.304 V Eo= -0.440 V Eo= +0.136 V Sn2+ + Fe The reaction in this case is non-spontaneous. That is, Fe2+/Fe electrode will act as 'anode' and Sn2+/Sn electrode will act as 'cathode,' the reverse case of what was assumed in the problem. To reverse the polarity:
Exercise 2: Solution: To reverse the polarity: ????????? ?????????? ?Sn2+?Fe ?Sn?Fe2+ ? = ?? ?? ??ln ? = 0.304 0.0591 l?? 2 Ecell = 0, At the turn of polarity. Therefore, 0.304 = 0.0591 ?Sn2+ ?Fe2+ l?? 2 ?Sn2+ ?Fe2+= 5.156 10 11
Exercise 3: The emf of a cell made of Zn (anode) and H2 electrode (cathode) immersed in 0.7 M ZnCl2 is +0.690 volts. What is the pH of the solution? Activity coefficient ?=0.6133. ????2 ??2++ 2?? 0.7M 0.7M 0.7M Zn2+ + 2e- Zn 2H+ + 2e- H2 ?=0.6133 Eo= - 0.763 V Eo= 0.0 V Zn2+ + 2e- Zn2+ + H2 Eo= +0.763 V Eo= 0.0 V Eo =+0.763 V Anode: Cathode: Final: Zn + 2H+ Zn 2H+ + 2e- H2
Exercise 3: ????????? ?????????? ? = ?? ?? ??ln +0.690 = +0.763 0.0591 ?Zn2+?H2 ?Zn ?2 (0.7 0.6133 ?2 l?? 2 H+ +0.690 = +0.763 0.0591 l?? 2 H+ 0.073 = 0.0591 (0.42931 ?2 l?? 2 H+ 2.4703 = l?? 0.42931 ??? ?2 2.8376 = 2?? H+ pH= 1.4187
Exercise 4: Calculate the theoretical tendency of nickel to corrode in deaerated water of pH = 8. Assume the corrosion products are H2 and Ni (OH)2 and the solubility product is 1.6x 10-16? Solution: Ni2+ + 2e- Ni 2H+ + 2e- H2 Anode: Ni Ni2+ + 2e- Cathode: 2H+ + 2e- Final: Ni + 2H+ Ni2+ + H2 pH + pOH = 14 pOH = 14- 8 = 6 [OH-] = 10 6 [H+] = 10 8 Eo= - 0.25 V Eo= 0.0 V Eo= +0.25V Eo= 0.0 V Eo =+0.25 V H2
Exercise 4: Ni2+OH 2 ?? ??2 Ni2+10 6 2 ???= 1.6 10 16= 1 Ni2+=1.6 10 4 ????????? ?????????? ?Ni2+?H2 ?Ni ?2 ? = ?? ?? ??ln ? = 0.25 0.0591 l?? 2 H+ 1.6 10 4 [10 8]2 ? = 0.25 0.0591 l?? 2 E= - 0.11063 V
Exercise 5: Calculate the pressure of H2 required to stop corrosion of iron immersed in 0.1 M FeCl2and =0.58, pH = 4. Solution: Fe 2+ + 2e- Fe 2H+ + 2e- H2 Anode: Fe Fe2+ + 2e- Cathode: 2H+ + 2e- Final: Fe + 2H+ Fei2+ + H2 Eo= - 0.440 V Eo= 0.0 V Eo= +0.440 V Eo= 0.0 Eo =+0.440 V H2 V ????2 ??2++ 2?? 0.1 M 0.1 M 2(0.1 M) ?=0.75 ????????? ?????????? ? = ?? ?? ??ln
Exercise 5: ? = 0.44 0.0591 ?Fe2+?H2 ??? ?2 l?? 2 H+ ??2 ?? ??= 1 ???., standard state Ecell =0.0 V 0.0 = 0.44 0.0591 ??2= stop corrosion 0.1 0.75 ??2 [10 4]2 l?? 2 ??2 = 103503648 atm. ??2= 10+8 ???
Exercise 6: Calculate if silver would corrode when immersed in 0.5 M CuCl2 and =0.42 to form solid AgCl. What is the corrosion tendency? Solution : 2AgCl + 2e- 2Ag + 2Cl- Cu 2+ + 2e- Cu Anode: 2Ag + 2Cl- Cathode: Cu 2+ + 2e- Final: 2Ag +Cu 2+ + 2Cl- ????2 ??2++ 2?? 0.5 M 0.5 M 2(0.5 M) ? = ?? ?? ??ln Eo= + 0.222 V Eo= + 0.34 V Eo= - 0.222 V Eo= + 0.34 V 2AgCl + 2e- Cu 2AgCl + Cu Eo =+0.118 V ?=0.47 ????????? ??????????
Exercise 6: ?2 AgCl?Cu ? = 0.118 0.0591 l?? ?2?????2+?2?? 2 ? = 0.118 0.0591 1 l?? 0.5 0.472 2 0.5 0.472 2 E= 0.061451 V silver will corrode
