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Potentiometry

Potentiometry. Potentiometry. Ecell=ERt-ELf. Eleft= E0+0.0591/n log OX/RED. Eright= E0+0.0591/n log OX/RED. Types Of Electrodes. Reference Electrode. Indicator Electrode. Reference Electrodes. Primary. Secondary. NHE PPE SHE. Ag/AgCl/Cl -. SCE. Electrode of 1st Kind.

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Potentiometry

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  1. Potentiometry

  2. Potentiometry Ecell=ERt-ELf Eleft= E0+0.0591/n log OX/RED Eright= E0+0.0591/n log OX/RED

  3. Types Of Electrodes Reference Electrode Indicator Electrode

  4. Reference Electrodes Primary Secondary NHEPPESHE Ag/AgCl/Cl- SCE

  5. Electrode of 1st Kind Metal rod immersed in its solution M0 / M n+(xM) Examples • Cu / CuSO4 • Cd / CdSO4 • Zn / ZnSO4 ُE = E0 + (0.0591 / n) log [Mn+] / [M0]

  6. Electrode of 2nd Kind Metal rod / sparingly soluble salt / anion solution. Examples • Ag / AgCl (sat’d) / KCl (xM) • Hg / Hg2Cl2 (sat’d) / KCl (xM). Ac Bd + ne-↔ cA0 + dB- ُE = E0 + (0.0591 / n) log [AcBd ] / [A0]c* [B-]d

  7. Inert Type Electrode Pt, Au, C electrodes are used for electrical contact in the following systems:

  8. Applications Determination of E cell Determination of unknown concentration

  9. Determination of potential of electrochemical cell

  10. "1" Cu0 / Cu 2+(0.1M) // Ag+ (0.2M) / Ag 0 Anode Cathod Left Side Right Side Cu0 ↔Cu 2+ + 2e- E0Cu2+/Cu0 = 0.337volt E left = E0Cu2+/Cu0+ 0.0591 / n log [Cu+2] / [Cu0] = 0.337 + 0.0591 / 2 log (0.1 / 1) = 0.307 volt.

  11. Cu0 / Cu 2+(0.1M) // Ag+ (0.2M) / Ag 0 Anode Cathod Left Side Right Side Ag+ +e- ↔ Ag 0 E0Ag+ / Ag0 = 0.799 volt E right = E0Ag+/Ag0 + (0.0591 / n) log [Ag+] / [Ag0] = 0.799 + 0.0591 / 1 log (0.2 / 1) = 0.757 volt.

  12. Cu0 / Cu 2+ // Ag+ (0.2M) / Ag 0 Anode Cathod Left Side Right Side E cell = E right – E left E cell = 0.757 – 0.307 = + 0.45 volts Galvanic Cell The reaction proceeds in the written direction.

  13. Overall Equation Cu0 / Cu 2+ // Ag+ (0.2M) / Ag 0 Cu0 ↔ Cu 2 + + 2e- 2Ag+ +2e- ↔ 2Ag 0 Cu 0 + 2 Ag + ↔ Cu2+ + 2 Ag0

  14. "2" Ag 0 / Ag+ (0.2M) // Cu 2+(0.1M) / Cu0 Anode Cathod Left Side Right Side E cell = 0.307 –0.757 = - 0.45 volts Electrolytic Cell The reaction proceeds in the opposite direction. Cu 0 + 2 Ag + ↔ Cu2+ + 2 Ag0

  15. "3" Pt0 / UO22+(0.015M),U4+(0.2M),H+(0.03M) // Fe2+ (0.01M) , Fe3+(0.025M) / Pt0 Left Side U4+ + 2H2O↔ UO22++ 4H+ + 2e- E0UO22+,H+/U4+ = 0.334volt E left = E0UO22+,H+/U4++ 0.0591 / n log [UO22+] [H+]4 / [U4+] = 0.334 + 0.0591 / 2 log [0.015* (0.03)4/ 0.015] = 0.12 volt.

  16. "3" Pt0 / UO22+(0.015M),U4+(0.2M),H+(0.03M) // Fe2+ (0.01M) , Fe3+(0.025M) / Pt0 Right Side Fe3+ +e- ↔ Fe2+ E0 Fe3+/Fe2+ = 0.771volt E right = E0Fe3+,Fe2++ 0.0591 / n log [Fe3+] / [Fe2+] = 0.771 + 0.0591 / 1 log [0.025/ 0.01] = 0.795 volt.

  17. Pt0 / UO22+(0.015M),U4+(0.2M),H+(0.03M) // Fe2+ (0.01M) , Fe3+(0.025M) / Pt0 E cell = E right – E left E cell = 0.795 – 0.12 = + 0.675 volts Galvanic Cell The reaction proceeds in the written direction.

  18. Overall Equation Pt0 / UO22+(0.015M),U4+(0.2M),H+(0.03M) // Fe2+ (0.01M) , Fe3+(0.025M) / Pt0 U4+ + 2H2O↔ UO22++ 4H+ + 2e- 2Fe3+ + 2e- ↔ 2Fe2+ U4+ +2Fe3+ +2H2O↔ UO22+ + 2Fe2+ +4H+

  19. "4" Pt0/VO2+(0.25M) , V3+ (0.1M), H+ (0.001M) // Tl+(0.05M), Tl3+(0.1M),Pt0

  20. "5" Pb0 / PbSO4 (sat’d),SO42-(0.2M) // Sn4+(0.25M) / Sn2+(0.15M) Left Side PbSO4 ↔ Pb2+ + SO42- Pb2+ + 2e-↔ Pb0 PbSO4+ 2e- ↔ Pb0 + SO42- E left = E0PbSO4/Pb0,SO4+ 0.0591 / n log [PbSO4] / [Pb0][SO42-] =-0.35 + 0.0591 / 2 log [1/1*0.2] = -0.33 volt.

  21. Pb0 / PbSO4 (sat’d),SO42-(0.2M) // Sn4+(0.25M) / Sn2+(0.15M) Right Side Sn4++2e-↔ Sn2+ E right = E0Sn4+/Sn2++ 0.0591 / n log [Sn4+] / [Sn2+] = 0.154 + 0.0591 / 2 log [0.25/ 0.15] = 0.16 volt.

  22. Pb0 / PbSO4 (sat’d),SO42-(0.2M) // Sn4+(0.25M) / Sn2+(0.15M) E cell = E right – E left E cell = 0.16 – (-0.337) = + 0.49 volts Galvanic Cell The reaction proceeds in the written direction.

  23. Overall Equation Pb0 / PbSO4 (sat’d),SO42-(0.2M) // Sn4+(0.25M) / Sn2+(0.15M) Pb0 + SO42- ↔ PbSO4+ 2e- Sn4++2e-↔ Sn2+ Pb0 + SO42-+ Sn4+↔ PbSO4 + Sn2+

  24. "6" Pt0/ I2(S) , I- (0.02M) // Fe2+(0.04M), Fe3+(0.04M)/Pt0

  25. "7" IMPORTANT Cu0 / CuSO4 (0.05M) // CdSO4 (0.02M) / Cd0

  26. "8" Ag0 / AgBr(sat’d), Br-(0.04M) // H+(10 4-M),H2(gas)(0.9atm),Pt0

  27. "9" 0.09M CE // V2+(0.2M) , V3+(0.3M)/Pt0 Calomel Electrode Hg0l , Hg2Cl2(sat’d) , KCl (X M = 0.09M)

  28. Hg0, Hg2Cl2 (sat’d) ,Cl-( 0.09M) // V2+(0.2M) , 3+(0.3M)/Pt0 Left Side Hg2Cl2 ↔ Hg22+ + 2Cl- Hg22+ + 2e-↔ 2Hg0 Hg2Cl2+ 2e- ↔ 2Hg0 + 2Cl- E left = E0CE+ 0.0591 / n log [Hg2Cl2] / [Hg0]2[Cl-]2 =0.268 + (0.0591 / 2) log [1/1*(0.09)2] = 0.329 volt.

  29. Hg0, Hg2Cl2 (sat’d) ,Cl-( 0.09M) // V2+(0.2M) , 3+(0.3M)/Pt0 Right Side V3++e-↔ V2+ E right = E0V3+/V2++ (0.0591 / n) log [V3+] / [V2+] = - 0.256 + (0.0591 / 1) log [0.3/ 0.2] = - 0.245 volt.

  30. Hg0, Hg2Cl2 (sat’d) ,Cl-( 0.09M) // V2+(0.2M) , 3+(0.3M)/Pt0 E cell = E right – E left E cell = -0.245 – 0.329 = - 0.574volts Electrolytic Cell The reaction proceeds in the OPPOSITE direction.

  31. Overall Equation 2Hg0 + 2Cl- ↔ Hg2Cl2+ 2e- 2V3++ 2e- ↔ 2V2+ 2Hg0 + 2Cl- +2V3+↔ Hg2Cl2+ 2V2+

  32. Determination of unknown concentration of some analyte

  33. Determine the conc. Ag+ Pt0 / Fe2+(0.05M),Fe3+(0.25) // Ag+ (xM) / Ag0 Left Side Right Side Ecell = -0.106 volt E0 Fe3+,Fe2+ = 0.771 volt E0 Ag+ / Ag0 = 0.799 volt

  34. "1" Pt0 / Fe2+(0.05M),Fe3+(0.25) // Ag+ (xM) / Ag0 Left Side Right Side Fe3+ +e- ↔ Fe2+ E0 Fe3+/Fe2+ = 0.771volt E left = E0Fe3+,Fe2++ 0.0591 / n log [Fe3+] / [Fe2+] = 0.771 + 0.0591 / 1 log [0.025/ 0.05] = 0.8123 volt.

  35. Pt0 / Fe2+(0.05M),Fe3+(0.25) // Ag+ (xM) / Ag0 Left Side Right Side Ag+ +e- ↔ Ag 0 E0Ag+ / Ag0 = 0.799 volt E right = E0Ag+/Ag0 + (0.0591 / n) log [Ag+] / [Ag0] = 0.799 + 0.0591 / 1 log ( x / 1) Ecell = E right – E left -0.106 = {0.799 + 0.0591 log x} – 0.8123 Log Ag+= - 1.56 [Ag+] = 0.027 M

  36. "2" Determine the conc. Cu2+ SCE // Cu2+ (xM) / Cu0 E SCE = 0.246 volt

  37. "3" Determine pH by quinhydrone electrode Equimolar ratio p-benzoquinone [Q] Hydroquinone [H2Q]

  38. E QE = E0QE + (0.0591/n) log [Q]*[H+]2 / [H2Q] = 0.699 + 0.0591/2 log [H+]2 = 0.699 + 0.0591 log [H+] =0.699 - 0.0591 pH

  39. "3" Determine pH by quinhydrone electrode SCE // QE (sat’d), H+(xM)/Pt0 Ecell = 0.313 volt E0QE = 0.699 volt E SCE = 0.246 volt

  40. "3" Determine pH in quinhydrone electrode SCE // QE (sat’d), H+(xM)/Pt0 E cell = E QE – E SCE 0.313 = 0.699- 0.0591 pH -0.246 …………………………. pH = 2.36

  41. Determination of pH by glass electrode Na+ Na+ Combination Electrode Na+ Na+ Glass Electrode

  42. E glass electrode = L - 0.0591pH

  43. "4" Determination of pH by glass electrode SCE // (H+) = xM , GE

  44. SCE // (H+) = xM , GE E cell = E GE – E SCE = L - 0.0591 pH -0.246 0.2094 = L – 0.0591(4.006) – 0.246 -0.3011 = +L – 0.0591 (pH) – 0.246 0.2094+0.3011=0.0591(-4.006+pH) pH = 12.6 + + + -

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