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Example Find the potential of a 50 mL solution of a 0.10 M Fe 2+ after addition of 0, 25, 40, 50, and 75 mL of 0.1 M Ce 4+ . Fe 3+ + e D Fe 2+ E o = 0.771 V Ce 4+ + e D Ce 3+ E o = 1.70 V Solution 1. After addition of 0 mL Ce 4+

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  1. Example Find the potential of a 50 mL solution of a 0.10 M Fe2+ after addition of 0, 25, 40, 50, and 75 mL of 0.1 M Ce4+. Fe3+ + e D Fe2+ Eo = 0.771 V Ce4+ + e D Ce3+ Eo = 1.70 V Solution 1. After addition of 0 mL Ce4+ The solution contains Fe2+ only and attempts to apply Nernst equation will not be possible since assumably no Fe3+ will be present. EFe = EFeo – (0.0592/nFe) log [Fe2+]/[Fe3+]

  2. EFe = EFeo – (0.0592/nFe) log [Fe2+]/[Fe3+] EFe = EFeo – (0.0592/nFe)log 0.1/0 This means that as Fe3+ approaches zero, the electrode potential approaches negative infinity. This, in fact, is not practically true since it suggests that a Fe2+ solution will have unbelievably high reducing power. The way out of this problem is to always assume the presence of very small Fe3+ concentration in equilibrium with Fe2+, even in very pure Fe2+ solutions. The conclusion with regards to calculation of electrode potential after addition of 0 mL Ce4+ is that we can not calculate it using direct Nernst equation.

  3. 2. After addition of 25 mL Ce4+ Initial mmol Fe2+ = 0.10 x 50 = 5.0 Mmol Ce4+ added = 0.10 x 25 = 2.5 Mmol Fe2+ left = 5.0 – 2.5 = 2.5 [Fe2+] = 2.5/75 mmol Fe3+ formed = 2.5 [Fe3+] = 2.5/75 Application of Nernst equation gives: EFe = EFeo – (0.0592/nFe)log {(2.5/75)/(2.5/75)} EFe =0.771 V Therefore, the standard electrode potential for Fe2+/Fe3+ couple can be calculated at half the way to the equivalence point.

  4. 3. After addition of 40 mL Ce4+ Initial mmol Fe2+ = 0.10 x 50 = 5.0 mmol Ce4+ added = 0.10 x 40 = 4.0 mmol Fe2+ left = 5.0 – 4.0 = 1.0 [Fe2+] = 1.0/90 mmol Fe3+ formed = 4 [Fe3+] = 4.0/90 Application of Nernst equation gives: EFe = EFeo – (0.0592/nFe) log {(1.0/90)/(4.0/90)} EFe = 0.771 – 0.0592 log 1.0/4.0 EFe = 0.807 V

  5. 4. After addition of 50 mL Ce4+ Initial mmol Fe2+ = 0.10 x 50 = 5.0 mmol Ce4+ added = 0.10 x 50 = 5.0 mmol Fe2+ left = 5.0 – 5.0 = ?? This is the equivalence point mmol Fe3+ formed = 5.0 [Fe3+] =5.0/100 = 0.05 M At equivalence point, we have from the above discussion: Eeq pt = (nFe EFeo + nCe ECeo)/ (nFe + nCe) Eeq pt = (1 * 0.771 + 1 * 1.70)/(1 + 1) Eeq pt =1.24 V

  6. 3. After addition of 75 mL Ce4+ Initial mmol Fe2+ = 0.10 x 50 = 5.0 mmol Ce4+ added = 0.10 x 75 = 7.5 mmol Ce4+ excess = 7.5 – 5.0 = 2.5 [Ce4+] = 2.5/125 mmol Ce3+ formed = 5.0 [Ce3+] = 5.0/125 Application of Nernst equation gives: ECe = ECeo – (0.0592/nCe) log [Ce3+]/[Ce4+] ECe =1.70 – (0.0592/1)log {(5.0/125)/(2.5/125)} ECe = 1.68 V

  7. Example Find the electrode potential of a 50 mL of a 0.10 M solution of Fe2+ after addition of 0, 5, 10, and 20 mL 0f 0.10 KMnO4, at pH 1. Solution MnO4- + 8H+ + 5 Fe2+D Mn2+ + 5 Fe3+ + 4 H2O Fe3+ + e D Fe2+ Eo = 0.771 V MnO4- + 8H+ + 5e D Mn2+ + 4 H2O Eo = 1.51 V

  8. 1. After addition of 0 mL MnO4- The solution contains Fe2+ only and attempts to apply Nernst equation will not be possible since assumably no Fe3+ will be present. EFe = EFeo – (0.0592/nFe) log [Fe2+]/[Fe3+] EFe = EFeo – (0.0592/nFe)log 0.1/0 This means that as Fe3+ approaches zero, the electrode potential approaches negative infinity. This, in fact, is not practically true since it suggests that a Fe2+ solution will have unbelievably high reducing power. The conclusion with regards to calculation of electrode potential after addition of 0 mLMnO4- is that we can not calculate it using direct Nernst equation.

  9. 2. After addition of 5 mL MnO4- mmol Fe2+ left = initial mmol Fe2+ - mmol Fe2+ reacted mmol Fe2+ reacted = 5 mmol MnO4- mmol Fe2+ left = initial mmol Fe2+ - 5 mmol MnO4- mmol Fe2+ left = 0.1*50 – 5 *(0.1*5) mmol Fe2+ left = 5.0 – 2.5 = 2.5 , [Fe2+] = 2.5/55 mmol Fe3+ formed = 2.5, [Fe3+] = 2.5/55 Application of Nernst equation gives: EFe = EFeo – (0.0592/nFe)log {(2.5/55)/(2.5/55)} EFe =0.771 V Therefore, the standard electrode potential for Fe2+/Fe3+ couple can be calculated at half the way to the equivalence point.

  10. 3. After addition of 10 mL MnO4- mmol Fe2+ left = initial mmol Fe2+ - mmol Fe2+ reacted mmol Fe2+ left = initial mmol Fe2+ - 5 mmol MnO4- mmol Fe2+ left = 5.0 – 5 * (0.1*10) = ?? This is the equivalence point At the equivalence point we have: Electrode potential for the iron oxidation = electrode potential for permanganate reduction EMnO4-/Mn2+ = Eo MnO4-/Mn2+– (0.0592/n MnO4-/Mn2+) log [Mn2+]/[MnO4-][H+]8 EFe = EFeo – (0.0592/nFe) log [Fe2+]/[Fe3+]

  11. EMnO4-/Mn2+ = Eo MnO4-/Mn2+– (0.0592/n MnO4-/Mn2+) log [Mn2+]/[MnO4-][H+]8 EFe = EFeo – (0.0592/nFe) log [Fe2+]/[Fe3+] Addition of the two equations gives: (nMnEep+ nFeEep) = (nMnEoMn + nFeEFeo) – 0.0592 log { [Mn2+][Fe2+]/[Fe3+][ [MnO4-][H+]8} Substitute for the following: [Fe2+] = 5 [MnO4-] [Fe3+] = 5[Mn2+] (n Mn + nFe) Eep = (n Mn Eo Mn + nFe EFeo) – 0.0592 log { [Mn2+] * 5 [MnO4-] /5 [Mn2+][ [MnO4-][H+]8} (5 + 1) Eep = (5 * 1.51 + 1 * 0.771) – 0.0592 log 1/(0.10)8 Eep = 1.31 V

  12. 4.After addition of 20 mL MnO4- mmol MnO4- excess = mmol MnO4- added - mmol MnO4- reacted mmol MnO4- reacted = 1/5 mmol Fe2+ mmol MnO4- excess = mmol MnO4- added - 1/5 mmol Fe2+ mmol MnO4- excess = 0.1*20 – 1/5* (0.1*50) = 1 [MnO4-]= 1/70 mmol Mn2+ formed = 1 [Mn2+] = 1/70 EMnO4-/Mn2+ = Eo MnO4-/Mn2+– (0.0592/n MnO4-/Mn2+) log [Mn2+]/[MnO4-][H+]8 EMnO4-/Mn2+ = 1.51 – (0.0592/5) log (1/70)/{(1/70)(0.1)8} EMnO4-/Mn2+ = 1.42 V

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