1. 1 Oxidation Reduction Equilibria and Titrations Lecture 38

2. 2 Oxidation - Reduction reactions (Redox rxns) involve the transfer of electrons from one species of the reactants to another. This results in an increase in oxidation number (O.N.) of a specific species and a complementary decrease in oxidation number of another species. Example: Ce 4+ + Fe 2+  Ce 3+ + Fe 3+ The O.N. of cerium was decreased while that of iron was increased. Cerium is reduced while iron is oxidized. A process that involves an increase in O.N. is an oxidation process and vice versa

3. 3 Usually, a Redox reaction can be separated into two halves. Ce 4+ + e  Ce 3+ Reduction Fe 2+  Fe 3+ + e Oxidation Electrons appear in each half reaction while they do not show up in the overall equations.

4. 4 Identification of a Redox Reaction It is a good practice to indicate the O.N. of each species in a chemical reaction in order to check if it is a Redox reaction or not. If the O.N. of any species changes, then it is a definite indication of a Redox reaction. Example, 2 KMnO 4 + 5 H 2 C 2 O 4 + 6 HCl  2 MnCl 2 + 2KCl + 10 CO 2 + 8 H 2 O It is observed that in the left-hand part of the equation, manganese has an O.N. equals 7 and carbon has an O.N. equals 3. In the right-hand part, the O.N. of manganese is 2 and that of carbon is 4. Therefore, permanganate is reduced while oxalic acid is oxidized.

5. 5 An example of a non-Redox reaction can be written where no change in O.N. occurs, Na 2 CO 3 + 2 HCl = 2 NaCl + CO 2 + H 2 O +1 +4 -2 +1 -1 +1 -1 +4 -2 +1 -2 There is no change in O.N. of any species involved in the reaction, which indicates that this is not a Redox reaction.

6. 6 Balancing Redox Reactions Balanced chemical equations are the basis of any reasonable quantitative calculations. Therefore, it is very important to learn balancing chemical equations, especially Redox equations that can be manipulated through definite steps. For example, dichromate reacts with iron yielding Fe 3+ and Cr 3+ Cr 2 O 7 2- + Fe 2+  Cr 3+ + Fe 3+ To balance this equation, the following steps should be followed:

7. 7 1. Split the equation into two half reactions Fe 2+  Fe 3+ Cr 2 O 7 2-  Cr 3+ 2. Proceed with each half reaction separately starting with mass balance. Let us start with first half reaction Fe 2+  Fe 3+ One mole of Fe 2+ yields one mole of Fe 3+ which is balanced.

8. 8 3. Balance the charges on both sides. Fe 2+  Fe 3+ It is clear that an electron (e) should be added to the right side in order for the charges to be equal Fe 2+  Fe 3+ + e This is a straightforward process but now consider the second half reaction, which contains species that are not mass balanced

9. 9 Cr 2 O 7 2-  Cr 3+ 1. Adjust the number of moles of chromium on both sides. Cr 2 O 7 2-  2 Cr 3+ 2. For each oxygen atom place an H 2 O on the other side Cr 2 O 7 2-  2 Cr 3+ + 7 H 2 O 3. Adjust the number of hydrogen atoms on both sides of the equation by adding H + 14 H + + Cr 2 O 7 2-  2 Cr 3+ + 7 H 2 O 4. The charges on both sides should be balanced at this point. This can be done by addition of 6 electrons to the left side of the equation 6 e + 14 H + + Cr 2 O 7 2-  2Cr 3+ + 7H 2 O

10. 10 5. This is the case if the reaction is carried out in acidic solution. The combination of the two half reactions necessitates the multiplication of the half reaction involving the iron by a factor of 6 in order to cancel the electrons involved 6 Fe 2+  6 Fe 3+ + 6 e 6 e + 14 H + + Cr 2 O 7 2-  2 Cr 3+ + 7 H 2 O _________________________________________ 6 Fe 2+ + 14 H + + Cr 2 O 7 2-  6 Fe 3+ + 2Cr 3+ + 7 H 2 O This is the balanced equation assuming acidic conditions. In basic solutions, balancing Redox equations requires extra effort where all previous steps are needed in addition to other steps.

11. 11 Example C 2 O 4 2- + MnO 4 -  Mn 4+ + CO 2 First, proceed as in case of acidic solution. 1. Split the equation into two half reactions C 2 O 4 2-  CO 2 MnO 4 -  Mn 4+ 2. Balance the first half reaction by adjusting the number of atoms on both sides (mass balance) C 2 O 4 2 ‑  2 CO 2

12. 12 3. Adjust the charges on both sides (charge balance) C 2 O 4 2- = 2 CO 2 + 2 e The first half reaction is completely balanced. Now, turn to balance the second half reaction MnO 4 -  Mn 4+ 1. Mass balance shows 1 mole of Mn on both sides, therefore Mn is adjusted. 2. Adjust the oxygens on both sides by placing an H 2 O for each oxygen atom present. MnO 4 -  Mn 4+ + 4 H 2 O

13. 13 3. Adjust the number of hydrogen atoms on both sides by additon of H + 8 H + + MnO 4 -  Mn 4+ + 4 H 2 O 4. Adjusting the charges on both sides gives 3 e + 8 H + + MnO 4 -  Mn 4+ + 4 H 2 O Now, watch carefully. 5. Add an OH - on both sides for each H + present 8 OH - + 3 e + 8 H + + MnO 4 -  Mn 4+ + 4 H 2 O + 8 OH - 6. Combine the OH - and H + to form H 2 O 3 e + 8 H 2 O + MnO 4 -  Mn 4+ + 4 H 2 O + 8 OH -

14. 14 7. Adjust the number of H 2 O molecules on both sides 3 e + 4 H 2 O + MnO 4 -  Mn 4+ + 8 OH - 8. Combine the two half reactions 3 (C 2 O 4 2-  2 CO 2 + 2 e) 2 ( 3 e + 4 H 2 O + MnO 4 -  Mn 4+ + 8 OH - ) ______________________________________________ 3 C 2 O 4 2- + 8 H 2 O + 2 MnO 4 -  6 CO 2 + 2 Mn 4+ + 16 OH - The first half reaction was multiplied by 3 and the second was multiplied by 2 in order to cancel the electrons on both sides.

15. 15 Electrochemical Cells There are two types of electrochemical cells in the concept whether the cell generates potential (called a galvanic cell) or consumes potential (called an electrolytic cell). A cell is simply constructed from two electrodes immersed in solution. The electrode at which reduction occurs is called the cathode while that at which oxidation occurs is called the anode. In galvanic cells, the cathode is positive (+) while the anode is negative (-). The signs of the anode and cathode are the opposite in electrolytic cells.

16. 16 The cells we will study in this chapter are of the first type where electrons generated by one half reaction will be consumed by the other half reaction forcing the current to flow.

17. 17 The Standard Hydrogen Electrode (SHE) The standard hydrogen electrode is an electrode with an arbitrarily assigned of zero. It is also known as the normal hydrogen electrode (NHE), and is used in combination with other half cells at standard state in order to determine the standard electrode potential of the other half cell.

18. 18 Hydrogen electrode is based on the redox half cell: 2H + (aq) + 2e → H 2(g) This redox reaction occurs at platinized platinum electrode. The electrode is dipped in an acidic solution and pure hydrogen gas is bubbled through it. The concentration of both the reduced form and oxidized form is maintained at unity. That implies that the pressure of hydrogen gas is 1 bar and the activity of hydrogen ions in the solution is unity.

19. 19 Hydrogen electrode 2H + + 2e - = H 2 E o = 0.000 V. when P H2 = 1 atm, [H + ] = 1 M, 298K 2H + + 2e - = H 2 E o = 0.000 V. when P H2 = 1 atm, [H + ] = 1 M, 298K

20. 20 Determination of Standard Electrode Potential The standard electrode potential of a half reaction can be determined using a conventional two electrode cell but replacing one electrode with a standard electrode like the standard hydrogen electrode (SHE) for which an arbitrary potential zero is assigned. Therefore, the electrode potential is the reading of the voltmeter. For example, when a 1.00 M Cu 2+ solution is placed in contact with a Cu wire and the cell is completed with a SHE at standard conditions of temperature and pressure, the potential will read 0.342 V. This is due to reaction: Cu 2+ + 2e  Cu (s) E o = 0.342 V

21. 21 Cu 2+ + 2e = Cu(s) H 2 = 2H + + 2e

22. 22 On the other hand, when Zn 2+ solution is placed in contact with a Zn electrode in an electrochemical cell with a SHE as the second electrode, the potential will read - 0.762 V. The reaction is: Zn 2+ + 2e  Zn (s) E o = - 0.762 V From these results we can predict that Cu 2+ is a better oxidizing agent since the standard electrode potential is more positive than that for Zn 2+.

23. 23

24. 24 The more positive the E o, the better oxidizing agent is the oxidized form (e.g., MnO 4 - ). The more negative the E o, the better reducing agent is the reduced form (e.g., Zn). The more positive the E o, the better oxidizing agent is the oxidized form (e.g., MnO 4 - ). The more negative the E o, the better reducing agent is the reduced form (e.g., Zn).

25. 25 However, SHE is troublesome and more convenient reference electrodes are used. Saturated calomel electrode (SCE) and Ag/AgCl electrodes are most common.

26. Lecture 39 Redox Equilibria Cell Potential

27. 27 Commercial saturated calomel electrode The S.C.E. is a common reference electrode. The cell half-reaction is: Hg 2 Cl 2 + 2e - = 2Hg + 2Cl -. E = 0.242 V for saturated KCl. The S.C.E. is a common reference electrode. The cell half-reaction is: Hg 2 Cl 2 + 2e - = 2Hg + 2Cl -. E = 0.242 V for saturated KCl.

28. The potential of the following half cell reaction at standard state was measured versus SCE giving a value of 0.529V. Find E o of the half cell reaction: Fe 3+ + e  Fe 2+

29. 29 Schematic representation of Zn 2+ /Zn(s) electrode potential relative to different reference electrodes (E o SHE = 0.000V) Zn 2+ + 2e  Zn (s) E o = - 0.762 V

30. The potential of the following half cell reaction at standard state was measured versus SCE giving a value of - 0.103V. Find E o of the half cell reaction: Sn 4+ + 2e  Sn 2+

31. 31 Cell for potentiometric measurements A complete cell consists of an indicating electrode that responds to the analyte and a reference electrode of fixed potential. The potential difference between the two is measured. A complete cell consists of an indicating electrode that responds to the analyte and a reference electrode of fixed potential. The potential difference between the two is measured.

32. 32 Calculating the Cell Potential The process of calculating the cell potential is simple and involves calculation of the potential of each electrode separately, then the overall cell potential can be determined from the simple equation: E cell = E cathode - E anode Sometimes, this relation is written as: E cell = E right – E left This is true since the convention is to place the cathode to the right of the cell while the anode is placed to the left of the cell

33. 33 Cell Representation Zn(s) | ZnCl 2 (0.200 M) II CuSO 4 (0.100 M) I Cu(s)

34. 34 This cell is read as follows: a zinc electrode is immersed in a 0.200 M ZnCl 2 solution (this is the first half-cell, anode), 0.100 M CuSO 4 solution in which a Cu(s) electrode is immersed (this is the second half-cell, cathode). If the E cell is a positive value, the reaction is spontaneous and if the value is negative, the reaction is nonspontaneous in the direction written and will be spontaneous in the reverse direction.

35. 35 Effect of Concentration on Electrode Potential The IUPAC convention for writing half-cell reactions is to represent the process as a reduction. The more positive half-cell reaction is the oxidizing agent, and the less positive half-cell reaction is the reducing agent. The relationship between the concentration and the electrode potential for a half-cell reaction is represented by Nernst equation where for the half-cell reaction we have: aA + bB + ne  cC + dD E o = x V

36. 36 The Nernst equation can be written as: E = E o – (RT/nF) ln {[C] c [D] d /[A] a [B] b } Where, R is the molar gas constant (R = 8.314 J mol -1 K -1 ), T is the absolute temperature in Kelvin ( T = o C + 273) and F is the Faraday constant ( F = 96485 Coulomb.eq -1 ) and n is the number of electrons per mole. One may write after substitution: E = E o – (0.0592/n) log {[C] c [D] d /[A] a [B] b }

37. 37 Example Calculate the electrode potential for the half- cell below if the solution contains 0.100 M Cu 2+. The half-cell reaction is: Cu 2+ + 2e  Cu (s) E o = 0.342 V Solution E = E o – (0.0592/n) log [Cu(s)]/[Cu 2+ ] E = 0.342 – (0.0592/2) log 1/0.100 E = 0.312 V

38. 38 Example Calculate the electrode potential of a half-cell containing 0.100 M KMnO 4 and 0.0500 M MnCl 2 at pH 1.000. Solution MnO 4 - + 8 H + + 5 e  Mn 2+ + 4 H 2 O E o = 1.51 V E = E o – (0.0592/n) log [Mn 2+ ]/[MnO 4 - ][H + ] 8 E = 1.51 – (0.0592/5) log 0.0500/0.100*(0.100) 8 E = 1.42 V

39. Lecture 40 Redox equilibria and Titrations, Cont… Redox Titrations (1)

40. 40 Redox Indicators Redox visual indicators are of two types. The first type is called specific while the other type is called nonspecific. Nonspecific redox indicators are very weak oxidants or reductants which have different colors of the oxidized and reduced forms. When all the analyte is consumed in a redox reaction, the first drops of excess titrant will react with the indicator, thus changing its color.

41. 41 An example of a specific indicator is starch where it forms a blue complex with iodine but not with iodide. When iodine is consumed, the blue complex disappears and the solution turns almost white indicating the end point. Another specific indicator is the permanganate ion where it is also called a self indicator. The purple color of the permanganate ion is converted to the colorless Mn 2+.

42. 42 The redox indicator equilibrium can be represented by the equation below: In OX + n e = In RED E o = x V E = E In o – (0.0592/n In ) log [In RED ]/[In OX ] The color of the oxidized form can be clearly distinguished when 10 [In RED ] = [In OX ] and the color of the reduced form can be clearly distinguished when [In RED ] = 10 [In OX ].

43. 43 Substituting into the electrode potential equation we get:  E Color Change = E In o + (0.0592/n In ) Therefore, the color change of the indicator occurs around E o In and, in fact, very close to it. For a real titration, E o In should be as close as possible to the electrode potential at the equivalence point, E ep, and should thus be at the potential range of the sharp break of the titration curve

44. 44 Examples of some redox indicators

45. 45 Redox Titrations The idea of redox titrations is that the potential of a redox reaction will change as the concentration changes. Therefore, consider the following redox reaction, Fe 2+ + Ce 4+  Fe 3+ + Ce 3+ If we titrate Fe 2+ with Ce 4+, the concentration of Fe 2+ will decrease upon addition of Ce 4+ and thus the potential of an electrode immersed in a Fe 2+ solution will start to change. This means that the potential of the cell will be a function of Fe 2+ concentration. Thus, if the volume of Ce 4+ added is plotted against the electrode potential, a titration curve will be obtained.

46. 46 Titration Curves Oxidation-reduction titrations and titration curves can be followed by measuring the potential of an indicator electrode (e.g. platinum) as compared to a reference electrode (has a fixed constant potential, like Saturated Calomel Electrode, SCE). The potential of the cell measured is: E cell = E solution – E reference electrode

47. 47

48. 48 E cell = E solution – E reference electrode This means that: E solution = E cell + E reference electrode Assume the process of titration of Fe 2+ with Ce 4+, we have Fe 2+ + Ce 4+  Fe 3+ + Ce 3+ Fe 3+ + e  Fe 2+ E o = 0.771 V Ce 4+ + e  Ce 3+ E o = 1.70 V We have three cases to consider:

49. 49 1. Before equivalence point we have Fe 2+ /Fe 3+ couple and we can use the Nernst equation to calculate the electrode potential where: E Fe = E Fe o – (0.0592/n Fe ) log [Fe 2+ ]/[Fe 3+ ] Therefore, we calculate the concentration of Fe 2+ and Fe 3+ and insert the values in the Nernst equation to find the resulting potential of the electrode.

50. 50 We can use the other half reaction to calculate the potential of the cell but it is easier to use the Fe 2+ /Fe 3+ couple since the concentration of Ce 4+ is difficult to calculate at this stage. The Nernst equation for such potential calculation is: E Ce = E Ce o – (0.0592/n Ce ) log [Ce 3+ ]/[Ce 4+ ]

51. 51 2. At the equivalence point, we may combine the two Nernst equations for the two half reactions by adding them up. We get: n Fe E Fe + n Ce E Ce = n Fe E Fe o + n Ce E Ce o – (0.0592) {log [Fe 2+ ]/[Fe 3+ ] + log [Ce 3+ ]/[Ce 4+ ]} n Fe E Fe + n Ce E Ce = n Fe E Fe o + n Ce E Ce o – (0.0592) {log [Fe 2+ ] [Ce 3+ ]/[Fe 3+ ] [Ce 4+ ]} But [Fe 3+ ] = [Ce 3+ ] and [Fe 2+ ] = [Ce 4+ ] at equivalence point.

52. 52 Therefore, the equation simplifies to: n Fe E Fe + n Ce E Ce = n Fe E Fe o + n Ce E Ce o Also, at equivalence point we have E Fe = E Ce = E eq pt, therefore: (n Fe + n Ce ) E eq pt = n Fe E Fe o + n Ce E Ce o The final equation can be written as: E eq pt = (n Fe E Fe o + n Ce E Ce o )/ (n Fe + n Ce )

53. 53 3. After equivalence point, it is easier to calculate the cell potential from the cerium half-cell using the equation: E Ce = E Ce o – (0.0592/n Ce ) log [Ce 3+ ]/[Ce 4+ ] It is easier to calculate the Ce 3+ and Ce 4+ concentrations and it will be difficult to calculate the Fe 2+ concentration since it is totally consumed, except for an equilibrium concentration.

54. 54 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  Fe 2+ E o = 0.771 V Ce 4+ + e  Ce 3+ E o = 1.70 V Solution 1. After addition of 0 mL Ce 4+ The solution contains Fe 2+ only and attempts to apply Nernst equation will not be possible since assumably no Fe 3+ will be present. E Fe = E Fe o – (0.0592/n Fe ) log [Fe 2+ ]/[Fe 3+ ]

55. 55 E Fe = E Fe o – (0.0592/n Fe ) log [Fe 2+ ]/[Fe 3+ ] E Fe = E Fe o – (0.0592/n Fe )log 0.1/0 This means that as Fe 3+ approaches zero, the electrode potential approaches negative infinity. This, in fact, is not practically true since it suggests that a Fe 2+ solution will have unbelievably high reducing power. The way out of this problem is to always assume the presence of very small Fe 3+ concentration in equilibrium with Fe 2+, even in very pure Fe 2+ solutions. The conclusion with regards to calculation of electrode potential after addition of 0 mL Ce 4+ is that we can not calculate it using direct Nernst equation.

56. 56 2. After addition of 25 mL Ce 4+ Initial mmol Fe 2+ = 0.10 x 50 = 5.0 mmol Ce 4+ added = 0.10 x 25 = 2.5 mmol Fe 2+ left = 5.0 – 2.5 = 2.5 [Fe 2+ ] = 2.5/75 M mmol Fe 3+ formed = 2.5 [Fe 3+ ] = 2.5/75 M Application of Nernst equation gives: E Fe = E Fe o – (0.0592/n Fe )log {(2.5/75)/(2.5/75)} E Fe =0.771 V Therefore, the standard electrode potential for Fe 2+ /Fe 3+ couple can be calculated at half the way to the equivalence point.

57. 57 3. After addition of 40 mL Ce 4+ Initial mmol Fe 2+ = 0.10 x 50 = 5.0 mmol Ce 4+ added = 0.10 x 40 = 4.0 mmol Fe 2+ left = 5.0 – 4.0 = 1.0 [Fe 2+ ] = 1.0/90 M mmol Fe 3+ formed = 4 [Fe 3+ ] = 4.0/90 M Application of Nernst equation gives: E Fe = E Fe o – (0.0592/n Fe ) log {(1.0/90)/(4.0/90)} E Fe = 0.771 – 0.0592 log 1.0/4.0 E Fe = 0.807 V

58. 58 4. After addition of 50 mL Ce 4+ Initial mmol Fe 2+ = 0.10 x 50 = 5.0 mmol Ce 4+ added = 0.10 x 50 = 5.0 mmol Fe 2+ left = 5.0 – 5.0 = ?? This is the equivalence point mmol Fe 3+ formed = 5.0 [Fe 3+ ] =5.0/100 = 0.05 M, [Ce 3+ ] = 0.05 M At equivalence point, we have from the above discussion: E eq pt = (n Fe E Fe o + n Ce E Ce o )/ (n Fe + n Ce ) E eq pt = (1 * 0.771 + 1 * 1.70)/(1 + 1) E eq pt =1.24 V

59. 59 3. After addition of 75 mL Ce 4+ Initial mmol Fe 2+ = 0.10 x 50 = 5.0 mmol Ce 4+ added = 0.10 x 75 = 7.5 mmol Ce 4+ excess = 7.5 – 5.0 = 2.5 [Ce 4+ ] = 2.5/125 M mmol Ce 3+ formed = 5.0 [Ce 3+ ] = 5.0/125 M Application of Nernst equation gives: E Ce = E Ce o – (0.0592/n Ce ) log [Ce 3+ ]/[Ce 4+ ] E Ce =1.70 – (0.0592/1)log {(5.0/125)/(2.5/125)} E Ce = 1.68 V

60. Lecture 41 Redox Equilibria and Titrations, Cont… Redox Titrations (2)

61. 61 Example Find the electrode potential of a 50 mL of a 0.10 M solution of Fe 2+ after addition of 0, 5, 10, and 20 mL 0f 0.10 KMnO 4, at pH 1. Solution MnO 4 - + 8H + + 5 Fe 2+  Mn 2+ + 5 Fe 3+ + 4 H 2 O Fe 3+ + e  Fe 2+ E o = 0.771 V MnO 4 - + 8H + + 5e  Mn 2+ + 4 H 2 O E o = 1.51 V

62. 62 1. After addition of 0 mL MnO 4 - The solution contains Fe 2+ only and attempts to apply Nernst equation will not be possible since assumably no Fe 3+ will be present. E Fe = E Fe o – (0.0592/n Fe ) log [Fe 2+ ]/[Fe 3+ ] E Fe = E Fe o – (0.0592/n Fe )log 0.1/0 This means that as Fe 3+ approaches zero, the electrode potential approaches negative infinity. This, in fact, is not practically true since it suggests that a Fe 2+ solution will have unbelievably high reducing power. The conclusion with regards to calculation of electrode potential after addition of 0 mL MnO 4 - is that we can not calculate it using direct Nernst equation.

63. 63 2. After addition of 5 mL MnO 4 - mmol Fe 2+ left = initial mmol Fe 2+ - mmol Fe 2+ reacted mmol Fe 2+ reacted = 5 mmol MnO 4 - mmol Fe 2+ left = initial mmol Fe 2+ - 5 mmol MnO 4 - mmol Fe 2+ left = 0.1*50 – 5 *(0.1*5) mmol Fe 2+ left = 5.0 – 2.5 = 2.5, [Fe 2+ ] = 2.5/55 M mmol Fe 3+ formed = 2.5, [Fe 3+ ] = 2.5/55 M Application of Nernst equation gives: E = E Fe o – (0.0592/n Fe )log {(2.5/55)/(2.5/55)} E =0.771 V Therefore, the standard electrode potential for Fe 2+ /Fe 3+ couple can be calculated at half the way to the equivalence point.

64. 64 3. After addition of 10 mL MnO 4 - mmol Fe 2+ left = initial mmol Fe 2+ - mmol Fe 2+ reacted mmol Fe 2+ left = initial mmol Fe 2+ - 5 mmol MnO 4 - mmol Fe 2+ 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 E MnO4 - / Mn 2+ = E o MnO4 - / Mn 2+ – (0.0592/n MnO4 - / Mn 2+ ) log [Mn 2+ ]/[MnO 4 - ][H + ] 8 E Fe = E Fe o – (0.0592/n Fe ) log [Fe 2+ ]/[Fe 3+ ]

65. 65 E eq = E o MnO4 - / Mn 2+ – (0.0592/n MnO4 - / Mn 2+ ) log [Mn 2+ ]/[MnO 4 - ][H + ] 8 E eq = E Fe o – (0.0592/n Fe ) log [Fe 2+ ]/[Fe 3+ ] Addition of the two equations gives: (5E ep + E ep ) = (5E o Mn + E Fe o ) – 0.0592 log { [Mn 2+ ][Fe 2+ ]/[Fe 3+ ][MnO 4 - ][H + ] 8 } Substitute for the following: [Fe 2+ ] = 5 [MnO 4 - ] [Fe 3+ ] = 5[Mn 2+ ] 6E ep = (5E o Mn + E Fe o ) – 0.0592 log { [Mn 2+ ] * 5[MnO 4 - ]/5 [Mn 2+ ] [MnO 4 - ][H + ] 8 } (5 + 1) E ep = (5 * 1.51 + 1 * 0.771) – 0.0592 log 1/(0.10) 8 E ep = 1.31 V

66. 66 4. After addition of 20 mL MnO 4 - mmol MnO 4 - excess = mmol MnO 4 - added - mmol MnO 4 - reacted mmol MnO 4 - reacted = 1/5 mmol Fe 2+ mmol MnO 4 - excess = mmol MnO 4 - added - 1/5 mmol Fe 2+ mmol MnO 4 - excess = 0.1*20 – 1/5* (0.1*50) = 1 [MnO 4 - ]= 1/70 M mmol Mn 2+ formed = 1 [Mn 2+ ] = 1/70 M E = E o MnO4 - / Mn 2+ – (0.0592/n MnO4 - / Mn 2+ ) log [Mn 2+ ]/[MnO 4 - ][H + ] 8 E = 1.51 – (0.0592/5) log (1/70)/{(1/70)(0.1) 8 } E = 1.42 V