1. Version 2012 Updated on 0510 Copyright © All rights reserved Dong-Sun Lee, Prof., Ph.D. Chemistry, Seoul Women’s University Chapter 17 Applications of Standard Electrode Potentials

2. Gustav Robert Kirchhoff(1824-1877) was a German physicist who made many important contributions to physics and chemistry. In addition to his work in spectroscopy, he is known for Kirchhoff’s laws of current and voltage in electrical circuits. These laws can be summarized by the following equations:  I = 0 and  E = 0. These equations state that the sum of the currents into any circuit point (node) is zero and the sum of the potential differences around any circuit loop is zero.

3. Calculating potentials of electrochemical cells E cell = E right – E left Ex. Cu Cu 2+ (0.0200M) Ag + (0.0200M) Ag Cu (s) + 2Ag + = Cu 2+ (aq) + 2Ag(s) Ag + + e = Ag(s) E o = + 0.799 V Cu 2+ + 2e = Cu(s) E o = + 0.337 V 2Ag + + 2e = 2Ag(s) E o = + 0.799 V E = 0.799 – (0.05916/1) log (1/0.0200) = 0.6984 V Cu 2+ + 2e = Cu(s) E o = + 0.337 V E = 0.337 – (0.05916/2) log (1/0.0200) = 0.2867V Cu (s) + 2Ag + = Cu 2+ (aq) + 2Ag(s) E o cell = + 0.799 V – (+ 0.337 V) E cell = E right – E left = 0.6984 – 0.2864 = + 0.462 V = + 0.412 V  G = – nFE = – 2 ×96485 C × 0.412 V = 79,503 J (18.99 kcal)

4. EX Calculate the potential of the following cell and indicate the reaction that would occur spontaneously if the cell were short-circuited. Pt U 4+ (0.200M), UO 2 2+ (0.0150M), H + (0.300M) Fe 2+ (0.0100M), Fe 3+ (0.0250M) Pt

5. The two half-reactions are Fe 3+ + e _ ↔ Fe 2+ E o = + 0.771V UO 2 2+ + 4H + + 2e - ↔ U 4+ + 2H 2 O E o = + 0.334V The electrode potential for the right-hand electrode is E right = 0.771 – 0.05916 log [Fe 2+ ]/[Fe 3+ ] = 0.771 – 0.05916 log 0.0100/0.0250 = 0.771 – (– 0.0236) = 0.7946 V The electrode potential for the left-hand electrode is E Left = 0.334 – (0.0592 / 2) log [U 4+ ] / [UO 2 2+ ][ H + ] 4 = 0.334 – (0.0592 / 2) log 0.200 / (0.0150) (0.0300) 4 = 0.334 – 0.2136 = 0.1204 V And E cell = E right – E left = 0.7946 – 0.2136 = 0.674V The positive sign means that the spontaneous reaction is the oxidation of U 4+ on the left and the reduction of Fe 3+ on the right, or U 4+ +2 Fe 3+ + 2H 2 O ↔ UO 2 2+ +Fe 2+ + 4H +

6. EX Calculate the cell potential for Ag AgCl (sat’d), HCl (0.0200M) H 2 (0.800atm ), Pt Cell without liquid junction

7. Note that this cell does not require two compartments (nor a salt bridge) because molecular H 2 has little tendency to react directly with the low concentration of Ag + in the electrolyte solution. This is an example of a cell without liquid junction. The two half- reactions and their corresponding standard electrode potentials are 2H + + 2e - ↔ H 2 E o H+/H2 = 0.000V AgCl(s) + e - ↔ Ag (s) + Cl – E o AgCl/Ag = 0.222V The two electrode potentials are E right = 0.000 – (0.0592/2)log p H2 / [H + ] 2 = – (0.0592/2) log 0.800 / (0.0200) 3 = – 0.0977 V E left = 0.222 – 0.0592log [Cl - ] = 0.222 – 0.0592 log 0.0200 = 0.3226V The cell potential is thus E cell = E right – E left = -0.0977 – 0.3226 = – 0.420V The negative sign indicates that the cell reaction as considered 2H + + 2 Ag(s) ↔ H 2 + 2AgCl(s) is nonspontaneous. To get this reaction to occur, we would have to apply an external voltage and construct an electrolytic cell.

8. Ex Calculate the potential for the following cell using (a) concentration and (b) activities: Zn ZnSO 4 (5.00 ×10 –4 M), PbSO 4 (sat’d) Pb (a) [SO 4 2– ] = C ZnSO4 = 5.00 ×10 –4 PbSO 4 (s) + 2e  Pb (s) + SO 4 2– E o = – 0.350 V Zn 2+ + 2e  Zn (s) E o = – 0.763 V E right = E o – (0.05916 / 2) log [SO 4 2– ] = – 0.350 – (0.05916 / 2) log (5.00 ×10 –4 ) = – 0.252 V E left = E o – (0.05916 / 2) log (1 / [Zn 2+ ] = – 0.763 – (0.05916 / 2) log {1 / (5.00 ×10 –4 )} = – 0.860 V E cell = E right – E left = – 0.252 – (– 0.860) = 0.608 V

9. (b) Ionic strength for 5.00 ×10 –4 M ZnSO 4 :  = (1/2) {(5.00 ×10 –4 ) ×(+2) 2 + (5.00 ×10 –4 ) ×(– 2) 2 } = 2.00 ×10 –3 Debye-Huckel Eq. : – log  = (0.51 Z 2  ) / ( 1+ 3.3   )  SO42– = 0.4 nm,  Zn2+ = 0.4 nm   SO42– = 0.820,  Zn2+ = 0.825 Activity =  [C] E right = E o – (0.05916 / 2) log {  SO42– [SO 4 2– ] } = – 0.350 – (0.05916 / 2) log (0.820 × 5.00 ×10 –4 ) = – 0.250 V E left = E o – (0.05916 / 2) log {1 / (  Zn2+ [Zn 2+ ])} = – 0.763 – (0.05916 / 2) log {1 / (0.825 × 5.00 ×10 –4 )} = – 0.863 V E cell = E right – E left = – 0.250 – (– 0.863) = 0.613 V

10. Redox systems in the respiratory chain, P=phosphate ion. (From P.Karlson, Introduction to modern Biochemistry.)

11. Calculation Redox Equilibrium Constant Cu Cu 2+ (x M) Ag + (y M) Ag Cu (s) + 2Ag + = Cu 2+ (aq) + 2Ag(s) K eq = [Cu 2+ ] / [Ag + ] 2 2Ag + + 2e = 2Ag(s) E o = + 0.799 V Cu 2+ + 2e = Cu(s) E o = + 0.337 V E cell = E right – E left = E Ag – E Cu = 0 or E right = E left = E Ag = E Cu E o Ag – (0.05916/2) log (1/[Ag + ] 2 ) = E o Cu – (0.05916/2) log (1/[Cu 2+ ]) E o Ag – E o Cu = (0.05916/2) log (1/[Ag + ] 2 ) – (0.05916/2) log (1/[Cu 2+ ]) E o Ag – E o Cu = (0.05916/2) log (1/[Ag + ] 2 ) + (0.05916/2) log ([Cu 2+ ]/1) 2 (E o Ag – E o Cu ) / 0.05916 = log ([Cu 2+ ]/[Ag + ] 2 ) = log K eq ln K eq = –  G o /RT= – nFE o cell / RT ln K eq = – nE o cell / 0.05916 = – n ( E o right – E o left ) / 0.05916

12. Titration curve : potentiometric titration of Fe 2+ with Ce 4+. Titration reaction: Ce 4+ + Fe 2+  Ce 3+ + Fe 3+ Cell: Hg  Hg 2 Cl 2  Cl – || Ce 4+, Ce 3+, Fe 3+, Fe 2+ | Pt Apparatus for potentiometric titration of Fe 2+ with Ce 4+. Reaction at the SCE reference electrode: 2Hg(l) + 2 Cl –  Hg 2 Cl 2 (s) + 2 e E o = 0.241V Reaction at the Pt indicator electrode: Fe 3+ + e  Fe 2+ E o = 0.767V (in 1M HClO 4 ) Ce 4+ + e  Ce 3+ E o = 1.70V (in 1M HClO 4 ) Cell reaction: 2Fe 3+ + 2Hg(l) + 2 Cl –  2 Fe 2+ + Hg 2 Cl 2 (s) 2Ce 4+ + 2Hg(l) + 2 Cl –  2 Ce 3+ + Hg 2 Cl 2 (s) E Ce = E Fe = E solution E cell = E cathode – E anode = E solution – E SCE E o cell = (0.767)– (0.241) = 0.526 V The equilibium constant for this titration reaction K = 10 nEo/0.05916 log K = n E o cell / 0.05916 = (1) (0.526) / (0.05916) = K = 1.7  10 17

13. Theoretical titration curve for titration of 100ml of 0.050M Fe 2+ with 0.100M Ce 4+ in 1M HClO 4. ½VeVe Initial Fe 2+ [Ce 4+ ] = [Ce 3+ ] [Fe 2+ ] = [Fe 3+ ] [Ce 3+ ] = [Fe 3+ ] [Ce 4+ ] Region Major constituents Comment Before the titration begins Fe 2+ No calculation possible Before the equivalence point Fe 2+, Fe 3+, Ce 3+ Use the Nernst equation for the analyte half reaction At the equivalence point Fe 3+, Ce 3+ Use e.p equation After the equivalence point Fe 3+, Ce 3+, Ce 4+ Use the Nernst equation for the titrant half reaction

14. 1) Before the equivalence point (amount Fe 2+ remaining) = (amount Fe 2+ initial) – (amount Ce 4+ added  amount Fe 2+ used) (amount Fe 3+ produced) = (amount Fe 2+ used) [Fe 2+ ] = amount Fe 2+ (mmol) / vol(ml), [Fe 3+ ] = amount Fe 3+ (mmol) / vol(ml) E cell = E cathode – E anode = E solution – E SCE = (0.767 – 0.05916 log{[ Fe 2+ ]/[ Fe 3+ ]}) – (0.241) = 0.526 – 0.05916 log{[ Fe 2+ ]/[ Fe 3+ ]} When V = (1/2) Ve, [ Fe 2+ ] = [ Fe 3+ ], E cell = 0.526.

15. 2) At the equivalence point [Ce 3+ ] = [Fe 3+ ] [Ce 4+ ] = [Fe 2+ ] E eq = E o Ce4+ – 0.05916 log[Ce 3+ ]/ [Ce 4+ ] E eq = E o Fe3+ – 0.05916 log[Fe 2+ ]/ [Fe 3+ ] 2E eq = E o Ce4+ + E o Fe3+ – 0.05916 log[Ce 3+ ]/ [Ce 4+ ] – 0.05916 log[Fe 2+ ]/ [Fe 3+ ] = E o Ce4+ + E o Fe3+ – 0.05916 log[Ce 3+ ] [Fe 2+ ] / [Ce 4+ ] [Fe 3+ ] = E o Ce4+ + E o Fe3+ = 1.70 + 0.767 = 2.467 E eq = (E o Ce4+ + E o Fe3+ ) / 2 = 1.23 V E (cell) = E right – E calomel = 1.23 – 0.241 = 0.99V

16. 3) After the equivalence point [Fe 2+ ] = 0 E (cell) = E + – E calomel = ( 1.70 – 0.05916 log [Ce 3+ ]/[Ce 4+ ]) – 0.241 amount Ce 4+ remaining = amount Ce 4+ added – amount Ce 4+ used = amount Ce 4+ added – (amount Fe 2+ initial × reacting ratio )

17. Titration curves for 0.1000M Ce 4+ titration, Curve A: Titration of 50.00mL of 0.05000 M Fe 2+. Curve B: Titration of 50.00 ml of 0.02500 M U 4+.

19. General approach to redox titration ( titration with an oxidizing agent) T + A  T – + A + titrant analyte T + e  T – E = E o T – 0.05916 log [T – ] / [T] [T – ] / [T] = 10 ( EoT – E) / 0.05916 [T – ] =  [T] A + + e  A E = E o A – 0.05916 log [A ] / [A + ] [A ] / [A + ] = 10 ( EoA – E) / 0.05916 [A ] =  [A + ] Mass balance [T – ] + [T] = T total [T – ] + (1/  ) [T – ] = T total [T – ] = (  T total ) / (1+  ) [A + ] = A total / (1+  ) [T – ] = [A + ] 1:1 mole ratio (  T total ) / (1+  ) = A total / (1+  ) Fraction of titration  = T total / A total ( = 1 at eq point)  = (1+  ) / {  ( 1+  )}

20. Titration of a mixture The titration of two species will exhibit two breaks if the standard potentials of the redox couples are sufficiently different. Example : Titration reactions First : IO 3 – + 2Sn 2+ + 2Cl – + 6H +  ICl 2 – + 2Sn 4+ + 3H 2 O Second : IO 3 – + 2Tl + + 2Cl – + 6H +  ICl 2 – + 2Tl 3+ + 3H 2 O Half reactions IO 3 – + 2Cl – + 6H + + 4e  ICl 2 – + 3H 2 O E o = 1.24 V  = [ICl 2 – ] /[IO 3 – ] = 10 [4(1.24 – E) / 0.05916] – 2 pCl – 6 pH Sn 4+ + 2e  Sn 2+ E o = 0.139 V  1 = [Sn 2+ ]/[Sn 4+ ] = 10 2 (0.139 – E) / 0.05916 Tl 3+ + 2e  Tl + E o = 0.77 V  2 = [Tl + ]/[Tl 3+ ] = 10 2 (0.77 – E) / 0.05916

21. Theoretical curve for titration of 100 ml of 0.01M Tl + with 0.01M IO 3 – in 1M HCl Theoretical curves for 0.01M Tl + plus 0.01M Sn 2+ titrated with 0.01M IO 3 –. The initial volume of analyte is 100ml and all solutions contain 1M HCl.

22.  plot for the Fe 2+ /Ce 4+ system. Totration curve calculated using the inverse master equation approach.  =  Fe3+ /  Ce3+

23. Factors affecting the shape of titration curves : 1) concentration 2) completeness of reaction V(ml) E(V) concentrated diluted Higher E o titrant Lower E o titrant

24. Effect of titrant electrode potential on reaction completeness.

25. Redox indicator In (oxidized) + ne  In (reduced) E = E o – (0.05916 / n) log [In (reduced) ] / [In (oxidized) ] [In (reduced) ] = [In (oxidized) ] E = E o ± (0.05916 / n) [In (reduced) ] / [In (oxidized) ]  (10/1) [In (reduced) ] [In (reduced) ] / [In (oxidized) ]  (1/10) [In (oxidized) ] Indicator transition range = transition range – E(calomel) vs SCE vs SHE Color changes for general redox indicators depend only on the potential of the system. The range of potentials over which a color change occurs (the transition potential) is often pH dependent.

26. Iron complexes of 1,10-phenanthrolines (orthophenanthrolines)

28. Starch-Iodine complex Starch solution(05~ 1%) is not redox indicator. The active fraction of starch is amylose, a polymer of the sugar  -D-glucose ( 1,4 bond). The polymer exists as a coiled helix into which small molecules can fit. In the presence of starch and I –, iodine molecules form long chains of I 5 – ions that occupy the center of the amylose helix. [I I I I I] – [I I I I I] – Visible absorption by the I 5 – chain bound within the helix gives rise to the characteristic starch-iodine color.

29. Structure of the repeating unit of the sugar amylose. Schematic structure of the starch-iodine complex. The amylose chain forms a helix around I 6 unit. View down the starch helix, showing iodine, inside the helix.

30. Potentiometric end points for redox titrations: reference electrode || analyte solution | Pt Reference electrode : SCE The end point can be determined from a plot of the measured potential as a function of titrant volume. E(vs SCE) Titrant volume

31. Summary Potential of electrochemical cells Redox equilibrium constant Redox titration curve Redox indicator Iodine starch indicator