For a Pt|Fe3+, Fe2+ half-cell, find the potential for the filling ratios of [Fe3

Question

For a Pt|Fe3+, Fe2+ half-cell, find the potential for the filling ratios of [Fe3+]/[Fe2+]. I need help with 18-27 and 18-28.

18-27. For a PeFe, Fe half-cell, find the potential for the following ratios of [FeFe:0.001, 0.0025, 0.005, 0.0075, 0.010, 0.025, 0.050, 0.075, 0.100, 0.250, 0.500, 0.750, 1,00, 1.250, 1.50, 1.75, 2.50, 5.00, 10.00, 25.00, 75.00, and 100.00 18-28. For a 18-28. Fot a PtCe Ce h alf-cell, tind the potential for the same ratios of [Ce4+)/Ice+1 as given in Problem 18-27 for [Fe3+)/[Fe2+].

Explanation / Answer

18-27 : Using Nernst equation,

E = 0.771 + 0.0592 log([Fe3+]/[Fe2+])

[Fe3+]/Fe2+] = 0.001

E = 0.771 + 0.0592log(0.001) = 0.593 V

[Fe3+]/Fe2+] = 0.0025

E = 0.771 + 0.0592log(0.0025) = 0.617 V

[Fe3+]/Fe2+] = 0.005

E = 0.771 + 0.0592log(0.005) = 0.635 V

[Fe3+]/Fe2+] = 0.0075

E = 0.771 + 0.0592log(0.0075) = 0.645 V

[Fe3+]/Fe2+] = 0.010

E = 0.771 + 0.0592log(0.010) = 0.653 V

[Fe3+]/Fe2+] = 0.025

E = 0.771 + 0.0592log(0.025) = 0.676 V

[Fe3+]/Fe2+] = 0.050

E = 0.771 + 0.0592log(0.050) = 0.694 V

[Fe3+]/Fe2+] = 0.075

E = 0.771 + 0.0592log(0.075) = 0.704 V

[Fe3+]/Fe2+] = 0.100

E = 0.771 + 0.0592log(0.100) = 0.712 V

[Fe3+]/Fe2+] = 0.250

18-28 : Using Nernst equation,

E = 1.61 + 0.0592 log([Ce4+]/Ce3+])

[Ce4+]/Ce3+] = 0.001

E = 1.61 + 0.0592log(0.001) = 1.432 V

[Ce4+]/Ce3+] = 0.0025

E = 1.61 + 0.0592log(0.0025) = 1.456 V

[Ce4+]/Ce3+] = 0.005

E = 1.61 + 0.0592log(0.005) = 1.474 V

[Ce4+]/Ce3+] = 0.0075

E = 1.61 + 0.0592log(0.0075) = 1.484 V

[Ce4+]/Ce3+] = 0.010

E = 1.61 + 0.0592log(0.010) = 1.492 V

[Ce4+]/Ce3+] = 0.025

E = 1.61 + 0.0592log(0.025) = 1.515 V

[Ce4+]/Ce3+] = 0.050

E = 1.61 + 0.0592log(0.050) = 1.533 V

[Ce4+]/Ce3+] = 0.075

E = 1.61 + 0.0592log(0.075) = 1.543 V

[Ce4+]/Ce3+] = 0.100

E = 1.61 + 0.0592log(0.100) = 1.551 V

[Ce4+]/Ce3+] = 0.250

E = 1.61 + 0.0592log(0.250) = 1.574 V

[Ce4+]/Ce3+] = 0.500

E = 1.61 + 0.0592log(0.500) = 1.592 V

[Ce4+]/Ce3+] = 0.750

E = 1.61 + 0.0592log(0.750) = 1.603 V

[Ce4+]/Ce3+] = 1.0

E = 1.61 + 0.0592log(1.0) = 1.61 V

[Ce4+]/Ce3+] = 1.250

E = 1.61 + 0.0592log(1.250) = 1.616 V

[Ce4+]/Ce3+] = 1.50

E = 1.61 + 0.0592log(1.50) = 1.620 V

[Ce4+]/Ce3+] = 1.75

E = 1.61 + 0.0592log(1.75) = 1.624 V

[Ce4+]/Ce3+] = 2.50

E = 1.61 + 0.0592log(2.50) = 1.633 V

[Ce4+]/Ce3+] = 5.00

E = 1.61 + 0.0592log(5.00) = 1.651 V

[Ce4+]/Ce3+] = 10.00

E = 1.61 + 0.0592log(10.0) = 1.669 V

[Ce4+]/Ce3+] = 25.0

E = 1.61 + 0.0592log(25.0) = 1.693 V

[Ce4+]/Ce3+] = 75.0

E = 1.61 + 0.0592log(75.0) = 1.721 V

[Ce4+]/Ce3+] = 100.0

E = 1.61 + 0.0592log(100.0) = 1.728 V

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