Question

An aqueous solution contains 0.100 M Ce3+; 1.00 × 10-1 M Ce4+; 1.00 × 10-4 M Mn 2+; 0.100 M MnO4 - and 1.00 M HClO4 at 298 K.

You are given:

Ce4+ + e-  Ce3+ E° = 1.70 V

MnO4 - + 8H+ + 5e-  Mn2+ + 4H2O E° = 1.507 V 5.

(a) Write a balanced net reaction that can occur between the species in this solution and calculate its E° value.

(b) Calculate ΔG° and K for the reaction at 298 K.

(c) Calculate the total E for the net reaction in part (a) at 298 K.

(d) At what pH would the concentrations of Ce3+, Ce4+, Mn2+ and MnO4 - listed above be in equilibrium at 298 K?

Answer 1

reductions:
Ce⁴⁺. + e^- → Ce³⁺. (aq). Eo = + 1.70 volts
MnO₄^- + 8 H⁺ + 5 e^- → Mn²⁺ + 4 H₂O   Eo = 1.507 volts

so the reaction that can occur is
reduction 5Ce⁴⁺. + 5e^- → 5Ce³⁺. (aq). Eo = + 1.70 volts
oxidation: Mn²⁺ + 4 H₂O -->  MnO₄^- + 8 H⁺ + 5 e^- Eo = -1.507 volts
then Eo = 0.193 volts

a)Write a balanced net reaction that might occur between species in this solution.
5Ce⁴⁺. +  Mn²⁺ + 4H₂O → 5 Ce³⁺. +  MnO₄^- + 8H⁺


b) Calculate ΔG˚
ΔG˚ = - nFEo
ΔG˚ = - (5) (96.5 kJ/mol e-volt)(0.193 volt)
ΔG˚ = - 93.15 kJ

b)Calculate K for the reaction
ΔG˚ = -RT ln K

-93150 Joules = -(8.314 J/K-mol)(298K) ln K

lnK = -93150 Joules / -(8.314 J/K-mol)(298K)

ln K = 37.6

K = 2.135 e16


c)Calculate E for the conditions given.

E = Eo - (0.0592 /n) (log [products] / [reactants])

5Ce4+. +  Mn+2 + 4H2O → 5 Ce3+. +  MnO4- + 8H+

E = 0.193 - (0.0592 / 5) (log [Ce3]⁵ [MnO4-] [H]⁸ / [Ce4]⁵ [Mn²⁺])

E = 0.193 - (0.01184) (log [0.1]⁵ [0.1] [1]⁸ / [10^-1]⁵ [10^-4])

E = 0.193 - (0.01184) (log 10³)

E = 0.193 - 0.01184 *3

E = 0.193 - 0.01184 * 3

E = 0.193 - 0.0355

d) At what pH would be the given concentrations of Ce4+, Ce3+, Mn2+, and MnO4- be in the equilibrium at 298 K?

E = Eo - (0.0592 /n) (log [products] / [reactants])

@ equilibrium becomes:

0 = Eo - (0.0592 /n) (log [products] / [reactants])

&

K = 2.135 e16 for
5Ce4+. & Mn+2 &4H2O → 5 Ce3+. & MnO4- & 8H+

2.135 e16 = [Ce3]^5 [MnO4] [H]^8 / [Ce4]^5 [Mn])

2.135 e16= [0.1]^5 [0.1] [H+]^8 / [0.1]^5 [10^-4])

2.135 * 10¹⁶= 10³ [H+]^8

[H+]^8 = 2.135 * 10¹⁶ / 10³

[H+]^8 = 213500 * 10⁸

[H+] = 46.3

pH = -log[H+]
- log 46.3

E = 0.157