— Physical Chemistry — Electrochemistry Chemistry Question
Electrochemical equilibria
Calculate the standard electrode potential for the aqueous couple [Fe(CN)6] 3– / [Fe(CN)6] 4– from the following data:
E°(Fe 3+(aq) | Fe2+(aq)) = + 0.770 V
Fe3+(aq) + 6 CN–(aq) [Fe(CN)6] 3–(aq) log10Kc = 43.9
Fe2+(aq) + 6 CN–(aq) [Fe(CN)6] 4–(aq) log10Kc = 36.9
Model Answer
Fe3+(aq) + –e → Fe2+(aq) ; E° = + 0.770 V ; ∆G° = −74.3 kJ mol −1
Fe3+(aq) + 6 CN−(aq) → Fe(CN)3 6−(aq) ; Kc = 7.9×1043 ; ∆G° = −250.4 kJ mol−1
Fe2+(aq) + 6 CN−(aq) → Fe(CN)6 4−(aq) ; Kc = 7.9×1036 ; ∆G° = −210.5 kJ mol−1
Hence, from the cycle:
Fe(CN)6 3−(aq) + –e → Fe(CN)6 4−(aq) ; E° = +0.356 V ∆G° = −34.4 kJ mol−1 ;
The following standard electrode potentials have been reported:
In+(aq) + e– In(s) E° = – 0.13 V
In3+(aq) + 3 e– In(s) E° = – 0.34 V
Tl+(aq) + e– Tl(s) E° = – 0.34 V
Tl3+(aq) + 3 e– Tl(s) E° = + 0.72 V
Calculate the equilibrium constant for the disproportionation reaction
3 M+(aq) → M3+(aq) + 2 M(s) for In and Tl. Comment on the result.
Model Answer
b) (1) In+(aq) + –e → In(s) E° = − 0.13 V, ∆G° = 12.5 kJ mol−1
(2) In3+(aq) + 3 e− → In(s) E° = −0.34 V, ∆G° = 98.4 kJ mol−1
To balance 3×(1) – (2).
3 In+(aq) + 3 –e → 2 In(s) + In3+(aq) Kc = 4.5×1010 ∆G° = −60.8 kJ mol−1
(1) Tl+(aq) + –e → Tl(s) E° = −0.34 V, ∆G° = 32.8 kJ mol−1
(2) Tl3+(aq) + 3 –e → Tl(s) E° = +0.72 V, ∆G° = −208.4 kJ mol−1
3Tl+ (aq) + 3 –e → 2 Tl(s) + Tl3+(aq) ∆G° = +306.8 kJ mol−1