2Fe3+ + 2e- = 2Fe2+ E0 = 0.77 V
Fe = Fe2+ + 2e- E0 = 0.41 V
E0cell = Ered – Eox = 0.77 V – 0.41 V = 0.36 V
delG0 = -nFE0cell= (-2 mol e-)(96458 C/mol e-)(0.36 V) = -69449.76 J = -69.45 kJ
delG0 = -RTlnKeq
-69449.76 J = (-8.314 J/mol K-1)(298 K)(lnKeq)
lnKeq = 28
Keq = 1.45
Answer: -69.45 kJ
K for this reaction would be greater than one (1.45).
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