Question

Use standard reduction potentials to calculate the standard free energy change in kJ for the reaction:

2Fe3+(aq) + Fe(s)2Fe2+(aq) + Fe2+(aq)

Answer: ______ kJ

K for this reaction would be ______ (greater, less) than one.

Answer 1

2Fe³⁺ + 2e^- = 2Fe²⁺    E₀ = 0.77 V

Fe = Fe²⁺ + 2e^-           E₀ = 0.41 V

E⁰_cell = E_red – E_ox = 0.77 V – 0.41 V = 0.36 V

delG⁰ = -nFE⁰_cell= (-2 mol e^-)(96458 C/mol e^-)(0.36 V) = -69449.76 J = -69.45 kJ

delG⁰ = -RTlnK_eq

-69449.76 J = (-8.314 J/mol K^-1)(298 K)(lnK_eq)

lnK_eq = 28

K_eq = 1.45

Answer: -69.45 kJ

K for this reaction would be greater than one (1.45).