Question

1. Use standard reduction potentials to calculate the standard free energy change in kJ for the reaction:
3I₂(s) + 2Cr(s) -----> 6I^-(aq) + 2Cr³⁺(aq)

Answer: _______ kJ

K for this reaction would be greater or less than one.

2.

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

2Cu²⁺(aq) + Sn(s) ----> 2Cu⁺(aq) + Sn²⁺(aq)

Answer: ______ kJ

K for this reaction would be greater or less than one.

Answer 1

1)

Oxidation half reaction (at anode)

2Cr(s) --------> 2Cr³⁺ (aq) + 6e E°_red = -0.74V

Reduction half reaction (at cathode)

3I2(s) + 6e ---------> 6I^-(aq) E°_red= +0.54V

E°_cell = E°_red,cathode- E°_red,anode

E°_cell = +0.54 - (- 0.74V)

E°_cell = 1.28V

number of electrons transfered, n = 6

∆G° = -nFE°_cell

F = Faraday constant ,96485J/mol

∆G° = - (6× 96485J/mol × 1.28V)

∆G° = - 741kJ

K for this reaction would be greater than 1

2)

Oxidation half reaction

Sn(s) -------> Sn²⁺(aq) + 2e E°_red= -0.13V

Reduction half reaction

2Cu²⁺(aq) + 2e -------> 2Cu⁺(aq) E°_red= +0.159V

E°_cell = +0.159V - (-0.130V)

E°_cell = 0.289V

n = 2

∆G° = -(2× 96485C/mol × 0.289V)

∆G° = -55.8kJ

K for this reaction would be greater than one