Question

1) Fe2+(aq) + 2 Al(s) <==> 2 Al3+(aq) + 3 Fe(s). What is the voltage for this cell?

2) What is the Gibbs Free Energy for the above reaction assuming standard conditions?

3. What is the Gibbs Free Energy for the cell in question 1 if the Fe2+ (aq) is [.15] and Al3+ (aq) is [3.80]?

4. Describe what happens within the salt bridge between the two cells.

Please show work

Answer 1

1.

E⁰_Fe2+/Fe = - 0.44 V

E⁰_Al3+/Al = - 1.66 V

E⁰_cell = E⁰_Fe2+/Fe - E⁰_Al3+/Al = - 0.44 - ( - 1.66 ) = 1.22 V

2.

deltaG⁰ = - n F E⁰_cell

deltaG⁰ = - 6 * 96500 * 1.22

deltaG⁰ = - 706380 J = - 706.4 kJ

3.

K = [Al³⁺]² / [Fe²⁺]³

K = (3.80)² / (0.15)³

K = 4278.5

Therefore,

deltaG = deltaG⁰ + RT lnK

deltaG = - 706380 + (8.314 * 298.15*ln(4278.5)

deltaG = - 685653.7 J = - 685.6 kJ

4.

Anions will migrate towards anodic compartment (Al/Al³⁺) and cations will migrate towards cathodic compartment (Fe²⁺/Fe) through salt bridge to maintain electrical neutrality.