Question

A) Calculate the standard free-energy change at 25 ∘C for the following reaction:

Answer 1

Answer:

A) ΔG° = -3,80x10⁵ kJ

B) E° = 2,85V

Explanation:

A) It is possible to answer this problem using the standard ΔG's of formation. For the reaction:

Mg(s) + Fe²⁺(aq) → Mg²⁺(aq) + Fe(s)

The ΔG° of reaction is:

ΔG° = ΔGFe(s) + ΔGMg²⁺(aq) - (ΔGFe²⁺(aq) + ΔGMg(s) (1)

Where:

ΔGFe(s): 0kJ

ΔGMg²⁺(aq): -458,8 kJ

ΔGFe²⁺(aq): -78,9 kJ

ΔGMg(s): 0kJ

Replacing in (1):

ΔG° = 0kJ -458,8kJ - (-78,9kJ + okJ)

ΔG° = -3,80x10² kJ ≡ -3,80x10⁵ kJ

B) For the reaction:

X(s) + 2Y⁺(aq) → X²⁺(aq) + 2Y(s)

ΔG° = ΔH° - (T×ΔS°)

ΔG° = -629000J  - (298,15K×-263J/K)

ΔG° = -550587J

As ΔG° = - n×F×E⁰

Where n are electrons involved in the reaction (2mol), F is faraday constant (96485 J/Vmol) And E° is the standard cell potential

Replacing:

-550587J = - 2mol×96485J/Vmol×E⁰

E° = 2,85V

I hope it helps!