Question

± Cell Potential and Free Energy Free-energy change, ΔG∘, is related to cell potential, E∘, by the equation ΔG∘=−nFE∘ where n is the number of moles of electrons transferred and F=96,500C/(mol e−) is the Faraday constant. When E∘ is measured in volts, ΔG∘ must be in joules since 1 J=1 C⋅V.

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

Mg(s)+Fe2+(aq)→Mg2+(aq)+Fe(s)

Express your answer to three significant figures and include the appropriate units.

Part B Calculate the standard cell potential at 25 ∘C for the reaction

X(s)+2Y+(aq)→X2+(aq)+2Y(s) where ΔH∘ = -751 kJ and ΔS∘ = -247 J/K .

Express your answer to three significant figures and include the appropriate units.

Answer 1

PART A:

Mg(s) + Fe²⁺ Mg²⁺ + Fe(s)

Mg(s) Mg²⁺ + 2e- ; E⁰ = +2.38 V
Fe²⁺ + 2e- Fe(s) ; E⁰ = -0.41 V
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Mg(s) + Fe²⁺ Mg²⁺ + Fe(s) E⁰ =   +2.38 V + ( -0.41 V ) = +1.97 V

It's the same two electrons being "lost" and "gained".

So, n = 2

From the half reactions you can see that two electrons are transferred.
Hence,
G⁰ = - n F E⁰

= - (2 mol e^-) (96500 C /∙mol e^-) ∙ (1.97 V)

= - 380210 CV

= - 380210 J

= - 380.21 kJ

PART B:

X(s) + 2Y⁺(aq) X²⁺(aq) + 2Y(s)

Here 2 electrons are exchanged, that is two electrons being "lost" and "gained."

T = 25 ^oC = (273 + 25) K = 298 K

G⁰ = H⁰ - T S⁰

= (- 751 kJ) - (298 K) (-247 J/K)

= - 751000 J + 73606 J

= - 677394 J  

Again

G⁰ = - n F E⁰

E⁰ = G⁰ / (-n F)

= (- 677394 J ) / [ - (2 mole e^-) (96,500 C / mole e^-) ]

= 3.51 V