Question

# ± Cell Potential and Free Energy Free-energy change, ΔG∘, is related to cell potential, E∘, by...

± 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.

PART A:

Mg(s) + Fe2+ Mg2+ + Fe(s)

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

#### Earn Coins

Coins can be redeemed for fabulous gifts.