Free-energy change, delta G degree, is related to cell potential, E degree, by t

Question

Free-energy change, delta G degree, is related to cell potential, E degree, by the equation delta G degree = -nF E degree where n is the number of moles of electrons transferred and F = 96, 500 C/(mol e^-) is the Faraday constant. When E is measured in volts, delta G degree must be in joules since 1 J = 1 C middot V. Calculate the standard free-energy change at 25 degree C for the following reaction: Mg(s) + Fe^2+ (aq) rightarrow Mg^2+ (aq) + Fe(s) Express your answer to three significant figures and include the appropriate units. Calculate the standard cell potential at 25 degree C for the reaction X(s) + 2Y^+ (aq) rightarrow X^2+ (aq) + 2Y(s) where delta H degree = -763 kJ and delta S degree = -307 J/K. Express your answer to three significant figures and include the appropriate units.

Explanation / Answer

A)

For:

Fe2+ + 2 e Fe(s) 0.44

Mg2+ + 2 e Mg(s) 2.372

E°cell = Ered - Eox = -0.44 - -2.372 = 1.932V

so

dG° = -nF*E°cell = -2*96500*1.932 = -372876J/mol

dG° = -372.87 kJ/mol

B)

Find dG°

dG° = dH - T*dS = -763*10^3 - 298*-307 = -671514 J/mol

so..

dG° = -nF*E°cell

E°cell = dG°/(n*F) = 671514 /(2*96500 )

E°cell = 3.4793471 V

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