deltaG?=-nFE? part a: Alculate the standard free-energy change at 25 ?C for the

Question

deltaG?=-nFE?

part a:

Alculate 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.

answer: -3.71*10^5J

part B:

Calculate the standard cell potential at 25 ?C for the reaction:

X(s)+2Y + (aq)?X 2+ (aq)+2Y(s)

where ?H = -663kJ and ?S = -173J/K

Free-energy change, delta G?Delta G ^ is related to cell potential, E?, by the equation delta G?=-nFE? where n n is the number of moles of electrons transferred and F=96500 C/(mole^-) F = 96,500; m c/(mol e^ -) is the Faraday constant. When E E ^ circ is measured in volts, delta G ^ Delta G^ circ must be in joules since 1J=1C*V. 1 m J = 1 C cdot V. part a: Alculate 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. answer: -3.71*10^5J part B: Calculate the standard cell potential at 25 ?C for the reaction: X(s)+2Y + (aq)?X 2+ (aq)+2Y(s) where ?H = -663kJ and ?S = -173J/K

Explanation / Answer

dGo = dHo -TdS

dGo = -663 kJoules - (298K)(-0.173kJ/K)
dGo = -611.446kJ
-------------------------

dGo = -nFEo
-611.446 kJ = -2 moles of electrons (96.5kJ/mol-ev)(Eo)
Eo = 3.17 volts

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