Free-energy change, G , is related to cell potential, E , by the equation G = n
ID: 923742 • Letter: F
Question
Free-energy change, G, is related to cell potential, E, by the equation
G=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, G must be in joules since 1 J=1 CV.
1. 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.
2. Calculate the standard cell potential at 25 C for the reaction
X(s)+2Y+(aq)X2+(aq)+2Y(s)
where H = -675 kJ and S = -357 J/K .
Express your answer to three significant figures and include the appropriate units.
Explanation / Answer
(1) Mg(s)+Fe2+(aq)Mg2+(aq)+Fe(s)
The reduction potentials are Fe2+ + 2e- ---> Fe ; Eo = -0.41 V
Mg2+ + 2e- ---> Mg ; Eo = -2.37 V
Since the reduction potential of Fe is more its acts as cathode
So standard potential of the cell , Eo = Eocathode - Eoanode
= EoFe2+/Fe - EoMg2+/Mg
= -0.41 - (-2.37) V
= +196 V
Standard free energy change is G=nFE
Where n = number of electrons transferred = 2
F = Faraday = 96500 C
Plug the values we get G=(2x96500x1.96) = 378.3x103 J
= 378.3 kJ
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(2) We know that G= H- TS
Given H = -675 kJ = -675x1000 J = -675000J
T = 25 oC = 25+273 = 298 K
S = -357 J/K
Plug the values we get G=-675000 - (298x(-357)) J
= -568.6x103 J
We know that G=nFE
E=- G/(nF)
= 568.6x103 / ( 2x96500)
= 2.95 V
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