Free-energy change, G , is related to cell potential, E , by the equation G = n

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 = -783 kJ and S = -363 J/K .

The answer to 1. is 3.71x10^5 J

What is the answer to 2 and please show work?

Explanation / Answer

1. Mg(s)+Fe2+(aq) - ---> Mg2+(aq)+Fe(s)

E0cell = E0cathode - E0anode

    cathode(reduction half reaction) = E0Fe2+/Fe = -0.44 v

   anode (oxidation half reaction)= E0Mg/Mg2+ = -2.37 V

E0cell = -0.44 - (-2.37) = 1.93 v

DG0 = -nFE0cell

      = -2*96500*1.93

     = 372490 joule = 3.72*10^5 joule

     = 372.49 kj

2. DG0 = DH0 - TDS0

       = (-783) - (298*-363*10^-3)

       = -674.83 kj

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