Answer:
a)[tex]\Delta G=372490 J[/tex]
b)[tex]\Delta G=-568614 J[/tex]
Explanation:
a) The reaction:
[tex]Mg(s) +Fe^{2+}(aq) \longrightarrow Mg^{2+}(aq) + Fe(s)[/tex]
The free-energy expression:
[tex]\Delta G=-n*F*E[/tex]
[tex]E=E_{red}-E_{ox][/tex]
The element wich is reduced is the Fe and the one that oxidates is the Mg:
[tex]-0.44V=E_{red}-(-2.37V)=1.93V[/tex]
The electrons transfered (n) in this reaction are 2, so:
[tex]\Delta G=-2mol*96500 C/mol * 1.93 V[/tex]
[tex]\Delta G=-372490 J[/tex]
b) If you have values of enthalpy and enthropy you can calculate the free-energy by:
[tex]\Delta G=\Delta H - T* \Delta S[/tex]
with T in Kelvin
[tex]\Delta G=-675 kJ*\frac{1000J}{kJ} - 298K*-357 J/K[/tex]
[tex]\Delta G=-568614 J[/tex]