Free-energy change, ΔG∘, is related to cell potential, E∘, by the equationΔG∘=−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, ΔG∘ must be in joules since 1 J=1 C⋅V.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 reactionX(s)+2Y+(aq)→X2+(aq)+2Y(s)where ΔH∘ = -675 kJ and ΔS∘ = -357 J/K .Express your answer to three significant figures and include the appropriate units.

Respuesta :

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]

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