Respuesta :
Answer:
Standard free-energy change at [tex]25^{0}\textrm{C}[/tex] is [tex]-3.80\times 10^{2}kJ/mol[/tex]
Explanation:
Oxidation: [tex]Mg(s)-2e^{-}\rightarrow Mg^{2+}(aq.)[/tex]
Reduction: [tex]Fe^{2+}(aq.)+2e^{-}\rightarrow Fe(s)[/tex]
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Overall: [tex]Mg(s)+Fe^{2+}(aq.)\rightarrow Mg^{2+}(aq.)+Fe(s)[/tex]
Standard cell potential, [tex]E_{cell}^{0}=E_{Fe^{2+}\mid Fe}^{0}-E_{Mg^{2+}\mid Mg}^{0}[/tex]
So, [tex]E_{cell}^{0}=(-0.41V)-(-2.38V)=1.97V[/tex]
We know, standard free energy change at [tex]25^{0}\textrm{C}[/tex]([tex]\Delta G^{0}[/tex]): [tex]\Delta G^{0}=-nFE_{cell}^{0}[/tex]
where, n is number of electron exchanged during cell reaction, 1F equal to 96500 C/mol
Here n = 2
So, [tex]\Delta G^{0}=-(2)\times (96500C/mol)\times (1.97V)=-380210J/mol=-380.21kJ/mol=-3.80\times 10^{2}kJ/mol[/tex]
Answer:
[tex]-3.72\times 10^5\ J[/tex]
Explanation:
The given reaction is:
[tex]Mg(s)+Fe^{2+}(aq)\rightarrow Mg^{2+}(aq)Fe(s)[/tex]
The two half reactions and their half potentials are as follows:
[tex]Mg^{2+}(aq)+2e^{-}\rightarrow Mg(s)[/tex] E° = -2.37 V
[tex]Fe^{2+}(aq)+2e^{-}\rightarrow Fe(s)[/tex] E° = -0.44 V
Half cell with more negative potential will act as anode and half cell with less negative potential will act as cathode.
Calculate cell potential as follows:
[tex]E\°_{cell}=E\°_{cathode}-E\°_{anode}\\=-0.44\ V-(-2.37\ V)\\=+1.93\ V[/tex]
Formula for the calculation of standard free energy is as follows:
[tex]\Delta G\°=-nFE\°_{cell}[/tex]
F = 96500 c/mol
n for the given reaction is 2.
[tex]\Delta G\°=-nFE\°_{cell}\\=-2 \times 96500\ C/mol \times 1.93\ V\\=-3.72\times 10^5\ J[/tex]
