Question

Calculate the standard free-energy change at 25 ∘C for the following reaction:

Answer 1

Answer:

Standard free-energy change at $25^{0}\textrm{C}$ is $-3.80\times 10^{2}kJ/mol$

Explanation:

Oxidation: $Mg(s)-2e^{-}\rightarrow Mg^{2+}(aq.)$

Reduction: $Fe^{2+}(aq.)+2e^{-}\rightarrow Fe(s)$

--------------------------------------------------------------------------------------

Overall: $Mg(s)+Fe^{2+}(aq.)\rightarrow Mg^{2+}(aq.)+Fe(s)$

Standard cell potential, $E_{cell}^{0}=E_{Fe^{2+}\mid Fe}^{0}-E_{Mg^{2+}\mid Mg}^{0}$

So, $E_{cell}^{0}=(-0.41V)-(-2.38V)=1.97V$

We know, standard free energy change at $25^{0}\textrm{C}$($\Delta G^{0}$): $\Delta G^{0}=-nFE_{cell}^{0}$

where, n is number of electron exchanged during cell reaction, 1F equal to 96500 C/mol

Here n = 2

So, $\Delta G^{0}=-(2)\times (96500C/mol)\times (1.97V)=-380210J/mol=-380.21kJ/mol=-3.80\times 10^{2}kJ/mol$

Answer 2

Answer:

$-3.72\times 10^5\ J$

Explanation:

The given reaction is:

$Mg(s)+Fe^{2+}(aq)\rightarrow Mg^{2+}(aq)Fe(s)$

The two half reactions and their half potentials are as follows:

$Mg^{2+}(aq)+2e^{-}\rightarrow Mg(s)$ E° = -2.37 V

$Fe^{2+}(aq)+2e^{-}\rightarrow Fe(s)$ 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:

$E\°_{cell}=E\°_{cathode}-E\°_{anode}\\=-0.44\ V-(-2.37\ V)\\=+1.93\ V$

Formula for the calculation of standard free energy is as follows:

$\Delta G\°=-nFE\°_{cell}$

F = 96500 c/mol

n for the given reaction is 2.

$\Delta G\°=-nFE\°_{cell}\\=-2 \times 96500\ C/mol \times 1.93\ V\\=-3.72\times 10^5\ J$