Respuesta :
Answer:- [tex]\Delta G^0[/tex] for the reaction is -58 kJ/mol.
Solution:- Oxidation half reaction taking place at the anode is--
[tex]3Sn(s)\rightarrow 3Sn^2^+(aq) + 6e^-[/tex] [tex]E^0[/tex] = 0.14 V
Reduction half equation taking place at the cathode is---
[tex]2Fe^3^+(aq) + 6e^-\rightarrow 2Fe(s)[/tex] [tex]E^0[/tex] = -0.04V
[tex]E^0[/tex] for the cell = [tex]E_(reduction) + E_(oxidation)[/tex]
[tex]E^0[/tex] for the cell = -0.04V + 0.14V = 0.10V
Now we could easily calculate [tex]\Delta G^0[/tex] by using the below formula--
[tex]\Delta G^0[/tex] = -nF[tex]E^0[/tex]
where n is the number of electrons transferred in the over all reaction. Looking at two half equations the value of n is 6.
F is Farady constant and it's value is 96500 C/mol.
Plugging in the values in the formula...
[tex]\Delta G^0[/tex] = -(6)(96500)(0.10)
[tex]\Delta G^0[/tex] = -57900 J
Since, 1000 J = 1 kJ
So, [tex]\Delta G^0[/tex] = -58 kJ/mol
Half-cell potentials are the potential of the half cell of the electrochemical cell. The standard free energy of the reaction is -58 kJ/mol.
What is the standard free energy?
The standard free energy of the reaction is the change of the free energy when the compound element forms 1 mole of the substance.
Reduction half-reaction at the cathode is given as,
[tex]\rm 2Fe^{3+} + 6e^{-} \rightarrow 2Fe, E^{\circ} = -0.04 \;\rm V[/tex]
Oxidation half-reaction at the anode is given as,
[tex]\rm 3Sn \rightarrow 3Sn^{2+} + 6e^{-}, E^{\circ} = 0.14 \;\rm V[/tex]
The cell potential can be calculated as:
[tex]\begin{aligned} \rm E^{\circ} &= \rm E_{reduction} + E_{oxidation}\\\\&= -0.04\;\rm V + 0.14\;\rm V \\\\&= 0.10\;\rm V\end{aligned}[/tex]
Now, energy change can be calculated by the formula,
[tex]\rm \Delta G^{\circ} = - \rm nFE^{\circ}[/tex]
Here,
Number of transferred electrons = 6
Farady constant (F) = 96500 C/mol
cell potential = 0.10 V
Substituting values in the above equation:
[tex]\begin{aligned} \rm \Delta G^{\circ} &= - \rm nFE^{\circ}\\\\&= -(6)(96500)(0.10)\\\\& = -57900 \;\rm J\end{aligned}[/tex]
Therefore, -57900 J or -58 kJ/mol is the standard free energy.
Learn more about free energy here:
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