Respuesta :
Answer: The [tex]K_{eq}[/tex] and [tex]\Delta G^o[/tex] of the reaction is [tex]1.13\times 10^{-19}[/tex] and -108080 J respectively.
Explanation:
For the given cell reaction:
[tex]Mg^{2+}(aq.)+K(s)\rightarrow Mg(s)+K^+(aq.)[/tex]
The half reaction follows:
Oxidation half reaction: [tex]K(s)\rightarrow K^++e^-;E^o_{K^+/K}=-2.93V[/tex]
Reduction half reaction: [tex]Mg^{2+}+2e^-\rightarrow Mg(s);E^o_{Mg^{2+}/Mg}=-2.37V[/tex]
Oxidation reaction occurs at anode and reduction reaction occurs at cathode.
To calculate the [tex]E^o_{cell}[/tex] of the reaction, we use the equation:
[tex]E^o_{cell}=E^o_{cathode}-E^o_{anode}[/tex]
Putting values in above equation, we get:
[tex]E^o_{cell}=-2.37-(-2.93)=0.56V[/tex]
- To calculate Gibbs free energy, we use the equation:
[tex]\Delta G^o=-nFE^o_{cell}[/tex]
where,
[tex]\Delta G^o[/tex] = Standard Gibbs free energy of the reaction = ?
n = number of electrons exchanged = 2
F = Faraday's constant = 96500
[tex]E^o_{cell}[/tex] = standard electrode potential of the cell = 0.56 V
Putting values in above equation, we get:
[tex]\Delta G^o=-2\times 96500\times 0.56=-108080J[/tex]
- To calculate the equilibrium constant of the reaction, we use the equation:
[tex]\Delta G^o=-RT\ln K_{eq}[/tex]
R = Gas constant = 8.314 J/K.mol
T = temperature of the reaction = [tex]25^oC=[273+25]=298K[/tex]
[tex]K_{eq}[/tex] = equilibrium constant of the reaction = ?
Putting values in above equation, we get:
[tex]-108080J=8.314J/mol.K\times 298K\times \ln K_{eq}\\\\K_{eq}=1.134\times 10^{-19}[/tex]
Hence, the [tex]K_{eq}[/tex] and [tex]\Delta G^o[/tex] of the reaction is [tex]1.13\times 10^{-19}[/tex] and -108080 J respectively.
