Respuesta :
Answer :
(a) The value of Gibbs free energy for the reaction is [tex]-6.50\times 10^5J/mol[/tex]
(b) The value of [tex]K[/tex] at 298 is, [tex]8.28\times 10^{113}[/tex]
Explanation :
The given cell reaction is:
[tex]Mg(g)+2VO_2^{+}(aq)+4H^+(aq)\rightarrow Mg^{2+}(aq)+2VO^{2+}(aq)+2H_2O(l)[/tex]
The half reaction will be:
Reaction at anode (oxidation) : [tex]Mg\rightarrow Mg^{2+}+2e^-[/tex]
Reaction at cathode (reduction) : [tex]2VO_2^{+}+4H^++2e^-\rightarrow 2VO^{2+}+2H_2O[/tex]
First we have to calculate the Gibbs free energy.
Formula used :
[tex]\Delta G^o=-nFE^o[/tex]
where,
[tex]\Delta G^o[/tex] = Gibbs free energy = ?
n = number of electrons = 2 (from the reaction)
F = Faraday constant = 96500 C/mole
[tex]E^o[/tex] = standard e.m.f of cell = 3.37 V
Now put all the given values in this formula, we get the Gibbs free energy.
[tex]\Delta G^o=-(2\times 96500\times 3.37)=-650410J/mol=-6.50\times 10^5J/mol[/tex]
Thus, the value of Gibbs free energy for the reaction is [tex]-6.50\times 10^5J/mol[/tex]
Now we have to calculate the value of equilibrium constant.
Formula used :
[tex]\Delta G^o=-2.303\times RT\times \log K[/tex]
where,
R = universal gas constant = 8.314 J/K/mole
T = temperature = [tex]25^oC=273+25=298K[/tex]
K = equilibrium constant = ?
Now put all the given values in this formula, we get the value of [tex]K[/tex]
[tex]-6.50\times 10^5J/mole=-2.303\times (8.314 J/K/mole)\times (298K)\times \log K[/tex]
[tex]\log K=113.918[/tex]
[tex]K=8.28\times 10^{113}[/tex]
Therefore, the value of [tex]K[/tex] at 298 is, [tex]8.28\times 10^{113}[/tex]
