Answer: The standard cell potential (E∘) for the reaction [tex]X(s)+Y^+(aq)\rightarrow X^+(aq)+Y(s)[/tex] is -0.121 V
Explanation:
The reaction is:
[tex]X(s)+Y^+(aq)\rightarrow X^+(aq)+Y(s)[/tex]
Relation between standard Gibbs free energy and equilibrium constant follows:
[tex]\Delta G^o=-RT\ln K[/tex]
where,
[tex]\Delta G^o[/tex] = Standard Gibbs free energy = ?
R = Gas constant = [tex]8.314J/K mol[/tex]
T = temperature = 298 K
K= equilibrium constant = [tex]8.97\times 10^{-3}[/tex]
Putting values in above equation, we get:
[tex]\Delta G^0=-(8.314J/Kmol\times 298K\times \ln (8.97\times 10^{-3})\\\\\Delta G^0=11678.9J/mol[/tex]
To calculate standard Gibbs free energy, we use the equation:
[tex]\Delta G^o=-nFE^o_{cell}[/tex]
Where,
n = number of electrons transferred = 1
F = Faradays constant = 96500 C
[tex]E^o_{cell}[/tex] = standard cell potential = ?
Putting values in above equation, we get:
[tex]11678.9J/mol=-1\times 96500\times E^0_{cell}[/tex]
[tex]\frac{11678.9J/mol}{-96500}=E^0_{cell}[/tex]
[tex]-0.121V=E^0_{cell}[/tex]
Thus standard cell potential (E∘) for the reaction [tex]X(s)+Y^+(aq)\rightarrow X^+(aq)+Y(s)[/tex] is -0.121 V