Answer: The reaction is non spontaneous at [tex]50^0C[/tex]
Explanation:
[tex]E_{cell}[/tex]= +ve, reaction is spontaneous
[tex]E_{cell}[/tex]= -ve, reaction is non spontaneous
[tex]E_{cell}[/tex]= 0, reaction is in equilibrium
[tex]Ag/Ag^{+}(0.0078M)//Ni^{2+}(0.023M)/Ni[/tex]
Using Nernst equation :
[tex]E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Ag^{+}]}{[Ni^{2+}]^2}[/tex]
where,
R = gas constant = [tex]8.314J/Kmol[/tex]
T= temperature in kelvin =[tex]50^0C=(5+273)K=323K[/tex]
n = number of electrons in oxidation-reduction reaction = 2
F= Faraday's constant = 96500 C
[tex]E^o_{cell}[/tex] = standard electrode potential =[tex]E^0_{cathode}- E^0_{anode}[/tex]
Where both [tex]E^0[/tex] are standard reduction potentials.
[tex]E^0_{[Ag^{+}/Ag]}= +0.80V[/tex]
[tex]E^0_{[Ni^{2+}/Ni]}= -0.23V[/tex]
[tex]E^0=E^0_{[Ni^{2+}/Ni]}- E^0_{[Ag^{+}/Ag]}[/tex]
[tex]E^0=-0.23-(+0.80)=-1.03V[/tex]
[tex]E_{cell}=-1.03 - \frac{2.303\times 8.314\times 323}{2\times 96500}\log \frac{[0.0078]}{[0.023]^2}[/tex]
[tex]E_{cell}=-1.067V[/tex]
Thus as [tex]E_{cell}[/tex] is negative , the reaction is no spontaneous.
