Respuesta :
Answer: [tex]3\times 10^{35}[/tex]
Explanation:
The balanced chemical equation will be:
[tex]2Ag(s)+Ni^{2+}(aq)\rightarrow 2Ag^{+}(aq)+Ni(s)[/tex]
Here Ag undergoes oxidation by loss of electrons, thus act as anode. Nickel undergoes reduction by gain of electrons and thus act as cathode.
[tex]E^0=E^0_{cathode}- E^0_{anode}[/tex]
Where both [tex]E^0[/tex] are standard reduction potentials.
[tex]E^0_{[Ag^{+}/Mg]}=+0.80V[/tex]
[tex]E^0_{[Ni^{2+}/Ni]}=-0.25V[/tex]
[tex]E^0=E^0_{[Ni^{2+}/Ni]}- E^0_{[Ag^{+}/Ag]}[/tex]
[tex]E^0=-0.25-(+0.80V)=-1.05V[/tex]
The standard emf of a cell is related to Gibbs free energy by following relation:
[tex]\Delta G=-nFE^0[/tex]
[tex]\Delta G[/tex] = gibbs free energy
n= no of electrons gained or lost =?
F= faraday's constant
[tex]E^0[/tex] = standard emf
[tex]\Delta G=-2\times 96500\times (-1.05)=202650J[/tex]
The Gibbs free energy is related to equilibrium constant by following relation:
[tex]\Delta G=-2.303RTlog K[/tex]
R = gas constant = 8.314 J/Kmol
T = temperature in kelvin =[tex]25^0C=25+273=298K[/tex]
K = equilibrium constant
[tex]\Delta G=-2.303RTlog K[/tex]
[tex]+202650=-2.303\times 8.314\times 298\times logK[/tex]
[tex]K=3\times 10^{35}[/tex]
Thus the value of the equilibrium constant at [tex]25^0C[/tex] is [tex]3\times 10^{35}[/tex]
