Answer: [tex]1.26\times 10^{-42}[/tex]
Explanation:
[tex]2Ag(s)+Fe^{2+}\rightarrow 2Ag^{+}+Fe[/tex]
Here Ag undergoes oxidation by loss of electrons, thus act as anode. iron 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_{[Fe^{2+}/Fe]}=-0.44V[/tex]
[tex]E^0_{[Ag^{+}/Ag]}=+0.80V[/tex]
[tex]E^0=E^0_{[Fe^{2+}/Fe]}- E^0_{[Ag^{+}/Ag]}[/tex]
[tex]E^0=-0.44-(+0.80V)=-1.24V[/tex]
The emf is related to equilibrium constant by following relation:
[tex]E^0=\frac{0.0592}{n}\times log K[/tex]
[tex]E^0[/tex] = standard emf = -1.24 V
n= no of electrons gained or lost = 2
K = equilibrium constant = ?
[tex]-1.24=\frac{0.0592}{2}\times log K[/tex]
[tex]-1.24=\frac{0.0592}{2}\times log K[/tex]
[tex]logK=-41.9[/tex]
[tex]K=antilog(-41.9)=1.26\times 10^{-42}[/tex]
The equilibrium constant for the reaction is [tex]1.26\times 10^{-42}[/tex]
