Answer:
Value of equilibrium constant is [tex]1.92\times 10^{19}[/tex]
Explanation:
Oxidation: [tex]Ni-2e^{-}\rightarrow Ni^{2+}[/tex]; [tex]E_{Ni^{2+}\mid Ni}^{0}=-0.23 V[/tex]
Reduction: [tex]Cu^{2+}+2e^{-}\rightarrow Cu[/tex]; [tex]E_{Cu^{2+}\mid Cu}^{0}=0.34V[/tex]
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Overall: [tex]Ni+Cu^{2+}\rightarrow Ni^{2+}+Cu[/tex]
[tex]E_{cell}^{0}=E_{Cu^{2+}\mid Cu}^{0}-E_{Ni^{2+}\mid Ni}^{0}=(0.34+0.23)V=0.57V[/tex]
We know, [tex]k=e^{\frac{nFE_{cell}^{0}}{RT}}[/tex]
where, k is equilibrium constant, n is no. of electron exchanged, 1F = 96500 C/mol, R is gas constant and T = 298 K
So, [tex]k=e^{\frac{2\times 96500\frac{C}{mol}\times 0.57V}{8.314\frac{J}{mol.K}\times 298K}}=1.92\times 10^{19}[/tex]
