Answer:
Keq = 5.33*10²⁶
Explanation:
Based on the standard reduction potential table:
E°(Fe2+/Fe) = -0.45 V
E°(Cu2+/Cu) = +0.34 V
Since the reduction potential of copper is greater than iron, the former acts as the cathode and the latter as anode.
The half reactions are:
Cathode (Reduction): [tex]Cu^{2+} + 2e^{-}\rightarrow Cu[/tex]
Anode (Oxidation):[tex]Fe\rightarrow Fe^{2+}+ 2e^{-}[/tex]
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Overall reaction: [tex]Cu^{2+}+Fe\rightarrow Fe^{2+}+Cu[/tex]
The Gibbs free energy change at 25 C is related to the standard emf (E°) of the cell as well as the equilibrium constant K as:
[tex]\Delta G^{0} = -RTlnK_{eq}=-nFE_{cell}^{0}[/tex]
here:
[tex]E_{cell}^{0}= E_{cathode}^{0}-E_{anode}^{0}=0.34-(-0.45)=0.79V[/tex]
R = 8.314 J/mol-K
T = 25 C = 25+273 = 298 K
n = number of electrons involved = 2
F = 96500 Coulomb/mol e-
[tex]8.314J/mol.K*298K*lnKeq= 2mole\ e^{-}*96500C/mole\ e^{-}*0.79V[/tex]
Keq = 5.33*10²⁶
