Question

A galvanic cell consists of a iron electrode in 1 M Fe(NO3)2 and a copper electrode in 1 M Cu(NO3)2. What is the equilibrium constant for this reaction at 25oC?

Answer 1

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): $Cu^{2+} + 2e^{-}\rightarrow Cu$

Anode (Oxidation):$Fe\rightarrow Fe^{2+}+ 2e^{-}$

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Overall reaction: $Cu^{2+}+Fe\rightarrow Fe^{2+}+Cu$

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:

$\Delta G^{0} = -RTlnK_{eq}=-nFE_{cell}^{0}$

here:

$E_{cell}^{0}= E_{cathode}^{0}-E_{anode}^{0}=0.34-(-0.45)=0.79V$

R = 8.314 J/mol-K

T = 25 C = 25+273 = 298 K

n = number of electrons involved = 2

F = 96500 Coulomb/mol e-

$8.314J/mol.K*298K*lnKeq= 2mole\ e^{-}*96500C/mole\ e^{-}*0.79V$

Keq = 5.33*10²⁶