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

Keq = 5.33*10²⁶

Step-by-step 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