Question

Calculate Δ_rG° and log K_C of the reaction:

\[\ce{Fe{^{2+}_{(aq)}} + Ag{^{+}_{(aq)}} -> Fe{^{3+}_{(aq)}} + Ag_{(s)}}\]

Given `E_(Ag^+//Ag)^° = 0.80  V, E_(Fe^(3+)//Fe^(2+))^° = 0.77  V`

[R = 8.314 J K⁻¹, F = 96500 C mol⁻¹]

Answer 1

Given: `E_(Ag^+//Ag)^° = 0.80  V`

`E_(Fe^(3+)//Fe^(2+))^° = 0.77  V`

Formula: The standard Gibbs free energy change formula is:

Δ_rG° = −nFE°_cell

where:

n = number of moles of electrons transferred (in this case, n = 1, as one electron is transferred in the reaction).

F = Faraday constant (96500 C mol⁻¹) 

E°_cell = standard cell potential

The standard cell potential (E°_cell) is calculated using the standard reduction potentials.

`E°_"cell" = E°_"cathode" − E°_"anode"`

For this reaction:

\[\ce{Fe{^{2+}_{(aq)}} + Ag{^{+}_{(aq)}} -> Fe{^{3+}_{(aq)}} + Ag_{(s)}}\]

Ag⁺ is reduced (cathode).

Fe²⁺ is oxidised to Fe3+ (anode).

So,

= `E_(Ag^+//Ag)^° - E_(Fe^(3+)//Fe^(2+))^° `

= 0.80 V − 0.77 V 

= 0.03 V

Now, substitute the respective values into the Gibbs free energy equation:

Δ_rG° = −1 × 96500 C/mol × 0.03 V

= −2895 CV/mol 

= −2895 J/mol 

= −2.895 kJ/mol

The correlation between the standard cell potential and the equilibrium constant is described by the Nernst equation:

Δ_rG° = −RT ln K_C

Rearranging for log K_C:

`log K_C = (nE_"cell"^°)/0.0591`

Substitute the known values:

`log K_C = (1 xx 0.03  V)/0.0591`

= 0.507