Question

A hypothetical electrochemical reaction, A2+ + 2B → A + 2B+, has a standard electrochemical cell potential that is temperature dependent and given by: Eocell = a + bT3 where a= 0.17 V and b = 1.0x10-9 V/K3 (T is temp in kelvin) For this reaction, calculate ΔrSo at 30 oC in J/mol

Answer 1

The relationship between the standard Gibbs free energy change for the reaction occurring in an electrochemical cell, and the standard potential for that reaction is:

ΔG° = -n*F*E°

where ΔG° is the standard free energy change
n is the number of moles of electrons transferred in the reaction
F is Faraday's constant = 96,485.34 C/mol
E° is the standard potential

In this case n = A2+ + 2B → A + 2B+

Take the derivative of the previous equation with respect to temperature (at constant pressure):

dΔG°/dT = -n*F*(dE°/dT)

For a system closed to the exchange of mass with the surroundings, the Gibbs free energy is defined as:

G = U + pV - TS

dG = dU + pdV + Vdp - TdS - SdT

From the first Law, dU = TdS - pdV, so:

dG = Vdp - SdT

For a process occurring at constant pressure, dp = 0, so:

dG = - SdT and dG/dT = -S

One can also write d(ΔG)/dT = -ΔS

So, going back to the expression for dΔG°/dT, we have:

dΔG°/dT = -n*F*(dE°/dT) = -ΔS°

E° = a +bT3’

Differentiating the above equaton we get,

dE°/dT = 3bT2 = 3 x 1.0x10-9 x (273+30) = 9.09 x 10-9

Plugging in the numbers for this problem:

2*(96,485.34 C/mol)*(-9.09 * 10^-9 V/K) = ΔS°

ΔS° = -0.175 J/(mol*K)

The relationship between the standard Gibbs free energy change for the reaction occurring in an electrochemical cell, and the standard potential for that reaction is:

ΔG° = -n*F*E°

where ΔG° is the standard free energy change
n is the number of moles of electrons transferred in the reaction
F is Faraday's constant = 96,485.34 C/mol
E° is the standard potential

In this case n = A2+ + 2B → A + 2B+

Take the derivative of the previous equation with respect to temperature (at constant pressure):

dΔG°/dT = -n*F*(dE°/dT)

For a system closed to the exchange of mass with the surroundings, the Gibbs free energy is defined as:

G = U + pV - TS

dG = dU + pdV + Vdp - TdS - SdT

From the first Law, dU = TdS - pdV, so:

dG = Vdp - SdT

For a process occurring at constant pressure, dp = 0, so:

dG = - SdT and dG/dT = -S

One can also write d(ΔG)/dT = -ΔS

So, going back to the expression for dΔG°/dT, we have:

dΔG°/dT = -n*F*(dE°/dT) = -ΔS°

E° = a +bT3’

Differentiating the above equaton we get,

dE°/dT = 3bT2 = 3 x 1.0x10-9 x (273+30) = 9.09 x 10-9

Plugging in the numbers for this problem:

2*(96,485.34 C/mol)*(-9.09 * 10^-9 V/K) = ΔS°

ΔS° = -0.175 J/(mol*K)