First of all what is exactly Gibbs free energy in regards to an electrochemical process? How can we visualise it? How does Gibbs free energy change in an electrochemical reaction? How is it related to the cell potential?

Isn't potential the amount of work done per unit charge to bring it from infinite to that point?

Now the ionisation in cell can be thought of as occurring in multiple steps, for example:

$$\begin{aligned} \ce{Zn(s)} & \ce{->Zn(g)} && (1) \\ \ce{Zn(g)} & \ce{->Zn+(g) + e-} &&(2) \end{aligned}$$

Then the work done to remove the charge should be the change in Gibbs free energy of second reaction and that energy should be equal to potential energy, but we take the combined change in Gibbs free energy as change in electric potential energy of the electron. Why is that so?

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1 Answer 1


The Gibbs energy is just a way to measure the thermodynamically favoured direction of a reaction. It doesn't tell you anything about the kinetics, so it might well be that the activation energy is too high for a reaction to happen, although it is thermodynamically favored.

The Gibbs free energy is related to the cell potential $\Delta E$ via $$ \Delta G = -nF\Delta E\; ,$$ where $n$ is the number of Electrons transferred per mole product and $F$ is the Faraday constant. It follows that the larger the cell potential is, the more negative the Gibbs free energy of the system.

The cell potential can be calculated using the Nernst equation: $$ \Delta E = \Delta E^0 - \frac{RT}{nF}\ln\left( \frac{[\text{Red}]}{[\text{Ox}]}\right) $$

$[\text{Red}]$ and $[\text{Ox}]$ refer to the concentrations of the reduced and oxidized species respectively.

Using this information allows you to determine in which direction a given redox reaction will develop.


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