how is the relation between Gibbs free energy and cell potential?

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?

• Welcome to Chemistry.SE. You are asking several questions in this post. We encourage one question per post, which means more posts and more opportunities for rep gain for everyone. see our help center page for more info. – Ben Norris Dec 16 '13 at 12:26
• Actually sir all these questions are interrelated with each other i want explanation in context to a specific topic thats why i have posted them together – user28804 Dec 16 '13 at 12:53
• Do you mind, then, if I edit your question to make that clearer? – Ben Norris Dec 16 '13 at 12:54
• No not at all i dont mind – user28804 Dec 16 '13 at 16:39

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.