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