The spontaneity of a redox reaction
The cell potential is related to the change in Gibbs free energy by this equation:

$$
\Delta G^\circ= -nFE^{\circ}_\text{cell}
$$

where $n$ is the number of moles of electrons transferred in the balanced reaction equation and $F$ is Faraday's constant.

The UC Davis chem wiki has a brief explanation of the connection between cell potential, the reaction quotient (via the Nernst Equation), and the equilibrium constant. You should be able to find a more thorough explanation in a general chemistry textbook. My copy of Tro's Chemistry has a derivation of all of these relations, but to summarize the first relation:

Work against any potential is given by force times distance, and in the case of moving a charge, that works out to:

$$
w = -qE^{\circ}_\text{cell}
$$

Faraday's constant $F$ gives the charge in coulombs on one mole of electrons, which means the total charge is given by:

$$
q = nF
$$

Where $n$ is the number of moles of electrons.

We can substitute this back into the work equation to get:

$$
w = -nFE^{\circ}_\text{cell}
$$

Since $\Delta G^{\circ}$ is equal to the maximum amount of work that can be done by a chemical reaction, 

$$
\Delta G^\circ = w = -nFE^{\circ}_\text{cell}
$$

In other words, saying that change in Gibbs free energy is negative is the same as saying that the cell potential is positive – in both cases we are simply looking at the total usable energy change and saying that "things move downhill" – the reaction proceeds in the direction that minimizes the thermodynamic potential energy of the system.