Derivation of relationship between Gibbs free energy and electrochemical cell potential
There are two ways to understand this equation.  

One is to realise that (reversible ideal case) $\Delta G = W_\text{non-exp}$ (non-expansion). Therefore, in an ideal chemical cell, if the potential difference between the electrodes is $E$, to move one mole electrons across the external circuit will be $FE$, which must be equal to the decrease in gibbs free energy of the system. Hence for $n$ mole electrons transferred at the same potential, $W_\text{non-exp} = \Delta G = -nFE$. 

The fact that $\Delta G = W_\text{non-exp}$ can be derived as under:
\begin{align}
\mathrm{d}S 
  &amp;= \frac{\delta q}{T}  &amp;&amp; \text{(reversible case)}\\
\mathrm{d}U 
  &amp;= \delta q + W_\text{non-exp} + W_\text{exp}\\
W_\text{non-exp} 
  &amp;= \mathrm{d}U - W_\text{exp} - \delta q \\
  &amp;= \mathrm{d}U + p\,\mathrm{d}V - T\,\mathrm{d}S 
  &amp;&amp; \text{(const. $p$ and $T$)} \\
  &amp;= \mathrm{d}H - T\,\mathrm{d}S \\
  &amp;= \mathrm{d}G
\end{align}

Another approach is to use 
$$\mu_\mathrm{electrochemical} = \mu_\mathrm{chemical} + zF\phi,$$ 
where $\phi$ is the electric potential at the point, $z$ is the charge on the species. Using this and 
$$\mathrm{d}G_{T,P} = \sum_{i=1}^n \mu_i\,\mathrm{d}n_i,$$
we get the required equation for $z = 1$ for an electron, and considering the electrochemical potential of both the electrodes. I am unsure of a rigorous derivation for the electrochemical potential but Wikipedia has some basics on it.