Gibbs free energy and electrical potential
Your $\Delta G$ is really $\Delta G_{\mathrm{reaction}}$, and the reaction does not happen alone: it is accompanied by charging of electrodes. The Gibbs free energy of charging is just the associated change in the electroststic energy. That is, it equals $zF$ (the amount of charge moving against the electric potential for 1 mole of reaction) times $E$. Therefore, at equilibrium, the total change of the Gibbs free energy satisfies
\begin{equation}
\begin{split}
\Delta G_{\mathrm{total}} &amp;= \Delta G_{\mathrm{reaction}} + \Delta G_{\mathrm{charging}}\\
&amp;= 0
\end{split}
\end{equation}
just as it should.

In fact, when deriving the expression for the cell potential $E$, the logic goes backwards. We know that the electrodes are charged until equilibrium is reached, in which case the sum of $\Delta G_{\mathrm{reaction}}$ and $\Delta G_{\mathrm{charging}} = zFE$ should vanish. Thus, the cell potential is given by
\begin{equation}
E = -\frac{\Delta G_{\mathrm{reaction}}}{zF}.
\end{equation}