Significance of sign of standard reduction potential
There are several questions here so let me start by defining the different types of cells and standard reduction potentials.

Definition of Electrochemical (Galvanic) cell: A cell that converts chemical energy into electrical energy. In these cells a redox reaction creates electrons that can do work.

Definition of Electrolytic cell: A cell that converts electrical energy into chemical energy. In these cells the electrical energy source provides the electrons to perform a reaction. 

Standard reduction potential: The tendency for a chemical species to be reduced and is measured in volts at STP. The more positive the potential is the more likely it will be reduced.

Now, here are your two reduction equations for $\ce{Zn}$ and $\ce{Cu}$:

$$\begin{alignat}{2}
\ce{Zn^2+(aq) + 2e- &amp;-&gt; Zn}\qquad&amp;{-}0.76\ \mathrm V\\
\ce{Cu^2+(aq) + 2e- &amp;-&gt; Cu}\qquad&amp;{+}0.34\ \mathrm V
\end{alignat}$$

The more positive the potential the more favorable the reaction as it is written will be. Remember that  $\Delta G = -nFE^\circ$ and that when $\Delta G$ is positive, the reaction is non-spontaneous, and when $\Delta G$ is negative, the reaction is spontaneous.  Positive values of $E^\circ$ will lead to negative values of $\Delta G$ and vice versa.

So, the reduction of $\ce{Cu^2+}$ to form $\ce{Cu}$ is more favorable than the reduction of $\ce{Zn^2+}$ to form $\ce{Zn}$. This means that  $\ce{Cu^2+}$ is a better oxidant than $\ce{Zn^2+}$. For an electrochemical cell, the cell potential can be calculated by the following equation: 

$$E^\circ_\text{cell}=E^\circ_\text{cathode}-E^\circ_\text{anode}$$ 

For a working electrochemical cell we need $E^\circ_\text{cell}$ to be positive. I would use $\ce{Zn}$ as the anode (oxidation) and $\ce{Cu^2+}$ as the cathode to give an $E^\circ_\text{cell}$ of $+1.10\ \mathrm V$.